Posted by **Jen** on Saturday, November 18, 2006 at 11:01pm.

this is my last question from applications of derivatives!

A 216m^2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed.

Thank you all for helping.

Let's suppose your dividing fence is vertical and has a length of x. Let the length of the other side be y.

Since the area must be 216, y = 216/x

Perimeter P = 3x + 2y which after substituting is equal to 3x + 432/x

Find derivative: dP/dx = 3 - 432/x^2. Set it to zero and solve for x.

3x^2 = 432

x^2 = 144

We can accept the positive solution only. Therefore, x = 12.

y = 18 (sub 12 for x into y = 216/x)

The dimensions are 18m and 12m

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