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In this review video, we'll work through a set of real exam style calculus 1 problems on derivative applications. These are the same types of problems I used to give my university students to review for their exams. And the examples come directly from my calculus one course videos, just the problem solving portion without the lecture, so you can stay focused on the examples. You can also download the free PDF of all the practice problems on my website,
0:27
understandthemath.com. Let's get started. We going to go over a related rates problem that involves the filling of a conicle tank. We are told that water is being pumped into an inverted conicle tank. I have that picture down here in the diagram so you can see what it looks like. And the water is being pumped in at a rate of 2 m cubed per minute. We're told that the tank has a height of 6 m and a base radius of 4 m. We want to find out how fast
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is the water level rising when the water is 2 m deep. Let's now write down what we're given in this problem. We are told that the water is being pumped into the tank at a rate of 2 m cubed per minute. So that means the rate of change of the volume is equal to 2 m cubed per minute. And that's positive because water is being pumped in. We're told that the tank has a height of 6 m. So we'll say h is 6 m and then a base radius of 4 m. So r is 4 m. And let's put that the radius is 4 m
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and then the height of the tank is 6 m. Now let's look at what we have to find. We want to find how fast is the water level rising. So when the water is rising, the height of the water is changing. So we want to find dhdt when we're told the water is 2 m deep. So that means the height is 2 m. Now we need to think about an equation. We need to find an equation that is going to relate the volume of our conicle tank to our height. That formula is the volume is 1/3 pi r 2h.
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Let's now look at this equation that we just wrote down. Our equation has a volume in it. It has a radius and it has a height. We want to find dhdt. We are given dv dt but we are not given any information about drdt. So what we have to do is we have to put r in terms of h and the only way that we know how to do that is to use similar triangles. So I am going to take just the cross-section here of our cone. We were given that the radius is four and the
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height is six. Now let's think about some water being pumped in. Let's say that this is how much water we have. So we can call this the radius and then this part here the height. We have then two similar triangles. Here's the first one where we have radius 4, height six. And then our second one, we have a radius and a height. And they share common angles. So we can use what's called similar triangles.
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Similar triangles tell us that the ratio of the size of the conicle tank over 6 is equal to the ratio of the radius to the height. We can then solve for r. So we find that r is equal to 4 / 6 * h which is 2 over 3 * h. Let me go ahead and highlight that because that's what we're going to plug in to our formula. Our formula is v is equal to 1/3 pi r 2h. Now we're going to substitute in that r is 2/3 h. So we have 1/3 pi 2/3 h that gets squared * h. Our volume then
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is 2 ^ 2 which is 4 and then we have 3 * 3 * 3 which is 27. So we have a 4 over 27 pih cubed. Now we can take the derivative. We have dv dt on the left and on the right we have 4 over 27<unk>i * 3h 2 * dhdt. Let's go ahead and cancel that three out. This then is 4 over 9 pi h2 dhdt. Now let's think about what we want to find. Let's go back up here. We want to find dhdt. So I am going to go ahead and highlight dhdt. That's what we want to solve for.
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So we can say that dhdt is going to be 9 * dv dt all divided by 4<unk> h^ 2. We want to find the rate of change of the height when our height is 2 m. So this is 9. And then dvdt, let's go back up. It is 2 m cubed per minute. 2 m cubed per minute. And we divide that by 4 *<unk> * 2 m squared. It's really important to put units in at this point so that you know what the units for your final answer are going to be.
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9 * 2 is 18 and then in our denominator we have 2 ^ 2 which is 4 * 4 which is 16. So we have a 16 pi and then our units are going to be me per minute. We can divide both the numerator and the denominator by two. So this is 9 over 8 pi m per minute. This tells us that our water level is rising at a rate of 9 over 8 pi m. Let's now look at a related rates problem that involves the tip of a shadow moving. We are told
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that a street light is mounted at the top of a 15t tall pole. I put a diagram down here so that you can visualize what this looks like. We're told that a person 6 feet tall walks away from the pole along a straight path at a speed of 4t per second. We want to find how fast is the tip of the person's shadow moving when the person is 30 ft from the pole. What we're going to do first is look at our diagram and put some variables in there. and then we'll write down what we're given,
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what we need to find, and then the equation that relates those two. We're told that the street light is on top of a 15 ft pole. So, we'll mark this length as 15. And then we have a person that is 6 ft tall. So, we'll label this as six. And they are walking away from the pole. So, let's just make an arrow like this so we can see the direction that they're walking. And they walk away from the pole at a speed of 4 ft per second. So, that means we need to set
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a distance here equal to x because the xdt will be the speed that the person is walking. We want to then find how fast is the tip of the person's shadow moving. This right here is the tip of the shadow and we want to see how fast that's moving. So, we need to assign a distance to this. And I'm just going to label that L. So let's say here that X is the distance from the person to the pole. And then L is our distance from the shadow or the tip of the shadow to the pole.
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Now let's go ahead and let's put down what we are given. We are told that the person walks away from the pole at a speed of 4 feet per second. So we can now say that dx dt is equal to 4t/s. And we want to find how fast the tip of the person's shadow is moving. So that is going to be dlddt. And we want to find that when the person is 30 ft from the pole. So that is when x is 30 ft. Now what we need to do is we need to find an equation that relates
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x to l. We are going to use what's called similar triangles to get this relationship.
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I'm going to go ahead and copy down our big triangle. And let's go ahead and let's label this. We have this side is 15 and this distance is L. And now we're going to have a smaller triangle and that's going to be the height of the person. And then this length right here. We're told that the person is 6 feet. So, we can label that as six. And then now we have to find this distance right here. So, let's go back over here to our picture.
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And I want to highlight this distance right here. That distance is this distance. And that is going to be L minus X. So let's come down here and let's write down the equation for our similar triangles. We have the ratio of 15 is to L as 6 is to L - X. Now what we're going to do is cross multiply. So we have 15 * L - X is equal to 6 L. So that is 15 L - 15X is equal to 6 L. So if we keep the L's
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on the left, we're going to have 15 - 6. So that is 9 L and we get a 15x on the right. And we want to find dlddt. So let's solve for L. So this gives us that L is 15 over 9x. And let's see, we can divide three into both of those. So we get 5 over 3 x. Similar triangles are hard, but once you have it, the equation is actually pretty simple. So we end up with the equation L is equal to 5 over 3 * X. Let's go back up here and see what we want to find. We want to find
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dlddt. So we are going to take the derivative with respect to time. So we have dl dt is equal to 5 over3 * dx dt. And we want to find dl dt. Looks like we have to go back up again when x is 30. So x is 30 ft. So this is 5 over3. Going back up once more, dxdt is 4 feet per second. So this gives us an answer of 20 over 3t per second. So we can now say that the tip of the person's shadow is moving at a rate of 20 / 3t per second. Before we finish,
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I just want to point out one thing which I think is kind of unique to this problem. Notice that we did not use x is equal to 30 ft. What that means is the rate of change of the tip of the person's shadow doesn't depend on how far the person is from the pole. That rate stays the same no matter where the person is located.
