Question

(1 point) Some airlines have restrictions on the size of items of luggage that passengers are allowed to take with them. Suppose that one has a rule that the sum of the length, width and height of any piece of luggage must be less than or equal to 192 cm. A passenger wants to take a box of the maximum allowable volume. If the length and width are to be equal, what should the dimensions be?

Answer 1

Here L = W, but H can be different.
 
The sum L+H+W  must be less than or equal to 192 cm.  

We can solve L + H + W = 192 for H:  H = 192 - W - L.  Remembering that W = L,   the formula for H becomes 192 - 2W.

The formula for volume would be V = L*W*H.

This becomes V = W*W*H, or  V = W^2*(192-2W)

Multiplying this out:  V = w^2*192 - 2W^3

Two ways of determining W:

1) graph V = 192W^2 - 2W^3 and determine the value of W at which V is at a max with the constraint W + L + H is equal to or smaller than 192.

2) Differentiate V with respect to W and set the result equal to zero:

384W - 6W^2 = 0.  Solving for W:  W(384 - 6W) = 0.

W = 0 is trivial, so just solve 384 - 6W = 0 for W:  6W = 384, and W = 64.

The width is 64 cm, the length is 64 cm also, and the height is (192-2W) cm, or 64 cm.

These dimensions produce the max volume.