Recall Tony made a rectangular box with 100 in^{2} 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)
= 6x^{2} + 2xz + 6zx
= 6x^{2} + 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(6x^{2} + 8xz)/dx
The left hand side of the equation, ds/dx is zero since the surface
area is constant at 100 in^{2}.
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 |