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We're now going to walk through two examples where we are going to find a linearization. We want to find the linearization L of X of the function at the point X is equal to A. The first function that we are going to be finding the linearization of is f(x)= x cub - 3x^2 + 4. And we are going to be finding that linearization at the point A is equal to 1. I've written down the formula for the
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linearization L of X and it's just f of A * frime of A * X - A. So let's start with that. Let's go ahead and let's write down what that formula is. L of X is F of A plus FP prime at A times X - A. When you find a linearization, what you have to actually find is f of a and then frime of a. Our f of a is equal to f at 1. So, we just plug 1 into our function up above and we have 1 - 3 + 4 and
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that is equal to 2. I'm going to highlight that so we can find it when we need it. Now, we need to find frime of a. So, we first have to find frime of x. So, frime of x looking at f ofx up above is 3x^2 - 6x. And now f prime of a is just f prime at 1 and that is 3 - 6 which is -3. Now we can write down l of x. l of x is f of a which is 2 plus frime of a which is -3 * x - a which is x -1. Typically, when you find linearizations, you don't multiply this answer out because
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this form is useful when you use L of X to approximate a function that has an X value that is close to one. Over here on the right, I have graphed our function f ofx and then our linearization L of X. And notice that these two functions are almost identical at the point a is equal to 1. And that's how L of x can be used to approximate f ofx. Let's now look at our second example. The function that we're going to be finding the linearization
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of is f ofx is equal to sinx. And we're going to be finding the linearization at the point a is p<unk> / 6. Just like we did with the first example, let's start by writing down our formula. L of x is equal to f of a plus f prime at a time x - a. Now let's go ahead and substitute in that a is equal to p<unk> / 6. This gives us L ofX is equal to F at p<unk> / 6 plus FP prime at<unk> / 6 * X -<unk> / 6. F at<unk> / 6 is just s of<unk> / 6 and that is equal to 12. And we'll highlight
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that so we can find it in a minute. Now we need to find f prime at p<unk> / 6. So we first have to find f prime. The derivative of s is cosine. So f prime is cossine x. Now we can find f prime at p<unk> / 6. That is cosine of p<unk> / 6 which is equal to the<unk> 3 / 2. And let's highlight that. And now we can write down L of X. L of X is f at p<unk> / 6 which is 12 plus the derivative at<unk> / 6 which is <unk>3 / 2 * x - a which is just x -<unk> / 6. And that's our linearization.
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I again went ahead and graphed f ofx and l ofx. And if we zoom in here, you can see that when x is pi / 6, that's this point right here, l of x and f ofx are very close to the same value. Let's now look at an application example where we're going to find some error estimates. We're told that the radius of a circular disc is 24 cm and it has a maximum error in measurement of 0.2 cm. We want to estimate the maximum error,
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the relative error and the percentage error in the calculated area of the disk. y is equal to f ofx is the function that we're going to be calculating the error in. So in our example, this is going to be the area function. The maximum error is given by dy. The relative error is then dy / y. And then the percentage error is just the relative error times 100%.
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I went ahead and drew a circle to represent our circular disc. It has a radius r and we're going to write down that the area is pi r². So we are going to be calculating the error in this area. Let's first find the maximum error. So the maximum error is going to be da. Taking that derivative we have 2 pi r * dr. We are told that r is 24 cm. And then the maximum error in measurement that's
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actually dr and that is 0.2. 2 cm. And now we can find da. So da is 2 pi * r which is 24 cm. Make sure to put units in there. And then our dr is 0.2 cm. And this is equal to 9.6<unk> cm squared. When we leave pi in our answer, our answer is an exact answer. Now let's find the relative error. That is da / a. da was 2 pi r. I'm just looking right up here for that. and then we divide it by a which is pi r 2. So one of the rs cancels, the pies cancel. So we end
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up with 2 dr in the numerator and then r in the denominator. We have 2 * dr which is 0.2 2 cm all divided by r which is 24 cm which gives us a relative error of 0.0167. The last thing that we need to calculate is the percentage error. So the percentage error is just our relative error 0.0167 0167 * 100%. So this is just 0.0167 * 100 and that gives us a value of 1.67%.
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So this calculation tells us that if the error in our radius is2 cm, the percentage error in our area is 1.67%. We're going to walk through some examples where we are going to find the critical number of a given function. I've written down a method that we can follow just to make things easier.
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The first function that we're going to find the critical numbers for is the function f(x) is x cub + x^2 - x. Our first step is to calculate frime at x. So let's do that. frime at x is = 3x^2 + 2x - 1. And then our second step is we're going to set that equal to zero. We can factor this. This is 3x -1 and x + 1 that's equal to zero. And then if we set both of these factors equal to zero, we get that x is equal to 1/3. And we get that x
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is equal to minus1. These two numbers then are our critical numbers. Over here on the right, I have graphed this function just so that you can see what it looks like when we have a critical number. When x is -1, the slope of our curve is equal to zero. And we have what's called a local max. And then when x is 1/3, we also have a slope of zero. And at that point, we have a local min.
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Let's now look at our second example. In this example, we're given that f ofx is equal to x cubed + x^2 + x. We're going to begin by taking the derivative f prime at x is = 3x^2 + 2x + 1. And we're going to set this equal to zero. Now, we have to factor this. When I look at this, I cannot see an easy way to factor it. So, we are going to have to use the quadratic formula. Let me just go ahead and put that over here. The quadratic formula tells us that x isb plus or
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minus the square<unk> of b^2 - 4 a c all / 2 a. Let's see what this gives us. X is equal to b. So that's -2 plus or minus the square<unk> of b^ 2. So that's 2^ 2. So that's 4 - 4 * a which is 3 * c which is 1. All divided by 2 * a which is 3. Let's look at this part right here. that is equal to 4 - 12 which is - 8 and that is less than zero. So this tells us that we have no real roots. And whenever we have no real roots, we have no critical numbers.
