Please Help!! Cannot solve this prblem.

Question

A cuboid container with a base length twice its width is made with 48 square meters of metal. Show that the height h=(8/x)-(2x/3), where x is the width of the base. Express the colume as a function of x and find the maximum volume of the container.

Answer 1

Given base length=L, base width=W, and the height=H.

The cuboid has a surface area A=2*(L*W)+2*(L*H)+2*(W*H). There are two of each face of size L*W, L*H, and W*H.

We are told L=2*W. Also x=W, so L=2*W=2*x. And finally A, the total surface area making up the cuboid is 48m^2.

48=2*(2x)*x+2*(2x)*H+2*(x)*H=4x^2+6xH

48=4x^2+6xH

Solve for H.

48-4x^2=6xH

48/6x-4x^2/6x=H

H=8/x-2x/3

The volume of the cuboid is represented by V=L*W*H

Using the relationships solved for:

V=(x)*(2x)*(8/x-2x/3)

V(x)=16x-4x^3/3

We expect the maximum volume to occur where dV/dx=0.

dV/dx=16-4x^2

16-4x^2=0 -> x=+/-2.

We expect x to be a positive value because the dimensions of the cuboid should all be positive (x=W).

x=2 where V is maximum.

The maximum volume is V(x=2)=16x-4x^3/3=16(2)-4(2)^3/3=20.33 m^3