A large soup can is to be designed so that the can will hold 54pi cubic inches of soup. The radius of the can is "x" inches and the height of the can is "h" inches. Find the values for the radius and the height such that the amount of metal needed is as small as possible.

Note: The side require 2pixh square inches of metal and each end requires pix^2 square inches.

Volume = πx^2h

h = 54/(πx^2)

Surface area = SA = 2πx^2 + 2πxh
SA = 2πx^2 + 2πx(54/(πx^2))
= 2πx^2 + 108/x
d(SA)/dx = 2πx - 108/x^2
= 0 for a min of SA
2πx = 108/x^2
x^3 = 108/(2π) = 54/π
x = (54/π)^(1/3) = appr. 2.58076
h = 54/(πª2.58076^2)) = appr. 2.58076

surprise! , the height should be equal to the radius.

Why did the 2ðx^2+108/x^2 become 2ðx-108/x^2

To find the values for the radius and height that will minimize the amount of metal needed for the can, we need to use optimization techniques.

Step 1: Write an equation for the volume of the can.
Since the volume of a cylinder is given by V = πr^2h, we can write the equation for the volume as follows:
πx^2h = 54π

Step 2: Solve the equation for one variable in terms of the other.
Solving for h in terms of x, we have:
h = 54 / x^2

Step 3: Write an equation for the amount of metal needed.
The amount of metal needed is the sum of the metal required for the sides and the metal required for the ends. The metal required for the sides is given by 2πxh, and the metal required for each end is given by πx^2. Therefore, the equation for the amount of metal needed is:
M = 2πxh + πx^2

Step 4: Substitute the expression for h from step 2 into the equation for the amount of metal needed.
Substituting h = 54 / x^2, we get:
M = 2πx(54 / x^2) + πx^2
M = (108π / x) + πx^2

Step 5: Find the derivative of the equation for the amount of metal needed with respect to x.
To find the values of x and h that minimize the amount of metal needed, we need to find the critical points of the equation. Taking the derivative of the equation with respect to x, we have:
dM/dx = -108π / x^2 + 2πx

Step 6: Set the derivative equal to zero and solve for x.
Setting dM/dx = 0, we have:
-108π / x^2 + 2πx = 0
-108π + 2πx^3 = 0
2πx^3 = 108π
x^3 = 54
x = ∛54

Step 7: Substitute the value of x back into the equation for h.
Using the equation h = 54 / x^2, we have:
h = 54 / (∛54)^2
h = 54 / 3∛2
h = 18/∛2

Therefore, the values for the radius (x) and height (h) that will minimize the amount of metal needed for the can are:
x = ∛54
h = 18/∛2