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Let's now look at the graph of this function so that we can see what a function looks like when you don't have any critical numbers. Notice that our function is always increasing. The slope of the function is never equal to zero and there are no max or mins. So there are no critical numbers. We are going to find absolute maximum and minimum values of our function on the given interval. Our first example is the function f(x) = x cub - 6 x^2 + 9 x + 2. And then the closed interval
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that we're given is from -1 to 2. The first step in our closed interval method is to find the critical numbers. So we're going to start by taking the derivative. Our derivative is going to be 3x^2 - 12x + 9. And we are going to set that equal to zero. We're going to need to factor this. But notice that there's a 3 common to each term. So let's go ahead and let's pull that out. So this is x^2 - 4x + 3 is equal to 0. And we can factor this as let's see we can have an x -1 and then
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an x -3. We set x -1 = 0 and we get x is = 1 and then we set x - 3 = 0 and we get x is= to 3. x= 1 and x= 3 is a critical number. Our second step is to check if each of these critical numbers is in the interval from minus1 to 2. X= 1 is in the interval from minus1 to 2. But X is = 3 is not in our interval from minus1 to two. Let's highlight this number because that is a number where we're
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going to potentially have a max or a min. Let's go to step three. Step three says that we have to find the value of f at the critical number in the interval and at the end points. So we are going to find f at one. You just go back up to the original function. Let me go ahead and highlight this in yellow. and we're going to substitute in that x is equal to 1. Notice that we are not substituting it
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in to that function. Up at the yellow, we have 1 cub. So that's just 1 - 6 + 9 + 2. Let's see. If we add up the positive terms, we have a 12. So this is 12 - 6, and we get a value of 6. Let's highlight that because that's one of the numbers we're going to have to compare. Now, we need to find our function at each end point. So, let's compute f at -1. Looking back up to the yellow highlight, we have -1 cubed, so that's -1. And then we have -6 - 9 + 2. Let's see. We
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have a - 10 - 6. So that's -16 + 2 gives us -14. That's a number that we have to compare. And now we have one more. Our other end point which is at 2. So f at 2 is 2 cubed which is 8 - 6 * 2 ^ 2 which is 4 + 2 * 9 which is 18 + 2. So let's see this is 28 minus 24. So that is four. Now our fourth step is to look at the three values that we got in step three and find the largest one and the smallest one. The largest value is six. So we are going to say that we have
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an absolute max equal to 6 at the point x is equal to 1. And now let's look at our pink values and see the smallest one. The smallest one is -14. So we have an absolute min equal to minus14 at x is equal to minus1. And those are our answers to our problem. Before we move on, I want to show you what the graph of this function looks like. It's graphed over here on the right. Notice that when x is equal to -1, that's our left hand endpoint, we have an absolute min.
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And then when x is equal to 1, that was one of our critical numbers, we have an absolute max. And and you can see that the derivative is equal to zero because we have a horizontal tangent. We are asked to verify that a function satisfies the three hypotheses of rolls theorem on the given interval. Then we are to find all numbers c that satisfy the conclusion of rolls theorem. The function f ofx that we're given is x cubed - 3x^2 + 2x + 1. And the closed interval that
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we're given is from 0 to 2. I've graphed this function over here on the right. And when we get done with this problem, we're going to come back to that graph and see how our answers that we're going to get make sense. The first hypothesis of rolls theorem is that f is continuous on the closed interval from a to b. Our function f ofx satisfies this because f ofx is a polomial and all polomials are continuous. So we say that f ofx is a polomial. So it is continuous
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on the closed interval from 0 to 2. Our second hypothesis says that f is differentiable on the open interval. Let's go ahead and let's take the derivative. frime of x is 3x^2 - 6 x + 2. Notice that frime of x is also a polomial. So let's write that down. frime of x is a polomial. So we can conclude that it meaning f ofx is differentiable. And here's where we have to be careful. It is differentiable on the open interval.
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It's always continuous on the closed interval and differentiable on an open interval. Let's now look at our third hypothesis. This says that f of a is equal to f of b. That just means the function is going to be the same value at the end points. So we are going to have to find the function at each end point. Our first end point x is zero. So we're going to plug in x is equal to 0. And looking back
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up here, let me go ahead and highlight it so you can see where we're looking. We're going to substitute x is equal to 0 into that. And you can see that we get 1. Now we're going to do the same thing for x is = 2. This is going to be 2 cubed which is 8 - 3 * 2 ^ 2 which is 4 + 4 + 1. Let's see this is 13 - 12 and it is equal to 1. So that hypothesis holds because the function has the same
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value at each end point. Now what we need to do is we want to find all numbers C that satisfy the conclusion of rolls theorem and the conclusion of rolls theorem is that there is a point C between A and B where FP prime at C is equal to zero. So what we have to do is we have to find the derivative which we've actually already found in step two and we have to set it equal to zero. So
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let's just copy it down again. Frime of x is 3x^2 - 6x + 2 and we set that equal to zero. Now when I look at this, we need to factor it. But there is no way that we can factor that um unless we use the quadratic formula. So let me just remind you what the quadratic formula is. I always think it helps to write it down. X is minus b plus or minus the square<unk> of b^ 2 - 4 a c all / 2 a. Let's
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now apply that. We have x is equal to b. So that's a - - 6 plus or minus the square t of b^ 2. So that's - 6^ 2 - 4 * a which is 3. And then c which is 2. And we divide all of this by 2 * a which is 2 * 3. Let's now see what we get. This is 6 plus or minus the square<unk> of 36 - 24 all / 6. So this is 6 plus or minus the square<unk> let's see 36 - 24 that is 12 all divided by 6. When we take
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the square<unk> of 12, that's just the square<unk> of 4 * 3. So, we can pull out the 4 as a 2 and we get x is = 6 plus or minus 2<unk> 3 all / 6. So this is 1 + or minus <unk>3 / 3. Now what we have to do is we have to determine if these points are in our interval from 0 to 2. Let's look at these individually. The first one is x is 1 + <unk>3 over 3. Putting that into our calculator, we get approximately 1.577.
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That is between 0 and 2. So that one will work. Now let's look at the second one. X is 1us <unk>3 over 3. That is approximately 0.423. So that one also works. It is between 0 and two. So we can write down that C is = 1 + <unk>3 over 3 and C is = 1 -<unk>3 over 3. Let's go ahead and let's highlight those answers. There's our first one and there is our second one. So, this tells us that at those two points, our function is going to have a slope of zero. Let's
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go back up to our graph at the top and let's see if that's true. Let's first look at the value of our function at the end points. At zero, we have a function value of this. And here is our function value at two. You can see that they are the same point. And then at this point we have a horizontal tangent. And at this point we have a horizontal tangent. This point is 1 - <unk>3 over 3. And
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then this point over here is 1 + <unk>3 over 3. I think it always helps to graph the function just so that you can really understand what it is you are calculating. I make all my graphs on Desmos. It's a free graphing calculator that is really easy to use and I highly recommend it. We are going to walk through some examples that require the use of the mean value theorem. Let's look at our first example. We are asked to verify that the function satisfies the hypothesis of the
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mean value theorem on the given interval. Then we are asked to find all numbers C that satisfy the conclusion of the mean value theorem. The function f ofx that we're given is x cubed + x^2 + 1 and the closed interval is from minus1 to 1. I put a graph of our function over here on the right. And we're going to look back at this graph after we finish all of our calculations.
