What is the optimal fencing for a triangular field?

[Chris L T521]

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:

Spoiler

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]

This week's problem was correctly answered by MarkFL and Sudharaka. You can find Sudharaka's solution below.

Spoiler

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

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

\[p=x+y+z~~~~~~~~~~(3)\]

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 \(x=y=z=\frac{p}{3}\).