Answer:
11.547 ft wide by 5.774 ft high
769.800 ft³ capacity
Step-by-step explanation:
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.
_____
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