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The first hypothesis of the mean value theorem is that our function is continuous on the closed interval. Our function f ofx is a polomial. So we can write down that f ofx is a polomial. So it is continuous.
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on the closed interval from minus1 to 1. Every polomial is continuous. Let's now look at the second hypothesis. F is differentiable on the open interval from a to b. It is a fact that all polomials are differentiable. But let's just go ahead and take the derivative. We're going to need it later anyway. frime of x is going to be equal to 3x^2 + 2x. Notice that the derivative is a polomial. So we can say that frime of x is a polomial. So f ofx is differentiable.
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on the open interval from minus1 to 1. Now what we have to do is find all numbers C. Here's C right here that satisfy the conclusion of the mean value theorem. The conclusion tells us that the derivative at some point C is going to be f of B minus F of A over B minus A. That's just saying that the instantaneous rate of change, which is frime of C, is equal to the average rate of change, which is the f of b minus f of a over b minus a. Let's start by writing down our function.
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We have f ofx is x cubed + x^2 + 1. And then we already found the derivative. That was 3x^2 + 2x. And now we're going to set the derivative 3x^2 + 2x = to f at 1 - f at -1 all / 1 -1. We have lots of negatives going on there. Let's just go ahead and let's work on the right hand side. f of 1 is going to be 1 + 1 + 1 and then we're going to subtract off f at min -1. In case you're wondering, I'm looking right here for f ofx. So this is going to be min -1 + 1 + 1. And then
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we're going to divide all of this by 2. Let's see what this gives us. This is 3 - 1 / 2 which is 2 / 2. So this actually is just equal to 1. We now need to solve the equation 3x^2 + 2x is = 1. Let's pull the one to the left. So we have 3x^2 + 2x - 1 is 0. We can factor this. Let's see. Okay, we're going to put a 3x here and an x here. So, this needs to be a + one and that can be a -1. That works. So, we end up with x is 1/3 and x is equal to -1. The conclusion
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of the mean value theorem says that frime of c is f of b minus f of a over b minus a. But that is only valid in the open interval. So let's notice down here that this point is not in the open interval from minus1 to 1. It is an end point. So this point will not work. And this is the point that satisfies the mean value theorem. So our answer is going to be that C is equal to 1/3. Now, let's take what we just found and let's go back up to our graph and see if it
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makes sense. Here is our function value at our left end point. That's at -1. And then at one, let's see, it's up here. That is our function value at our right end point. And the mean value theorem tells us that the average rate of change that's represented by the slope of the secant line is equal at some point to the instantaneous rate of change. We found the point 1/3 and that's about right here. And you can see that the slope of the tangent line at that point is the same as
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the slope of the secant line. So the number C that we got is definitely correct. We are going to find the intervals on which f ofx is increasing or decreasing. Just to make things easier, I've written down three steps that we can follow. We're first going to find the critical numbers. And then we're going to find the value of fp prime to the left and right of each of those critical numbers. And then step three, we can just write down the intervals for increasing that is
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when fprime is greater than zero or decreasing when frime is less than zero. Let's look at our first example. We're given the function f ofx= 3x 4 - 4x cubed - 12x^2 + 5. The first thing that we're going to do is find the critical numbers. So we are going to start by taking the derivative frime of x is equal to 12x cubed - 12x^2 - 24x. To find the critical numbers, we set that equal to zero. It looks like we can factor out a 12x and then we're left with x^2 - x - 2.
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We can then factor our quadratic. Let's see. We can factor it as x + 1 and then x - 2. And then we set all three of those terms equal to zero. And we get x is equal to 0 -1 and 2. Our second step is to find the values of f prime to the left and right of these three numbers. I'm going to highlight FP prime because that is what we're going to have to be looking at. Let's now make a number line. This is going to be x. And we're going to put our critical points on it. We
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have -1, we have zero, and then we have two. The spacing doesn't matter at all. We are now going to be looking at the value of fp prime to the left and right of these three points. When you're working with a polomial, the region to the left and the region to the right of all the critical points is determined by the leading term of f prime. The leading term has an x cubed in it. So it is going to be negative when x is negative and it's going to be positive when x is positive.
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That just makes it a little bit easier. Now what we have to do is look at values between min -1 and 0 and then between 0 and 2. For this part I think it's easiest if we look at the factored form of fp prime which is this part right here over here on the right. I am going to make a table. I'm going to put down x and then I'm going to put all of the factors that I highlighted in orange. So,
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we have x, x + 1, and x - 2. And we are going to be looking to see if those factors are positive or negative. The point that I'm going to choose between -1 and 0 is - one2. And then - 1/2 is negative. - 1/2 + 1 is positive. And then - one/2 - 2 is negative. Now let's pick a point between 0 and 2. I think the easiest point is 1. So 1 is positive. 1 + 1 is positive. And then 1 - 2 is negative. In the region between -1 and 0, we have a negative * a positive * a negative. So
51:39
that's going to give us a positive. And then in the region between 0 and 2, we have a positive positive and a negative. So that is going to be negative. I think before we go to step three and write down the intervals for increasing or decreasing, I've always found that it helps to draw an increasing or a decreasing line in each interval. So in our leftmost interval, we have decreasing. Then we have increasing, decreasing, and increasing. Now we can just
52:12
write down the intervals for increasing and decreasing. Our function is decreasing on the interval from minus infinity to minus1 and on the interval from 0 to 2. And then our function is increasing on the interval from minus1 to 0. and then our rightmost interval from 2 to infinity. And we'll go ahead and we'll highlight those answers. Before we go on to the next problem, let's look up here on the right at our graph. Our critical points are the points right here, here, and here.
53:02
They're the points where our function changes from increasing to decreasing or vice versa. And you can see on the graph that our function is decreasing in this interval and in this interval and then it's increasing in this interval and in that interval just like we calculated. Now let's look at our second example. We're given the function f ofx= 2 plus the quantity x - 5 cubed. Just like we did with our first example, we are going to start by taking the
53:40
derivative. We have frime at x is equal to 3 * x - 5^ 2ar and then the derivative of what's inside is just 1 and we set that equal to zero and we get a value x is equal to 5. For step two, we are going to draw our number line. And this time we only have one point to put on it and that is at x is equal to 5. Now we need to find the value of fp prime to the left and right of five. Let's go ahead and highlight this because that's what we have
54:20
to look at. Notice that that is a function that is squared. So it is always positive. So we're going to have a positive value for fprime to the left and a positive value to the right. So our function is increasing and increasing. One thing we have to keep in mind though at five the derivative is equal to zero. So the function has a horizontal tangent and is not increasing there. So for our answer for part three, we have to say that our function is increasing
55:04
on minus infinity to 5 and from 5 to infinity and we have to leave out the five. Here's our answer then. And before we go on to the next problem, let's look at our graph over here on the right. This is our critical point when x is equal to 5. You can see that there is a horizontal tangent there. And then to the left and right of that point, our function is increasing. We are going to use the first derivative test to find the local maximum and minimum values
55:45
of f ofx. Just to make things easier, I have written down some steps that we can follow. The first step is to find the critical numbers. The second step is to find the value of fprime to the left and right of each critical number. And then our third step is to just calculate the local maximum. That is going to be where fprime changes from positive to negative. And then find the local minimum. And that is where f prime changes from negative to positive. In our first example,
56:27
we're given the function f(x) is = xub - 6x^2 + 5. For our first step, we are going to find the critical values. So we're going to start by taking the derivative. Frime at x is 3x^2 - 12x. And then we set that equal to zero. We can factor out a 3x. So we're left with x - 4. And this gives us that x is equal to zero and x is equal to 4. Those are our critical values. In our second step, we are going to take these critical values and we are going to put them on our number line.
