le Exam 2 Sample Exam 2 6 Find the dimensions of the largest

le Exam 2 Sample Exam 2 6. Find the dimensions of the largest box with rectangular volume such that the sum of the lengths of its 12 edges is 56 cm

Solution

suppose the dimensions of the box are x,y,z

given that 4x+4y+4z = 56

thus if v as the volume of the box then v=xy(56/4-x-y) or v= 56xy/4 - x^2y - xy^2

Vx = 56 (y/4) - 2xy -y^2 = y (14-2x-y) = 0

=> y=0 or 14-2x-y=0

Vy = 56(x/4) -x^2 - 2xy = x( 14 -x -2y) = 0

=> x=0 or 14-x-2y = 0

By the geometry of the problem we must have x!=0 and y!=0

Thus the critical point equations reduced to a set of two equations in two unknowns.

2x+y = 14 and x+2y = 14

from here y=14-2x

x+2(14-2x) = 14

x+28-4x = 14

3x = 14

x= 14 / 3

on substituting this in y we get

y=14-2x = 14-2(14/3) = 14 - 28/3 = -14/3

The box of maximum volume is then a cube of side 14/3

 le Exam 2 Sample Exam 2 6. Find the dimensions of the largest box with rectangular volume such that the sum of the lengths of its 12 edges is 56 cm Solutionsup

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