(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?

The dimensions of the luggage should be 64 cm wide, by 64 cm long, by 64 cm high.

Since some airlines have restrictions on the size of items of luggage that passengers are allowed to take with them, supposing 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, and a passenger wants to take a box of the maximum allowable volume, to determine, if the length and width are to be equal, what should the dimensions be, the following calculation must be performed:

50 x 50 x 92 = 230,000 60 x 60 x 72 = 259,000 70 x 70 x 52 = 254,800 65 x 65 x 62 = 261,950 64 x 64 x 64 = 262,144 63 x 63 x 66 = 261,954

Therefore, the dimensions of the luggage should be 64 cm wide, by 64 cm long, by 64 cm high.

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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.

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