57:18
The nice thing is that spacing doesn't matter. So here's zero and here is four. And we are going to be looking at the value of fprime. Let me go ahead and highlight f prime so we can see what to look back at. Because fprime is a polomial. When we look at the regions to the left and right of our critical numbers, we only have to look at our leading term, which is x^2. That term is positive. So we know that to the left and right f prime is going to
57:57
be positive. And then all we have to do is now just pick a point in the middle. I'm going to choose x is equal to 1. And we're going to have to see what fprime is. fprime at 1 is just going to be 3 - 12 and that is negative. So we're going to put a negative here. And then let's go ahead and let's draw some lines. When frime is positive, our function is increasing, then it's decreasing,
58:26
and then it is increasing. Now let's look at our third step. We just have to find the value of the local maximum and local minimum. we are going to have a local maximum at zero. So let's calculate f at zero. Let me highlight this because this is what we're looking back up at to find f of 0. So f of 0 is just five. Now let's calculate f at 4. That's where our local minimum is going to be. And this is 4 cubed - 6 * 4^ 2ar which is 16 + 5 and that is let's see 4 cubed is
59:14
64 - 96 + 5 and that adds up to be - 27. So for our answer we can say that five is our local max and minus 27 is our local min. Before we go on to our next example, let's take a quick look at our graph. Our critical values are here and here. That's where our function changes from increasing to decreasing as well as where our local max and mins are. And you can see that right here at x is equal to 0 we have a local max and when x is equal to 4 we have a local min.
1:00:10
We are going to find the intervals of concavity and then we're going to find the inflection points for our function f ofx. Just to make things easier, I have written down four steps that we can follow. And we'll go through these as we walk through our example. For our first example, we have the function f ofx is equal to x 4 - 4x cubed. Our first step is going to be to find fp prime. Set it equal to zero and solve for x.
1:00:45
F prime of X is 4 X cubed - 12 X^2 and then FP prime of X is 12 X^2 - 24X. We are going to set that equal to zero and we're going to solve for X. Looks like we can pull out a 12x. So we're left with x - 2 and this gives us that x is equal to zero and x is equal to two. Now let's look up at our second step. We are going to find the value of fp prime to the left and right of each of these values of x. So I think the easiest thing to do is to go ahead and draw a number line.
1:01:40
Here's our number line. We're going to put an x there. And we're going to put in the points x is zero and x is 2. And we are going to be looking at the value of fp prime. So let me highlight fp prime and that is this value right here. Because fp prime is a polomial in this region and in this region to the left and right of our critical points. We can just look at the leading term to find the value of fp prime. It is an x squar so it is positive. So we can say it's positive here
1:02:20
and it's positive here. Now all we have to do is pick a point in the middle. I think the easiest point to choose would be that x is equal to 1. So we're going to find fp prime at 1. Looking back up to our formula in yellow, this is 12 - 24. That is less than zero. Doesn't matter the value. We just need to know if it's positive or negative. And let's put a minus right here. Now, let's look
1:02:52
back up at our steps and look at step three. We want to write down the intervals for concave up. that is when the second derivative is positive or concave down when the second derivative is negative. I think it's easiest if we put a little graph on the number line that we've made. If we have a positive, it's concave up and then we have concave down and then we have concave up. Now, it's really easy to write our intervals. So for step three, it is going to be concave up
1:03:33
on the intervals from minus infinity to 0 and from 2 to infinity and then it's going to be concave down on the interval from 0 to two. And let's go ahead and let's highlight those answers. There's our concave up and there is our concave down. Now let's go back up and look at step four. We want to write down the inflection points and those are just the points where the concavity changes. The concavity changes at zero and at two. So this means we need to find
1:04:16
the value of our function. Let me go ahead and highlight this in orange. At the points 0 and two. So let's write down number four. We are going to calculate f at 0. That's just 0. So that's nice. And then f at 2 that is going to be 2 4 - 4 * 2 cubed. This is 16 - 4 * 8. So this is 16 - 32 and that's -6. So we can say for f of 0 we have an inflection point equal to 0 0 and we'll highlight that. And then for x is equal to 2 we have an inflection
1:05:08
point equal to 2 - 16. and we'll highlight that. Before we go on to the next problem, let's look back up here at our graph. We found that we have inflection points at zero and at two. Those are those points right there. And then our concavity changes to the left and right of those points. We have concave up in this region. And then we have concave down and then we have concave up again. In these examples, we are going to use the second derivative test to find the local maximum and
1:06:01
minimum values of f ofx. I've written down some steps that we can follow. The first step is to find the critical numbers. And then the second step is to find fp prime and then check if fp prime is positive or negative at our critical numbers. And then step three is we just calculate the local max and ms. Let's see how this works. The first function that we're given is a polomial and it's f(x) is = xub - 6x^2 + 5. Our first step is to find our critical numbers. So we're
1:06:41
going to take the derivative frime of x is 3x^2 - 12x. We are going to set this equal to zero. we can factor out a 3x and we're left with x - 4. So this gives us that x is equal to 0 and x is equal to 4. Now in step two, we are going to find the second derivative. That second derivative is going to be based off of our first derivative. So we're going to differentiate that. So, we're going
1:07:20
to have a 6x - 12. Now, we don't set this equal to zero. Instead, we take our critical numbers, that's this number and this number, and we substitute it in for x. So, fp prime at 0 is -12 and that is less than zero. So that tells us that we are going to have a local max at x is equal to zero. Remember it's just opposite. If it's less than zero, you have a max. If it's greater than 0,
1:08:00
you have a min. Now let's find fp prime at 4. This is 6 * 4. So this is 24 - 12. That is greater than zero. The value doesn't matter. We're just checking if it's positive or negative. If it's greater than zero, that tells us that we're going to have a local min at x is equal to 4. Now, in step three, we actually have to find out what the minimum and maximum values are. So, we need to go back up to our original function and plug in those x values. We have f at zero.
