Question

Suppose you have a total surface area of 2400 inches squared of cardboard in order to construct a box with a square base. If x  represents the sidelength of the base of the box, write out a volume function to be maximized as a function of x. Assume that the box contains both a top and bottom. (PLEASE HURRY :))

V(x)=x^2(600−2x^2)

V(x)=2x^2+4x(2400/x2)

V(x)=x^2((2400−2x)/(24x))

The correct answer is not given.

V(x) = x^2 (2400-2x^2)

Answer 1

Let length of the box is x

width is x

height is y

Surface area of box

S=2x^2+4xy (because surface area of botton and top + 4 side area)

2400=2x^2+4xy

2(x^2+2xy)=2400

x^2+2xy=1200

2xy=1200-x^2

y=(1200 - x^2)/2x

Volume of the box

V=x^2y

=x^2((1200 - x^2)/2x) (Putting the value of y from above)

=x/2(1200-x^2)

=600x-x^3/2

Differentiating with respect to x

dV/dx=600 -3x^2/2

For critical points

dV/dx=0

600-3x^2/2=0

3x^2/2=600

x^2=400

x=20

Vmax = 600(20)-(20)^3)/2

=8000