A wire 40 text{ m} in length is to be cut int | vedclass

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(A) Let the length of the piece for the square be $x$ and the length of the piece for the circle be $40 - x$.Side of the square $s = \frac{x}{4}$. Are

Vedclass · Accepted Answer

$\frac{160}{\pi + 4}, \frac{40\pi}{\pi + 4}$

Vedclass · Answer

(A) Let the length of the piece for the square be $x$ and the length of the piece for the circle be $40 - x$.Side of the square $s = \frac{x}{4}$. Area of square $A_s = (\frac{x}{4})^2 = \frac{x^2}{16}$.Circumference of the circle $2\pi r = 40 - x$, so $r = \frac{40 - x}{2\pi}$. Area of circle $A_c = \pi r^2 = \pi (\frac{40 - x}{2\pi})^2 = \frac{(40 - x)^2}{4\pi}$.Total area $A(x) = \frac{x^2}{16} + \frac{(40 - x)^2}{4\pi}$.To minimize, find $A'(x) = \frac{2x}{16} + \frac{2(40 - x)(-1)}{4\pi} = \frac{x}{8} - \frac{40 - x}{2\pi} = 0$.$\frac{x}{8} = \frac{40 - x}{2\pi} \implies \pi x = 160 - 4x \implies x(\pi + 4) = 160$.$x = \frac{160}{\pi + 4}$.The other piece is $40 - \frac{160}{\pi + 4} = \frac{40\pi + 160 - 160}{\pi + 4} = \frac{40\pi}{\pi + 4}$.

$A$ wire $40 \text{ m}$ in length is to be cut into two pieces. One piece is formed into a square and the other piece into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum?

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