1:08:44
That one's easy. That's just equal to five. So that tells us that we have a local max equal to 5. And that of course is at x is equal to zero. Now let's plug in 4. This is going to be a little bit more work. f at 4 is 4 cubed - 6 * 4^ 2ar + 5. Let's see this is 4 cubed is 64 - 6 * 16 + 5. So let's see 64 - 96 + 5 that is actually - 27. So that tells us we have a local min equal to minus27 and that occurs where x is equal to 4. Let's now walk through some examples
1:09:49
so that we can understand how lital's rule works. We are going to evaluate the following limits. The first limit is the limit of x^2 + x - 6 all / x - 2 as x approaches 2. The first thing you always want to do when you are evaluating a limit is start by substituting the given value of x in our case x is approaching 2 into the expression that you're given. If we substitute x is equal to 2 on the numerator, we get 2^2 which is 4 + 2 - 6 and we divide all of that by 2 - 2. This
1:10:36
is equal to 0 / 0. When we get 0 / 0, we know that we can use lital's rule. I'm going to put a little lh over here each time I use lital's rule. Lat's rule tells us that this is equal to the limit as x approaches 2 of the derivative of the numerator which is 2x + 1 all divided by the derivative of the denominator which is 1. Now we go ahead and substitute in that x is 2 and we get 2 * 2 + 1
1:11:15
all divided by 1 and that gives us an answer of 5. So our limit is equal to 5. Let's now look at our second example. We are given the limit of cossine x - 1 all / x as x approaches zero. We'll start by substituting in that x is equal to 0. So we have cossine of 0 - 1 all / 0. Cosine of 0 is 1. So we have 1 - 1 / 0 and we get 0 / 0. This means we can use lital's rule and this is equal to the limit as x approaches zero. Now we're going to take the derivative of the
1:12:06
numerator. The derivative of cosine is minus s. So we have a minus sin x and then derivative of 1 is 0. And the derivative of the denominator is the derivative of x which is just 1. Now we're going to substitute in x is equal to 0. So this is minus sin of 0. Sine of 0 is just 0. So we get 0.
1:12:34
Here's our example. We're going to evaluate the following limit. Our limit is the limit of sinx * natural log x as x approaches zero from the right. x is approaching zero from the right because the natural log is only defined when x is greater than zero. We're going to start by just substituting in that x is approaching zero. So this is s of 0 from the right time natural log of 0 from the right. Sine of 0 is 0. And then the natural log of 0 from the right is minus infinity.
1:13:20
So we get the indeterminant form 0 * infinity. This means that we need to rewrite our function. Let's go ahead and let's do that. We're going to write this as the limit as x approaches zero from the right. Now we have to decide do we want to put 1 / sinx in the denominator or 1 / natural log x. 1 / sinx. Let me go ahead and write this over here. 1 / sinx is coseant x. We know how to take that derivative. We also know how to take the derivative of 1 / natural log x but it's a lot
1:14:03
more complicated. So I am going to choose to put the natural log x in the numerator and then in the denominator I have 1 / sinx. Let's see what this gives us when we let x go to 0 from the right. This is natural log 0 from the right all over 1 / sin 0 from the right and this is minus infinity / 1 / 0. So we get minus infinity over infinity and that is the indeterminant form that tells us that
1:14:49
we can apply lital's rule. So that's what we're going to do. I'm going to put a LH over here so that we know we're using Lital's rule. And this gives us the limit as x goes to zero from the right. Now we're going to take the derivative of natural log x which is 1 /x. And now we're going to take the derivative of our denominator. Let's just make a note here that this is coseant x. The derivative of cosecant is minus coseant xangent x. And now we need to know what that limit is.
1:15:31
I think the form that our limit is in makes it difficult to evaluate things. So I am going to change this back to be the limit as x goes to 0 from the right of 1 /x on the top. And then I'm going to write coseant as 1 / sinx. And then let's put in that coangent is cossine x over sin x. And we're going to rewrite this as the limit as x goes to 0 from the right. We're going to take the minus
1:16:09
sin^ 2 x up on top. And then in our denominator we have an x cossine x and that's much easier to evaluate. It's really common when you use lital's rule to have to simplify as you go along. Let's now substitute in that x is approaching zero. Because we don't have the natural log anymore, I am just going to write down 0 instead of 0 from the right. So this is - sin 0 all / 0 * cossine 0 sin of 0 is 0 and then we have 0 * cosine of 0 is 1. So we get 0 / 0 and that is the other
1:16:58
indeterminant form that tells us that we can use lital's rule. So we are going to use Low Tall's rule again and let's see what we get. This is the limit as x approaches zero from the right. We're going to take the derivative of minus sin^ 2 x. We're going to have to use the power rule. So this is - 2 sin x * cossine x. And then in the denominator, we are going to have to use the product rule because we have one function and a second function. The derivative
1:17:38
is our first function times the derivative of our second and that is a minus sinx. And then we have our second function cossine x times the derivative of our first. The derivative of x is 1.
1:17:56
Now let's go ahead and let x be equal to 0. So this is - 2 sin of 0 * cossine 0 all / 0 * - sin 0 + cossine 0. Sine of 0 is 0. So this is - 2 * 0 * 1 all / 0 + 1 in our denominator. And this gives us an answer of 0 over 1 which is equal to 0. And that's our final answer. The first limit that we're going to evaluate is the limit of x * e 1x - x as x goes to infinity. First thing you always want to do is substitute in what x is. So this is infinity * e^ 1 / infinity
1:18:56
minus infinity. This is infinity. 1 / infinity is 0. So we have e to the 0 - infinity. e to the 0 is 1. So we have infinity minus infinity. This is an indeterminate form. And we now need to put this form into 0 / 0 or infinity over infinity. Notice that with the function that we're given, let me go ahead and highlight this. Can you see that there is an x common to each term? Because of that, we can factor out an x. So let's go ahead and let's do that. This is the limit as
1:19:38
x goes to infinity. We're going to factor out an x and we're left with e to the 1 /x -1. Now we want to go ahead and substitute in that x is equal to infinity. So we have infinity * e to the 1 / infinity minus 1. This is infinity e to the 1 / infinity is e to the 0 - 1. So this is infinity to the 1 - 1 which is infinity * 0. This is an indeterminant form. It's not 0 / 0 or infinity over infinity. So we have one more step to do before we can use lital's rule. What we have
1:20:28
to do is rewrite this limit. Let's highlight this in such a way that we get 0 / 0 or infinity over infinity. So we have to take either the x or the e to the 1/x -1 to the denominator. It's going to be a lot easier if we put x in the denominator. So that's what we're going to do. This is the limit as x goes to infinity. We're gonna have e to the 1 /x minus one in the numerator and then
1:21:02
in the denominator we're going to have 1 /x. Let's see what that is equal to. We're going to let x go to infinity. So this is e to the 1 / infinity - 1 all / 1 / infinity. 1 / infinity is 0. So this is e to the 0 - 1 all over 0. So we get 1 - 1 / 0 and we get 0 / 0. Now finally we have a form that we can use lital's rule. I'm going to put an LH over here. So we know that we are using Lital's rule.
