# Optimization-Circle and Square

A piece of wire of length 60 is cut, and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to (a) minimize and (b) maximize the combined area of the circle and square?

So Area of circle and square= pir^2 +x^2 perimeter of circle and square = 2pir+4x

Solve for r: 2pir+4x=60 r=30-2x/pi

plug r into area equation

pi(30-2x/pi)^2+x^2=area (30-2x)^2/pi + x^2=

find derivative to get CP for min/max

A(X)=(30-2x)^2/pi + x^2 A'(x)= (-120+2x/pi)+2x

therefore, x=60 and x=0.

Now at this point I know that if I plug in 60 and 0 in the original area function, it will give me a number then I can compare those numbers and see where is it largest at what x value. But how would I find the minimum value exactly?

• You're almost there. You need to figure out which value of $x$ gives the maximum area, and which givess the minimum area. Then plug that $x$ value in and use the area equation you already calculated: $\pi(\frac{30-2x}{\pi})^{2} + x^{2}$. – ShallowBlue Nov 5 '14 at 21:36

You made a mistake in final calculations. You have the area as a function of $x$: $$A(x)=(30-2x)^2/\pi+x^2$$ So the derivative $$A'(x)=-120/\pi+8x/\pi+2x$$ $$A'=0 \to x=60/(\pi+4)$$ This is point of minimum since $A(x)$ is a parabola looking up. Maximum is obtained on 1 of edges $x=0,60$ which is easy to check. Obviously $x=60$ is a maximum. Plug corresponding $x_{min}$ and $x_{max}$ to find maximum/minimum values of the area.