The sum of the length and the girth (perimeter of a cross-section) of a package carried by a delivery service cannot exceed 108108 in. find the dimensions of the rectangular package of largest volume that can be sent.

x= 18

y= 18

z= 36

Step-by-step explanation:

The volume of the package = xyz

The girth (perimeter) of the package = 2(x+y)

Since the sum of the girth and length does not exceed 108, we have

z + 2(x+y) = 108

z = 108 - 2(x+y)

Put the value of z into V = xyz

V = xy(108 - 2(x+y)

V = xy (108 -2x - 2y)

V = 108xy - 2x^2y - 2xy^2

V= f(x,y)

Differentiate V with respect to x

dV/dx = 108y - 4xy - 2y^2 = 0

Differentiate V with respect to y

dV/dy = 108x - 4xy - 2x^2 = 0

Therefore we have

dV/dx = 108y - 4xy - 2y^2 = 0 and

dV/dy = 108x - 4xy - 2x^2 = 0

From the symmetry of both, x= y

Therefore;

108x - 4xy - 2x^2 = 0

108x - 4x(x) - 2x^2

108x - 4x^2 - 2x^2 = 0

108x - 6x^2 = 0

6(18x - x^2) = 0

18x - x^2 = 0

x(18 - x) = 0

Therefore, x= 0 or 18-x =0

x = 0 or x= 18

This means x=y=0 or x=y=18

V= f(x,y) has a critical point at (x, y) = (18,18)

Recall that z = 108 -2(x+y)

z = 108 - 2(18+18)

z = 108 - 2(36)

z = 108 - 72

z = 36

Therefore x=y=18 and z= 36

2

step-by-step explanation:

4 and 5)16.97 or 12√2

6)15.59 or 9√3

7)y = 15.59 or 9√3     x =27

8)17.80