Q:

Engineers are designing a box-shaped aquarium with a square bottom and an open top. The aquarium must hold 256 ft^3 of water. What dimensions should they use to create an acceptable aquarium with the least amount of glass?

Accepted Solution

A:
This is a calculus problem (in optimization).  Let the aquarium dimensions be L, W and H.  Then L*W*H = 256 ft^3.  But W=L here, so   W^2*H = 256 ft^3.

Write an expression for the area of the aquarium's glass sides and bottom.

                                                       256 ft^3                             256 ft^3
A= W^2 + 4(W*H) = W^2 + 4*W*(------------- ) = W^2 + 4*-------------------
                                                          W^2                                      W

So now we have A(W), the area as a function of W alone.

We want to minimize this area.  To do this, differentiate A(W) with respect to W and set the result = to 0.  We want to determine the "critical numbers."

dA/dW = 2W - 1024*W^(-2) = 0
                        1024
Then 2W = ---------------
                       W^2

2W^3 = 1024, so W^3 = 512, and  W = third root of 512 = 8

If W = 8 ft, then L = 8 ft also.     Since  L*W*H = 256 ft^3,

L*W*H = 256 ft^3   = (8 ft)^2*H = 256 ft^3.  Then H = 4

The acquarium dimensions are 8 by 8 by 4 feet.  This keeps the area of the aquarium to a minimum.