Calculus word problem

Question

Find the dimensions of a rectangle with area 216 m² whose perimeter is as small as possible.

smaller and larger values

Answer 1

Rectangle sides are x and y.

We are given x*y =216

The perimeter, P, is 2x + 2y.

To find a maximum or minimum, we need P as a function of just one variable. Let's choose x. From the first equation, y = 216/x. use this to replace y in the 2nd equation:

P = 2x + 216/x

For the next step, this is better written as P = 2x + 216x^-1

To get the minimum or maximum, take the 1st derivative of P and set = to zero:

P' = 2 - 216x^-2 = 0

216x^-2 = 2

multiply both sides by x² :

2x² = 216, x = ±√108 = ±10.39

The negative value is not useful, so we have x = 10.39

Use either of the original equations to find y