Originally Posted by

**ecMathGeek** 3)

The deminsions we will use are x = length, and y = width

We are given A = 32 = xy

We need P = 2x + 2y to be a minimum.

First, let's solve 32 = xy for x (we could solve it for y, too... we'd get the same answer)

32 = xy

x = 32/y

Second, plug in x = 32/y into P = 2x + 2y

P = 2(32/y) + 2y

P = 64/y + 2y

P = 64y^(-1) + 2y ... since 1/y = y^(-1)

Now, we need to make this a quadratic function (which we can do if we multiply everything by y)

yP = 64 + 2y^2

2y^2 - Py + 64 = 0

For the minimum area, all we need is y = -b/(2a)

y = -(-P)/(2*2)

y = P/4

Since P = 2y + 2x

2x = P - 2y

x = P/2 - y ... Plug in y = P/4

x = P/2 - P/4

x = P/4

Therefore, x and y are one forth the total perimater.