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?

Length = Width = Height = 64 ; Luggage Box is a CUBE

Step-by-step explanation:

Given : Length (L) + Width (W) + Height (H) ≤ 192 ; & Length = Width

Volume ( V )of luggage [Cuboid] = L.W.H

Since L = W :

V = L.L.H  → V = L^2H  

L + L + H = 192 →  2L + H = 192 → H = 192 - 2L

Putting value of 'H' in 'V'

V = L^2 (192 - 2L) → V = 192L^2 - 2L^3

Maximising volume function , differentiating it w.r.t 'L'

dV / dL = 2(192 L) - 3 (2L^2)  → = 384L - 6L^2

dV / dL = 384L - 6L^2  = 0

384L = 6L^2 → 384 = 6L

L = 384 / 6 → L = 64  ; W also = 64

Putting value of L in 1st given sum equation : 2L + H = 192

2 (64) + H = 192

128 + H = 192 → H = 192 - 128 → H = 64

Since L = W = H = 64 ; Luggage bag is a Cube

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

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