Rectangular Box

Recall Tony made a rectangular box with 100 in2 surface area in his arts and crafts class. The length of his box is 3 times of its width. He wondered how the height changes affects width's enlargement?

Let the length of the box be y, the width be x and the height be z.

Let the surface area of the box be s.

The surface area of the box can be calculated as

     s = 2xy + 2xz + 2yz

Since the length is 3 times of the box's width, substitute y = 3x into the equation.

     s = 2x(3x) + 2xz + 2z(3x)

        = 6x2 + 2xz + 6zx

        = 6x2 + 8xz

Because the height z is implicitly defined by its width x, the expression, the height z changes with respect to its width x (dz/dz), can be calculated using implicit differentiation method. Differentiate both side of surface area equation with respect to x

     ds/dx = d(6x2 + 8xz)/dx

The left hand side of the equation, ds/dx is zero since the surface area is constant at 100 in2.

The right hand side is


The left hand side equals the right hand side, so

     12x + 8z + 8xdz/dx = 0

Rearranging gives

     dz/dx = -(3x + 2z)/2x = -1.5 + z/x