Maximum-volume cone A cone is constructed by cutting a sector ...

Question

Maximum-volume cone A cone is constructed by cutting a sector from a circular sheet of metal with radius 20. The cut sheet is then folded up and welded (see figure). Find the radius and height of the cone with maximum volume that can be formed in this way.

Accepted Answer

Welcome back, everyone. A company plans to manufacture lampshades in the shape of a cone from circular pieces of fabric with a radius of 15 inches. What are the dimensions of the cone with the maximum volume that can be produced? Now, if we're gonna figure out what these dimensions are, OK, then let's try to visualize what our problem is telling us. Now, here we're producing lampshades in the shape of a cone. So let's just do a quick sketch of a cone on our right here, OK? And now for this cone, we know that it has, it's produced from circular pieces of fabric with a radius of 15 inches. So the slant length of our coin here, OK. It's gonna be 15 inches. We also know that the perpendicular length of our cone or the vertical height of our cone, we can call it H. And then we know that our cone will have a radius R. And now what's important to note is that our radius R is not the same as the radius of 15 inches that was used to manufacture our lampshades. That 15 inches instead would be the slant length of our cone because if we think about it, it's like we took our circular pieces of fabric and then folded it to make a cone. It's basically. Uh, this shaded part that I have here, uh, in our diagram, OK? So let's just keep that in mind as to how we have this as 15 inches. Now how can we use that to maximize our volume and thus solve for the dimensions of our cone, which would be H and R. Now here, if we think about it first, let's ask ourselves what do we know about the volume of our corn. Well, recall that the volume of a cone V is going to be equal to 1/3 of IR 2 H. Now, if we're going to maximize our volume, we'll need to differentiate V, and we want to differentiate V with respect to a given variable. So it would be best to differentiate V with respect to R to find a maximum volume. That means we'll need to express H in terms of R. How can we do that? Well, if we go back to our diagram here. Notice that we have a right triangle that's formed with HR and or slant length of 15 inches. So by the Pythagorean theorem, just clean this up a bit. By the Pythagorean theorem. Then R2 + 82 will be equal to 152. If we do some rearranging here, OK, then that means R2 + H2 equals. 225. Which means that 82 equals 225 minus R2d and thush will be equal to the square root of 225 minus r quad where H is going to be greater than 0, so we'll use the positive square root, which makes sense because H is at length, so it would have to be positive. Therefore, OK. Now that we have an expression for H, that means our volume as a function of the radius R is going to be equal to 1/3 of pi R2 multiplied by the square root of 225 minus R2. Know that we have V as a function of R, we can differentiate V of R with respect to R and set it to 0 to get the critical point of R and confirm if that critical point is our maximum. Now to differentiate VFR, notice that we're multiplying one term by another, so we'll need to use the product rule. So let's differentiate the. With respect to X. By using the power rule. OK. And know what that means if we can recall the poor rule. Remember that it basically tells the the product rule, sorry, the product rule, if we can recall the product rule. Remember it tells us that if we have a function as written as a product Y equals UV where U and V are functions of X, then the derivative of Y with respect to X equals U V plus V U. In other words, we're going to differentiate both terms U and V. And then multiply them by their original opposite terms, OK? So now in this case, let's let you. Be equal to 1/3 of pi R squared. And let's let V be equal to the square root of 225 minus R2. Now the derivative of you, OK, is going to be equal to. Every 2 of you is gonna be equal to 1/3, OK. Multiplied by. 2 Pi by 2 1/3 multiplied by 2R. We bring down the power we can differentiate it using the, the, the power rule of differentiation. So this is gonna be 2/3 of pi R. On the other hand, notice here that if we rewrite V as 225 minus R2 to the power of 1/2, then we can differentiate it using the chain rule, which means we're going to bring down the half. Subtract one from the power. So now that's gonna be of the alter function 225 minus R squared to the negative a half, and we're gonna multiply that by the derivative of our inner function, which would be negative2R. And now when we simplify that, this is going to be equal to negative R, OK. Negative R divided by the square root of that a little more uh a little, make it look a little better, divided by the square root of 225 minus R squared, OK? So now, in this case. Then that means the derivative of ZFR, OK. Is going to be equal to UV. So that's gonna be 1/3 or 2/3. Of IR multiplied by the square root of 225 minus R squared, OK. Plus, you multiplied by V. Remember, V is negative, so this is gonna be, we can put our minus sign here already. So now we're gonna be subtracting uh 1/3 of IR squad, OK, multiplied by negative R divided by the square root of 225 minus R squad. Now here We can go ahead and start to simplify, OK? I know when we do that, OK, then we should find that here. I'll leave you to work out the details in the interest of time, but we should find that this simplifies to 1/3 of pi, OK? We can factor 1/3 of pi. And rewrite this as 450 R minus 3 are cubed OK all divided by the square root of 225 minus R squad. That is if we were to factor out 1/3 of pi and then divide both terms by a square of 225 minus R2 to to to make them have the same denominator essentially, OK? And now, in this case, then. We could say that the derivative of the AR is gonna be equal to pi if we again further divide multiply through by 1/3. It's gonna be pi multiplied by 150 R minus R cubed divided by the square root of 225 minus R squad, OK, where R is not equal to 15 because if R is 15, then our derivative is going to be undefined. Now that we have the first derivative, if we set it to 0, OK. Then the derivative will be 0 and the numerator is 0. So if we set V of R to be zero, then we can say. The 150r minus r cubed will be equal to 0. Thus R cubed will be equal to 150 r. Let me make some space over here, which means that R squad will be equal to 150. Thus R equals the square root of 150. Which equals 5 route 6. OK. So the radius would be 5 root 6 inches or approximately 12.25 inches. This would be our critical point for our function. No, we need to test if it's a maximum, OK? And we can confirm that R equals 5 root 6 inches is our maximum by the first derivative test. So confirming that. The maximum By the first derivative test. The first derivative test helps us to understand the behavior of R before and after 5 route 6, OK? And if the sign changes from positive to negative, it will be a maximum. If the sign changes from negative to positive, it will be a minimum. So no, for R less than 5 root 6, let's understand the behavior before it gets to 5 root 6. Let's let R be equal to 12. If R is 12. Then the first derivative. Is going to be equal to 150 multiplied by 12 minus 12 cubed which equals 72. Our sign is positive. Next, if R is greater than 5 root 6, let's use another value. If we can let R be equal to 13, and when R is 13, then our first derivative is equal to 150 multiplied by 13 minus 13 cubed. Which equals -247. So since the derivative. Let me clean this up here. Since the derivative changes. From positive to negative. As you can see, That means Or value of R. Is a maximum. Changes from positive to negative, so we know it's a maximum. So we have found R or first dimension, and we know it's a maximum. Thus the maximum value of R will correspond to the maximum value of H, so we can solve for H. So that means then that H is gonna be equal to the square root of 225 minus 150, which is gonna be the square of 5 root 6. So H equals the square root of 75, which equals 5 root 3 inches. Therefore, the dimensions to maximize the volume of our lampshade would be for our cone to have a radius of 5 root 6 inches. And a height of 5 or 3 inches. Thanks a lot for watching everyone. I hope this video helped.

Maximum-volume cone A cone is constructed by cutting a sector from a circular sheet of metal with radius 20. The cut sheet is then folded up and welded (see figure). Find the radius and height of the cone with maximum volume that can be formed in this way. <IMAGE>

Key Concepts

Volume of a Cone

Optimization

Geometric Relationships

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