Math CentralQuandaries & Queries


Question from Kari, a student:

I have gotten stuck trying to solve this problem...please help:
A rectangular pen is to be built using a total of 800 ft of fencing. Part of this fencing will be used
to build a fence across the middle of the rectangle (the rectangle is 2 squares fused together so if you can
please picture it).
Find the length and width that will give a rectangle with maximum total area.

Hi Kari.

The pen is a rectangle, so there is a length and a width (L and W). Additionally, there is another Length or Width (it doesn't matter which, so we'll call it the Length) which is the middle dividing fence.

Thus, the fence length is 800 = 3L + 2W.

The area is, of course, A = LW.

Using the first equation, you can express L in terms of W (i.e. solve for L). Then you can substitute this into the second equation, thereby creating a combined third equation which has area A and an algebraic expression involving just W.

Now differentiate with respect to W. You'll end up with dA/dW on the left and an expression involving W on the right. Recall that at the maximum area, we have a critical point, so dA/dW = 0. So substitute 0 for dA/dW on the left.

Solve for W. Then use that to find L and you are done.

Stephen La Rocque.

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