Hello,

This is tricky, or I didn't get the thing

Let x be the length of the wire bent to a circle and y to form a square.

**Let's study the circle**
x is its perimeter, which is also $\displaystyle 2 \pi R$, where R is the radius. According to it, we have the equation $\displaystyle x=2 \pi R \Longleftrightarrow R=\frac{x}{2 \pi}$

The area enclosed by the circle is $\displaystyle A_1=\pi R^2$, which is also $\displaystyle A_1=\pi \left(\frac{x}{2 \pi} \right)^2 \Longleftrightarrow { \color{red} A_1 = \frac{x^2}{4 \pi} }$

**Let's study the square**
$\displaystyle y= {\color{blue}75} -x$ is its perimeter.

If q is the length of the side of the square, the perimeter is 4q.

Hence $\displaystyle y=4q \Longleftrightarrow q=\frac{y}{4}$

The area of a square is $\displaystyle A_2=q^2 \Longleftrightarrow A_2=\frac{y^2}{4^2} \Longleftrightarrow {\color{red} A_2=\frac{({\color{blue}75}-x)^2}{16}}$.

**Total area**
$\displaystyle A=A_1+A_2=\frac{x^2}{4 \pi}+\frac{({\color{blue}75}-x)^2}{16}=\frac{x^2}{4 \pi}+\frac{{\color{blue}5625 - 150 x} + x^2}{16}$ $\displaystyle = x^2 \left(\frac{1}{4 \pi}+\frac{1}{16} \right)+ \frac{{\color{blue}5625 - 150x}}{16}$

So find the derivative and see, in the interval [0,

75] (since the wire is

75 cm long) where the maximum is and where the minimum is