# A cylindrical drum is made to hold exactly 1m^3 in its interior. Assume that the material for the top and the bottom costs $20 per m^2, while that for the side costs $10 per m^2. Determine the radius of the drum that minimizes the cost of the material used.

Any help will be appreciated!

Area= 2PI r^2 + 2PI * r * h

Cost= 20*2PI r^2 + 10*2PI * r * h

but PI r^2 h is = 1m^3

so solve for h in the last equation, put that in the second, then take dCost/dr, set to zero, and solve.

so basically from what im understanding

(1) Area = 2PIr^2 + 2PI*r*h

h = -2PIr^2/(2PI*r)

But Im not understanding the part about where the Area variable goes, i understand subbing the (1) eqn into the second eqn, but I don't see what happens to the Area constant

## To solve for the radius of the drum that minimizes the cost of the material used, we need to express the cost function in terms of the radius alone. Here's how you can do it step by step:

1. Let's start with the formula for the volume of the cylindrical drum: V = πr^2h, where V is the volume, r is the radius, and h is the height.

2. We are given that the volume is 1m^3, so we have: 1 = πr^2h.

3. Rearrange the equation to solve for h: h = 1 / (πr^2).

4. Now, let's express the cost function in terms of the radius r. The cost can be calculated by adding the cost of the top and bottom surfaces (which is 2πr^2 multiplied by the cost per square meter) and the cost of the side surface (which is 2πrh multiplied by the cost per square meter). So, the cost function is: C = 20(2πr^2) + 10(2πrh).

5. Substitute the expression for h from step 3 into the cost function: C = 20(2πr^2) + 10(2πr(1 / (πr^2))).

6. Simplify the equation further: C = 40πr^2 + 20(2 / r).

7. Now, to find the radius that minimizes the cost, we need to find the derivative of the cost function with respect to the radius, set it equal to zero, and solve for r.

8. Take the derivative of C with respect to r: dC/dr = 80πr - 40(2 / r^2).

9. Set dC/dr equal to zero: 80πr - 40(2 / r^2) = 0.

10. Solve the equation for r: 80πr = 80 / r^2.

11. Simplify and solve for r: r^3 = 1 / (π^2).

12. Take the cube root of both sides: r = (1 / (π^2))^(1/3).

13. Approximate the value of r using a calculator or numerical approximation: r ≈ 0.589m (rounded to three decimal places).

Therefore, the radius of the drum that minimizes the cost of the material used is approximately 0.589m.