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## Saturday, 27 August 2011

### Math GRE - #26

A total of $x$ feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of $x$?

1. $\dfrac{x^2}{9}$
2. $\dfrac{x^2}{8}$
3. $\dfrac{x^2}{4}$
4. $x^2$
5. $2x^2$

Solution :

Let's call one side of the fence $a$ and the other side $b$. Since we are only looking at 3 sides of a rectangular fence (e.g. we're fencing in our back yard and one side is the wall of the house), let's call the side length that we have to fence twice. So we have: $x = 2a+b.$ The area of the fence is $A = ab = a(x-2a)$ where the last equality comes from our equation for $x$. To maximize the area, we take the derivative of $A$ with respect to $a$ and set equal to 0 to obtain: $0 = x-4a\implies a=\frac{x}{4}.$ We know this represents the maximum since the second derivative of $A$ is always negative. Using our equation for the perimeter, we now know that: $x=2a + b\implies b=\frac{x}{2}.$ Hence the maximum possible area is: $A=ab=\frac{x}{4}\cdot\frac{x}{2}=\frac{x^2}{8}.$

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