A model of carry-on luggage has a length that is 7 inches greater than its depth

Question

A model of carry-on luggage has a length that is 7 inches greater than its depth. Airline regulations require that the sum of the length, width, and depth cannot exceed 25 inches. These conditions, with the assumption that this sum is 25 inches, can be modeled by the function V(x) that gives the luggage's volume, in cubic inches, in terms of its depth, x, in inches. If its volume is 240 cubic inches, determine two possibilities for its depth. Volume = depth middot length middot width: 25 - (depth + length) V(x) = x middot (x + 7) middot [25 - (x + x + 7)] V(x) = x(x + 7)(18 - 2x) Two possibilities for its depth are inches.

Explanation / Answer

Let the depth of the carry-on luggage be x inches. Then its length is x +7 inches. Let its width be y inches. Then x + x +7 + y = 25 or, 2x + y = 18 or y = 18 -2x...(1) Then the volume V(x) of the carry-on luggage is depth*length*width = x(x+7)(18-2x). Thus x(x+7)(18-2x) = 240, or, 2x3 -4x2 -126x +240 = 0 or, x3 -2x2 -63x +120 = 0 or, (x -8)( x2 + 6x -15) = 0. Thus, either x = 8 or, ( x2 + 6x -15) = 0. If ( x2 + 6x -15) = 0, then x = [ -6 ± { (6)2 - 4*1*(-15)}]/ 2*1 = [-6 ± 96]/2 = ( -6 + 9.8)/2 as x cannot be negative. or, x = 3.8/2 = 1.9.

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