a piece of wire of length 53 is cut into two pieces. one piece is bent into a square and the other is bent…

Explanation:

Step1: Define variables

Let $x$ be the length of wire used for the square. Then the length of wire used for the circle is $53 - x$. The side - length of the square $s=\frac{x}{4}$, and the circumference of the circle $C = 53 - x$. Since $C = 2\pi r$, the radius of the circle $r=\frac{53 - x}{2\pi}$.

Step2: Express the sum of areas

The area of the square $A_{s}=s^{2}=(\frac{x}{4})^{2}=\frac{x^{2}}{16}$, and the area of the circle $A_{c}=\pi r^{2}=\pi(\frac{53 - x}{2\pi})^{2}=\frac{(53 - x)^{2}}{4\pi}$. The sum of the areas $A(x)=\frac{x^{2}}{16}+\frac{(53 - x)^{2}}{4\pi}$.

Step3: Differentiate the area function

Differentiate $A(x)$ with respect to $x$. Using the power rule, $A^\prime(x)=\frac{2x}{16}+\frac{2(53 - x)(- 1)}{4\pi}=\frac{x}{8}-\frac{53 - x}{2\pi}$.

Step4: Set the derivative equal to zero

Set $A^\prime(x) = 0$: [ \begin{align*} \frac{x}{8}-\frac{53 - x}{2\pi}&=0\ \frac{x\pi}{8\pi}-\frac{4(53 - x)}{8\pi}&=0\ x\pi-4(53 - x)&=0\ x\pi-212 + 4x&=0\ x(\pi + 4)&=212\ x&=\frac{212}{\pi + 4} \end{align*} ]

Step5: Find the length for the circle

The length of wire for the circle is $53 - x=53-\frac{212}{\pi + 4}=\frac{53(\pi + 4)-212}{\pi + 4}=\frac{53\pi+212 - 212}{\pi + 4}=\frac{53\pi}{\pi + 4}$.

Answer:

The length of wire for the square is $\frac{212}{\pi + 4}\approx29.12$, and the length of wire for the circle is $\frac{53\pi}{\pi + 4}\approx23.88$.