your company needs to design cylindrical metal containers with a volume of 16 cubic feet the top and bottom will be made of a steardy material that costs $2 per foot while the material for the sides costs $1 per square foot. find the radius, cost and height of the least expensive container.

The cost of the top and bottom will be the area multiplied by the cost per square foot. Of course, the area includes the area of both top and bottom. That cost is ...

cost for ends = ($2)(πr²)(2) = $4πr²

The cost for the side is the cost per square foot multiplied by the lateral area. That cost is ...

cost for sides = ($1)(2πrh) = $2πr(16/(πr²)) = $32/r

Then the total cost is ...

total cost = cost for ends + cost for sides

c = 4πr² +32/r . . . . . . dollars

This will be a minimum where its derivative is zero:

dc/dr = 0 = 8πr -32/r²

8πr³ = 32 . . . . . . add 32/r², multiply by r² . . . [eq3]

r = ∛(4/π) . . . . . . divide by the coefficient of r³ and take the cube root

h = 16/(π(∛(4/π))² = 16/∛(16π)

h = 4r

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We want a simple way to compute cost. Looking at [eq3] and comparing to the cost equation, we find ...

c = 16/r + 32/r = 48/r

c = 12∛(16π)

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So, the final dimensions and cost are ...

radius = ∛(4/π) ≈ 1.0839 . . . feet
height = 4r ≈ 4.3354 . . . feet
cost = 12∛(16π) ≈ 44.29 . . . dollars

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Comment on cylinder optimization

Note that the cost of the sides is exactly double the cost of the top and bottom, That will be true of the lowest-cost container, regardless of the relative prices of top/bottom material and side material.

When material is all the same cost, the optimum shape is height = diameter. Here, since the ends cost twice as much as sides per unit area, the height is twice the diameter to equalize the costs.