1:21:43
And now we're going to take the derivative of the numerator and the derivative of the denominator. The derivative of e^ 1x -1 is e^ 1x* the derivative of 1 /x. Let me go over here and let me write down in the left hand side the derivative of 1 /x. This is equal to -1x^2. If you haven't memorized that, that might be a good one to memorize. It just makes things go a little bit quicker if you have that memorized. So, back to our problem. We had e to the 1 /x. Now,
1:22:25
we're going to have to multiply by the derivative of 1 /x, which is - 1x^2. And then we have a 1 /x in the denominator. So, we're going to write down that derivative. And that is - 1x^2. One thing that is really nice with this problem is that the minus1x^2 cancel out and so we're just left with e to the 1 /x. This is the limit as x goes to infinity e to the 1 /x. Now we let x go to infinity. So this is e to the 1 / infinity which is e to the 0 and this is equal to 1.
1:23:09
So the value of our limit is equal to 1. In our example, we are going to take the limit of cossine of 3x raised to the 5x as x goes to 0. We're going to start by substituting in that x is equal to 0. So this is cosine of 0 to the 5 / 0. Cosine of 0 is 1. So this is 1 raised to the 5 / 0. 5 / 0 is infinity. So this is 1 to the infinity. This is an indeterminate power. You feel like it should
1:23:54
be one and sometimes it is, but often times it isn't. So you have to go through and evaluate it. Let's walk through our steps. Our first step is we're going to take y and we're going to set it equal to the function that's in our limit. And this is cossine 3x raised to 5 /x. Our second step is to take the natural log of both sides because that enables us to bring the 5 / x out in front. Natural log y is natural log cossine of 3x raised to the 5 /x. We can rewrite our right
1:24:42
hand side as 5 /x * natural log of cossine of 3x. Now in step three, we're going to take the limit. We were given that our limit is x approaching zero. So we're going to take the limit as x goes to zero of both sides natural log of y and then the limit as x goes to zero 5 natural log cossine 3x and we're going to write the 1 /x down here on the bottom. So we'll write the x like that. Now we
1:25:22
are going to evaluate this right hand side. We're going to let x go to zero. So this is 5 * natural log cosine of 0 is 1 and then we have a 0 on the bottom. Natural log of 1 is 0. So we get 0 / 0 and this is the indeterminant form that allows us to use lital's rule. So now we're going to use lital's rule. This is the limit as x goes to 0. We have 5 * the derivative of the natural log
1:26:02
that is 1 / what's inside the natural log which is cosine 3x* the derivative of what's inside which is a - 3 sin 3x. And we divide all of this by the derivative of x which is 1. Let's go ahead and let's simplify this. This is the limit as x goes to 0 - 15 and we have a s over a cosine. So that is tangent of 3x. Now we can let x be equal to 0. So we have -5 * tangent of 0 which is 0. And this
1:26:48
gives us 0. Now, here's where you have to be super careful. You are not done. That just tells us that the limit as x goes to 0 of the natural log of y is equal to zero. But we want to find what y is equal to. So, we have to write that y is equal to e to the 0. We're just solving this equation right here and this is equal to 1. So our limit is equal to one. In this optimization problem, we're finding the largest possible volume of a box with a square base and an open top. The
1:27:37
first step in solving this problem is to read through the problem carefully and figure out what information we're given and what we're asked to find. We're told that if 1,200 cm squared of material is available to make a box with a square base and an open top, we want to find the largest possible volume of the box. The second step is to fill in each of these four categories using the information from our problem. I went ahead and drew a box with a square base. Now,
1:28:16
what we have to do is label the sides. If it has a square base, this length and this length can both be x. And then we'll let this height be y. So, let's write down that x is our length and our width. You always want to write down what the variables are. and then y is our height. Now let's write down our equation. Our equation is going to relate the variables from our diagram to the information that we're given. So let's look back up in our
1:28:57
problem. It says that we have 1200 cm squared of material. So that means the base, the sides, and the top has to equal 1200 cm squared. So let's write down that we have a surface area of 1200 cm squared. And then let's also write down what our surface area is. It's equal to the base which is x^2. And then it's equal to all four sides. And those four sides are the same. So we're just going to have 4 x y. So our equation is going to be x^2 + 4x y is equal to 1,200. We'll leave out
1:29:45
the units for right now. Now let's go ahead and find if we are going to maximize or minimize a quantity. Going back up to the problem, it says we are to find the largest possible volume. That means that we're going to maximize the volume. So let's write down what the volume is. The volume of our box is the length * the width * the height. So that is x^2 * y. Now we need to put our volume in terms of one variable. So we need to come back to our original equation right here
1:30:24
and we need to solve for y. So let's go ahead and let's do that. We have 4xy is 1,200 - x^2 and that gives y is equal to 1200 - x^2 all / 4x. Now let's take that and put it back up into our equation for v. So we have x^2 * 1200 - x^2 all / 4x. Looks like one of the x's is going to cancel. So we end up with 14 x * 1200 - x^2. Let's go ahead and let's multiply this through. So we have our v is equal to 300x -4 x cubed and we are going to need that. So I'm going to go ahead and
1:31:27
highlight that. Now let's think about our domain. We are not told how big the length and the width of our box can be. But we do know that it can't be zero and it can't be negative. So we're going to say that our domain is x is greater than zero. And let's just make a note that this is an open domain. We'll need that in just a few minutes. Our third step is to take the derivative of our
1:31:55
volume and find its critical points. I'm going to go ahead and rewrite our equation for our volume. It is 300x -4 x cubed. Now let's go ahead and let's take the derivative. Our derivative is 300 - 3/4 x^2 and we're going to set that equal to zero. So we have 300 is equal to 3/4 x^2. If we multiply by 4 and divide by 3, we are going to get let's see 1,200 / 3 is equal to x^2. So this tells us that x2 is 400. And we take the square root of that and it's really nice and we get that
1:32:51
x is equal to 20. Now we need to put units. Let's go back up. We have cm squared for surface area. So x is 20 cm and we'll go ahead and highlight that. Now we need to find y. Our equation for y is right here. So let's use that. y is equal to 1,200 - x which is now 20. So we have 20^2 all / 4 * 20. Let's see this is 1,200 minus 400. So this is 800 / 80. So we get that y is equal to 10 cm. Now what we need to do is find our maximum volume. So let's write down our equation
1:33:48
for v. Let's see. We can pick two of them. V is X2 Y right here or the one we have below that. Let's use the X^2 Y. That one's going to be easier. So V is X^2 * Y. X is 20. So this is 20 cm squared. Y is 10 cm. So let's see. This is 400 * 10. So, this gives us 4,000 cm cubed. Now, it may seem that we're done, but we're not. Our next step is to justify that the answer that we got is a maximum. Our interval is an open interval. So, we can use the first derivative
1:34:40
test or the second derivative test. Let me go ahead and let's highlight our derivative. There's our derivative right there. That derivative is pretty easy. So, I think it's going to be easiest to use the second derivative test. So, let's write that down. We're going to use the second derivative test. It would not be wrong to use the first derivative test. It would just be a little bit harder. And let me go ahead and write down here what
1:35:09
our first derivative is so we can then take the second derivative.
