# You have 1000 yards of fencing and you plan to use the fencing to make 2 enclosures, one circular...

## Question:

You have 1000 yards of fencing and you plan to use the fencing to make 2 enclosures, one circular and one a square. How much of the 1000 yards of fencing should be used for each region if you want to maximize the combined area of both regions? Let x yards of fence be used for the circumference of the circle and the rest, (1000 - x) yards, be used for the perimeter of the square.

## Minima and Maxima

A point on the curve where the function has it's lowest or highest value is called the Minima or the maxima of the function and to find these minima and maxima we find the point where the derivative of the function becomes zero.

Here as given the perimeter of the square is taken to be (1000-x) and the perimeter of the circle is taken to be x

So, let's find the area of the corresponding area.

Circle,

{eq}\begin{align*} Area&=(\frac{1000-x}{4})^2 \end{align*} {/eq}

Square,

{eq}\begin{align*} Area&=2\pi(\frac{x}{2\pi})^2 \end{align*} {/eq}

Now let's minimize the total area by finding out the radius at which the total area is minimum

{eq}\begin{align*} TArea&=2\pi(\frac{x}{2\pi})^2\;+\;(\frac{1000-x}{4})^2\\ TArea&=\frac{x^2}{2\pi}\;+\;(\frac{1000-x}{4})^2\\ \frac{\mathrm{d} (TArea)}{\mathrm{d} x}&=\frac{x}{\pi}\;-\;2(\frac{1000-x}{4})\frac{1}{4}\\ Equating \;this\; to \;'0'\\ 0&=\frac{x}{\pi}\;-\;2(\frac{1000-x}{4})\frac{1}{4}\\ \frac{x}{\pi}&=\frac{1000-x}{8}\\ 8x&=1000\pi-x\pi\\ x&=\frac{1000\pi}{8+\pi} \end{align*} {/eq}

So, here we get the circumference of the circle to be {eq}x=\dfrac{1000\pi}{8+\pi}\;yards {/eq}