Question

A wire is 12 inches in length can be bent into a circle, a square, or cut to make both a circle and a square. How much wire should be used for the circle, if the total area enclosed by the figure(s) is to be a minimum? A maximum?

Answer 1

Total Area:
\[A = \pi r^2 + x^2\]
Total Length:
\[2 \pi r + 4x = 12\]
We want to solve for r when Area is max/min
solve length equation for x
\[x = 3 - \frac{\pi}{2} r\]
\[A = \pi r^2 +(3-\frac{\pi}{2}r)^2\]
set derivative equal to 0
\[\frac{dA}{dr} = 2 \pi r+ (-\frac{\pi}{2})(2)(3-\frac{\pi}{2}r) = 0\]
\[r = \frac{3}{2+\frac{\pi}{2}} = \frac{6}{\pi+4}\]
Next we need to determine if this "r" gives a max or min
To do this we can compare Area with endpoints along constraints, meaning what is area when r=0 or x=0
PLugging in our found "r" value we obtain a Area of :
\[A = 5.04\]
when r=0 , x =3
\[A_{\square} = 9\]
when x=0, r = 6/pi
\[A_{circle} = \frac{36}{\pi} = 11.46\]

From comparing these areas we find that the critical "r" value gives a min Area and max Area is obtained when all the wire is used for a circle.