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: Let variables

Let $x$ be the length of wire used for the circle. Then the length of wire used for the square is $53 - x$.

Step2: Express areas

The radius of the circle $r=\frac{x}{2\pi}$, and its area $A_{1}=\pi r^{2}=\pi(\frac{x}{2\pi})^{2}=\frac{x^{2}}{4\pi}$. The side - length of the square $s = \frac{53 - x}{4}$, and its area $A_{2}=(\frac{53 - x}{4})^{2}=\frac{(53 - x)^{2}}{16}$.

Step3: Define the total - area function

The total area $A(x)=\frac{x^{2}}{4\pi}+\frac{(53 - x)^{2}}{16}$.

Step4: Differentiate the function

$A^\prime(x)=\frac{2x}{4\pi}+\frac{2(53 - x)(- 1)}{16}=\frac{x}{2\pi}-\frac{53 - x}{8}$.

Step5: Set the derivative equal to zero

$\frac{x}{2\pi}-\frac{53 - x}{8}=0$. Multiply through by $8\pi$ to clear the fractions: $4x-\pi(53 - x)=0$. Expand to get $4x-53\pi+\pi x = 0$. Combine like terms: $(4 + \pi)x=53\pi$.

Step6: Solve for $x$

$x=\frac{53\pi}{4+\pi}\approx\frac{53\times3.14159}{4 + 3.14159}=\frac{166.50427}{7.14159}\approx23.315$.

Step7: Find the length for the square

The length for the square is $53 - x=53 - 23.315 = 29.685$.

Answer:

The length used for the circle is approximately $23.315$, and the length used for the square is approximately $29.685$.