1:35:17
Let's go ahead and let's find vp prime. It is equal to -6 over 4x which is -3 x. Now we're going to substitute in that x is equal to 20 cm. So we'll say vp prime at 20 is -3 * 20 and this is -30 and it's less than zero. We really don't care what the number is. We just need to know if it's positive or negative because it's negative. Our function is concave down and we have a max. We can then say that we have a max at
1:36:02
x is equal to 20 cm. And then we can also say that v is equal to 4,000 cm cubed is a maximum. Now we just have one more step. That step is to reread the problem and then clearly state our final answer in a complete sentence. Let's go back up and look at our problem. Our problem says to find the largest possible volume. So that's what we need to state. So we're going to write down that the largest possible volume
1:37:01
is 4,000 cm cubed. And that's our final answer. In this optimization problem, we're finding the largest rectangle that can fit inside a right triangle. The first thing we want to do in solving this problem is to read the problem carefully. figure out what information we've been given and then what we're asked to find. Our problem tells us that we are to find the area of the largest rectangle that can be inscribed in a
1:37:39
right triangle with legs of length 3 cm and 4 cm if two sides of the rectangle lie along the legs. What we want to do now is fill in all four of these categories that I've listed here. And notice that I have already drawn our picture just to make things easier. We have a right triangle and inscribed in a right triangle is a rectangle and it has legs of length 3 cm. So this distance here is 3 cm and 4 cm. So this distance here is 4 cm. Now what we want to do is assign variables.
1:38:31
Our variables need to be related to what we want to find. We want to find the area of our rectangle here in red. So, we need to make our variables the width and height of our rectangle. So, I'm going to let this distance be x and this be y. So, let's write that down. x is the width of our rectangle and then y is the height of our rectangle. Our second category is titled equation. What we need to do for this is we
1:39:16
need to find an equation that relates x to y. I think this one is difficult and a lot of students struggle with this. So, let me tell you exactly how we're going to do this. To find the equation that relates x and y, we're going to use similar triangles. If you studied related rates, you might remember doing this when you had water going into a cone. Now, what we're going to do is we're going
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to compare the sides of our two triangles. So, let me go ahead and let me highlight our second triangle. This right here is our second triangle and it has a leg of x. And then this distance right here is 3 - y. So using similar triangles we can say that 3 is to 4 as 3 - y is to x. Let's put that in our pocket for a minute and we'll come back to it. Let's now move on to what we want to
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maximize or minimize. We want to find the area of the largest rectangle. So that means that we are going to maximize the area of the rectangle. So we need to write down the formula for that area. The formula for the area of our rectangle is x * y. And now we need to get this formula for a just in terms of x. So, we're going to go back to our equation and we're going to solve this for y. Let's go ahead and let's cross multiply. This gives us 3x is = 12 - 4 y. So, we can say that
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4 y is = 12 - 3x and y is = 12 - 3x all / 4. Let's now go ahead and substitute that in for our area. A is now x * 12 - 3x all / 4. What we're going to do in step three is to take the derivative of a so that we can find its critical values. And it makes it easier to take the derivative if we write a in the simplest form possible. So I'm going to rewrite this and I'm going to say that a is equal
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to 12 / 4 which is 3. So we have 3x - 3/4 x^2. Then we can take the derivative easily without having to use the product rule. The last category that we have to fill out is the domain. Let's go back here and look at at our diagram. The domain is in terms of x and x can be between this point and this point. So we can say that x is between 0 and 4. And this is a closed interval. So I'm going to put that down because that's going to make a difference to us in a few minutes.
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For step three, we want to take the derivative of our area function and find its critical values and then find the maximum value of a. So let's go ahead and let's rewrite a. A is 3xus 3/4 x^2. Now we can go ahead and take the derivative. a prime is equal to 3 - 6 over 4 * x and we're going to set that equal to 0. So we have 3 is = 6 over 4 * x. So this means that 12 / 6 is equal to x. So we can say that x is equal to 2. And we need units with this. So, let's go
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back up and notice that in our original problem, our lengths are in terms of cime. I guess we did write that down on the diagram. So, x is equal to 2 cm. Now, let's go ahead and let's find y. Here's our equation for y, and that's what we're going to use. So let's say that y is 12 - 3 * 2 all divided by 4. This is 12 - 6 which is 6 / 4. We can say then that y is equal to 3 or 1.5 cm.
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Now, what we need to do is find our maximum area. And we have two equations for the area. Let's go back up here and look. We have this equation and this equation. They're the same thing. I think a is equal to xy is going to be a lot easier to use. So, let's use that to find what a is. So, I'll write down a is x * y. X was 2 cm. So this is 2 cm. And then Y was three cm. So we get that our area is 3 cm squared.
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It may now feel like we're done, but we're not quite done. What we have to do now is justify that this area that we got is a maximum. And to do that, we have to go back to our domain and see if it's open or closed. This domain is closed. So that means that to justify that our area is a maximum, we have to find a at our end points and then compare it to the 3 cm squared. To find our area at the end points, let's go back up and look at our
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diagram. At this point on our left, x is equal to zero. And then at this point on our right, y is equal to zero. We know that our area is equal to x * y. And we can use that. So a at 0 is going to be zero because x is 0. And then a at 4 is going to be zero because y is zero. And then we have a at our critical point which is 2 cm is equal to 3 cm squared. So this is our maximum area. Now we're almost done. Our last step is to reread our problem and clearly state our final answer
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in a complete sentence. Let's go back up to our problem. We want to find the area of the largest rectangle. So, we have to state what the area is. Let's state that the area of the largest rectangle is 3 cm squared. And that's our final answer. That's all for this exam review. I hope it helps you as you prepare for your exam. If this video has been helpful to you, please consider subscribing to my channel. Keep believing in yourself and have a great rest of your day.
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