How to find the minimum area?

Question

I am trying to solve the following problem, using functions and the application of derivatives to find maximums and minimums:

A wire of $L$ length is cut in two parts. The first one is used to form a square, and the second one to form a circumference. How should the wire be cut so that the sum of the two areas is minimum?

Considering that each side of the square is $x$, and $L$ the complete length of the wire, I was thinking of using two functions: $$f\left(x\right):\:A=x^2+\pi r^2$$ $$f\left(x\right):\:L=4x+2\pi r$$ The issue is that I am not sure how to solve this, considering that the functions I made are good in the first place.

Answer 1

Hint: You have $$A=x^2+\pi r^2$$ where $$r=\frac{y}{2\pi}$$ so we get

$$A=x^2+\frac{y^2}{4\pi}$$ Now we have the condintion: $$L=4x+y$$ with this equation we can eliminate $$y$$:

$$A(x)=x^2+\frac{1}{4\pi}(L-4x)^2$$ $$A'(x)=2x-\frac{8}{4\pi}(L-4x)$$ solving this we get

$$x=\frac{L}{4(\pi+4)}$$

and from here we get

$$y=L-4\frac{L}{4(\pi+4)}=L-\frac{L}{\pi+4}=\frac{L\pi}{\pi+4}$$

Answer 2

Note that $L$ is a given value, so solve your second equation for one of the variables and plug that into the first equation. That gives you $A$ as a function of the other variable. Now take the derivative of $A$ with respect to that variable, set to zero....

Answer 3

Suppose that we use $x$ units of wire for the square (so the side length is $x/4$) and $L-x$ units for the circle (so the radius is $\frac{L-x}{2\pi}$. Then the area can be expressed as $$ A(x)=\left(\frac{x}{4}\right)^2+\pi\left(\frac{(L-x)}{2\pi}\right)^2\quad (0\leq x\leq L) $$ Since $A$ is continuous and the domain is compact, a minimum exists. You can take it from here.