A garbage container is to be constructed out of 400 ft2 of material. The vessel will have a square base and an open top. Find the width and the height of the container that maximize its capacity and give the maximum capacity? Round your answers to 3 decimal places.

Volume is maximized for a given area by having the area of a pair of opposite sides equal the area of the bottom. That means the overall area of the container is 3 times the area of the bottom. Then the square bottom will have a width of ...

w = √(400/3) ≈ 11.547 . . . feet

The height is half that, so is ...

h = w/2 = 11.547/2 ≈ 5.774 . . . feet

The capacity is then ...

w²h = (11.547 ft)²(5.774 ft) = 769.800 ft³

The container is 11.547 ft wide by 5.774 ft high. It has a capacity of 769.800 cubic feet.

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You want to maximize w^2h subject to w^2 + 4wh = 400. Solving the constraint equation for h, we get h = (400 -w^2)/(4w) and the volume we want to maximize can be written as ...

V = w(400-w^2)/4

This will be an extreme when dV/dw = 0, so we want to solve ...

dV/dw = 0 = 100 -(3/4)w^2

w^2 = 400/3

w = √(400/3) . . . . . as above