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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes within restricted domains}}
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\textbf{Question}

A farmer is constructing an enclosure for some of his animals. He has
a 100m long length of straight fence which will be bent at equal angles at equal
distances from the ends of the fence. The resulting three segment
fence will be placed along an existing fence to make a trapezoid
shaped enclosure. What is the maximum area for the enclosure?


\textbf{Answer}

The dimensions of the enclosure are given by 

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And so $2x+y=100$ is the length of the fence. In order to obtain
maximum area, $x>0 \ \longrightarrow(a)$ and $0 < \theta < \pi/2 \ 
\longrightarrow(b)$. As $h=x\cos \theta$,
this means that the area $A$ is given by
\begin{eqnarray*}
A & = & xy\cos\theta + 2\times\frac{1}{2} (x\sin\theta)(x\cos\theta)\\
& = & x(100-2x)\cos\theta + x^2\sin\theta\cos\theta\\
& = & (100x-x2x^2)\cos\theta + \frac{1}{2}x^2 \sin 2\theta.
\end{eqnarray*}
We need a critical point that satisfies conditions (a) and (b), and so
\begin{eqnarray*}
0 = \frac{\partial A}{\partial x} & = & (100-4x)\cos\theta +
x\sin\theta\\
& \Rightarrow & \cos\theta (100-4x+2x\sin\theta) = 0\\
& \Rightarrow & 4x-2x\sin\theta = 100\\
\Rightarrow x & = & \frac{50}{2-\sin\theta}\\
& & \\
0 = \frac{\partial A}{\partial \theta} & = &-(100x-2x^2)\sin\theta +
x^2 \cos 2\theta\\
& \Rightarrow & x(1-2\sin^2\theta)+2x\sin\theta -100\sin\theta =0
\end{eqnarray*}
By combining these equations we get
$$\frac{50}{2-\sin\theta}(1-2\sin^2\theta+2\sin\theta) - 100\sin\theta
=0$$
$$50(1-2\sin^2\theta +2\sin\theta) = 100(2\sin\theta - \sin^2\theta)$$
$$50=100\sin\theta$$
And so $\sin\theta=1/2$, $\Rightarrow \theta=\pi/6$
\begin{eqnarray*}
\Rightarrow x & = & \frac{50}{2-(1/2)} = \frac{100}{3}\\
y & = & 100-2x = \frac{100}{3}
\end{eqnarray*}
This means that the maximum area for the enclosure is
$$A=\left (\frac{100}{3} \right )^2 \frac{\sqrt{3}}{2} + \left (
\frac{100}{3} \right )^2 \frac{1}{2} \frac{\sqrt{3}}{2} = \frac{2500}{\sqrt{3}}$$ 
square units. And so the three segments of the fences will all be of
equal length, with the bend angles at $120^o$.


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