# Problem of the Week #62 - June 3rd, 2013

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#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: A triangular field is to be enclosed by $p$ feet of fencing so as to maximize the area of the field. Find the lengths of the sides of this triangle.

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Hint:
Heron's formula for the area of a triangle with side lengths $x$, $y$ and $z$ is $A=\sqrt{s(s-x)(s-y)(s-z)}$, where $s=\frac{1}{2}(x+y+z)$ is the semiperimeter.

#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by MarkFL and Sudharaka. You can find Sudharaka's solution below.

$A=\sqrt{s(s-x)(s-y)(s-z)}~~~~~~~~(1)$

$s=\frac{1}{2}(x+y+z)~~~~~~~~~~(2)$

$p=x+y+z~~~~~~~~~~(3)$

Using (1), (2) and (3) we get,

$A=\frac{1}{4}\,\sqrt{p\,\left( p-2\,x\right) \,\left( p-2\,y\right) \,\left( 2\,x+2\,y-p\right) }$

Now we shall use the second partial derivative test to find the maximum of $$A$$ and the corresponding lengths of the sides.

Differentiating with respect to $$x$$ and $$y$$ we get,

$\frac{\partial A}{\partial x}=\frac{p\,\left( y+2\,x-p\right) \,\left( 2\,y-p\right) }{2\,\sqrt{p\,\left( 2\,x-p\right) \,\left( 2\,y-p\right) \,\left( 2\,y+2\,x-p\right) }}$

$\frac{\partial A}{\partial y}=\frac{p\,\left( 2\,x-p\right) \,\left( 2\,y+x-p\right) }{2\,\sqrt{p\,\left( 2\,x-p\right) \,\left( 2\,y-p\right) \,\left( 2\,y+2\,x-p\right) }}$

When $$\frac{\partial A}{\partial x}=0$$ we have,

$p=2y\mbox{ or }p=y+2x~~~~~(4)$

When $$\frac{\partial A}{\partial y}=0$$ we have,

$p=2x\mbox{ or }p=x+2y~~~~~~~~~(5)$

By (4) and (5) we get two possibilities,

$x=y=\frac{p}{2}\mbox{ or }x=y=\frac{p}{3}$

$$x=y=\frac{p}{2}\Rightarrow z=0$$. Hence these lengths do not form a triangle. So the only possibility is,

$x=y=z=\frac{p}{3}$

We can also show that,

$D\left( \frac{p}{3},\frac{p}{3}\right) = A_{xx}\left( \frac{p}{3},\frac{p}{3}\right) A_{yy} \left(\frac{p}{3},\frac{p}{3}\right) - \left( A_{xy}\left(\frac{p}{3},\frac{p}{3}\right) \right)^2=\frac{9}{4}>0$

and

$A_{xx}\left(\frac{p}{3},\frac{p}{3}\right)=-\sqrt{3}<0$

Therefore by the second partial derivative test, $$A$$ has a maximum at $$x=y=z=\frac{p}{3}$$.

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