Optimization A right triangle has legs of length h and r and a...

Question

Optimization A right triangle has legs of length h and r and a hypotenuse of length 4 (see figure). It is revolved about the leg of length h to sweep out a right circular cone. What values of h and r maximize the volume of the cone? (Volume of a cone = (1/3) πr²h.)

Accepted Answer

A company is planning to manufacture lamp shades in the shape of a right circular cone by spinning a right triangle around the shorter lake. If the hypotenuse of the triangle is fixed at 8 m, what dimensions of the legs will yield the maximum volume for the lampshade? We have 4 possible answers, which are all measures of height and radius of this cone. Now, to solve this, I'll try a quick image. Where hypotenuse is 8 m. We don't have R and we don't have H. We noticed though, this is a right triangle. So we can use Pythagorean theorem to make an equation. H2 plus R2 equals 82 or 64. Now, let's look at the volume of a cone. This formula is given by 1/3 by R2 H. So to solve this, we're going to turn this volume equation from two variables into the volume equation with 1 variable. Let's find a substitution. We have H2 plus R2 equals 64. If we subtract H2, get R2 equals 64 minus H2. Or R equals the square root. Of 64 minus 8 squared. And normally we would have a negative version as well, but because this is a radius, it should be a positive value. Now we have a substitution for R. Let's substitute this into our volume equation. You have 1/3 I. Multiplied. By R2, which is 64 minus H2, all multiplied by H. We end up gain 1/3 pi multiplied by 64H minus H cubed when simplified. Now, let's maximize this volume. We'll take this derivative with respect to H and set it equals to 0. So we end up getting. 1/3 spy Multiplied by 64 minus 3 H squared. We set this equals a 0. And solve for H. The 1/3 i will go away, so we get 0, equals 64. Minus 3 is 2. We subtract 64 To get 64 equals -32. Giving a 64 divided by 3 equals H2. And if we take the square root of both sides, We end up getting H. Equals 8 divided by the square root of 3. We can also verify if this is the maximum by taking the 2nd derivative. The second derivative, the prime of H, we look at our first derivative. Being 1/3 pi, 64 minus 3 squared. We would end up getting -2 pi H. This is negative When H squared than 0, which means that is our maximum. So now we have H, let's find R. Are Equals 64 Minus H2d underneath the square root. Yeah, the square root, 64 minus 8 divided by the square root of 3 square. If we were to simplify this, we end up getting 8 square of 6. Divided by 3. You now have R. 82 to 6 divided by 3 in meters. And H, which is 8 divided by square of 3. Or 8 squares of 3, divided by 3 in meters. And if we look at our puzzle answers, We see that this correspondence. To answer. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Optimization A right triangle has legs of length h and r and a hypotenuse of length 4 (see figure). It is revolved about the leg of length h to sweep out a right circular cone. What values of h and r maximize the volume of the cone? (Volume of a cone = (1/3) πr²h.) <IMAGE>

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