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This is a question in my textbook which I can't solve. Any help would be appreciated, thanks.

"A piece of wire 10 metres long is cut into two portions. One piece is bent to form a circle, and the other piece to form a square. Find the circumference of the circle if the sum of areas of the circle and square is to be a minimum. Give your answer in terms of $pi$."

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    $\begingroup$ You can use the same basic approach as in the window problem about which you asked earlier. $\endgroup$ – Brian M. Scott Aug 24 '13 at 7:26
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HINT:

If the radius of the circle is $=r$ meter, the perimeter of the circle $=2\pi r$ meter

If one side of the square is $=a$ meter, the perimeter of the sqaure $=4a$ meter

So, we have $2\pi r+4a=10=a=\frac{10-2\pi r}4$

and the sum of area $=\pi r^2+a^2=f(a)$ to be maximized

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It' fairly easy to solve a more general problem: You cut the wire of length $1$ in two pieces, $ x$ and $1-x$. From one piece you form a regular $n$-gon, from the other a regular $m$-gon. Minimize the sum of the area of both.

For any regular $n$-gon let $r_n$ the radius of its inscribed circle and $\pi_n$ the ratio of its circumference to its diameter. (We'll get $\pi_n=n\cdot\tan(\pi/n)$, so, e.g., $\pi_3=3\sqrt3$, $\pi_4=4$ and $\pi_6=2\sqrt3$ and so on. We may show that $\pi_\infty=\pi$.)

Now it's easy to verify that the circumference $L_n$ of the $n$-gon is $L_n=2\cdot\pi_n\cdot r_n$ and its area is $A_n=\pi_n\cdot r_n^2$. From $r_n=L_n/2\pi_n$ we conclude $A_n=L_n^2/4\pi_n$.

Thus the sum of the area equals $$\frac{x^2}{4\pi_n}+\frac{(1-x)^2}{4\pi_m}.$$ By standard methods it arrives its minimum in $$x=\frac{\pi_n}{\pi_n+\pi_m}.$$ So cut a wire of length $r$ in two pieces of length $\frac{r\pi_n}{\pi_n+\pi_m}$ and $\frac{r\pi_m}{\pi_n+\pi_m}$. The the minimal area is $$\left(\frac{r}{2}\right)^2\frac{1}{\pi_n+\pi_m}.$$

For $n=3$ and $m=6$ we'll get $0.6r$ and $0.4r$ with minimal area $(r/2)^2\cdot\frac{1}{5\sqrt3}$.

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Given Total length of wire $10$ meter, Now Let $x$ meter cut from it and form a Circle and

remaining $(10-x)$ form a Square.

So Here Radius of Square is $\displaystyle \frac{x}{2\pi}$ and Length of square is $\displaystyle \frac{(10-x)}{4}$

Now Area of Circle $\displaystyle \pi\cdot \frac{x^2}{4\pi^2} = \frac{x^2}{4\pi}$ and Area of Square $\displaystyle \frac{(10-x)^2}{16}$

So Total area $\displaystyle A = \frac{x^2}{4\pi}+\frac{(10-x)^2}{16}.$

Now here we have to minimize $A$

So using $\bf{Cauchy-Schwartz}$ Inequality, We get

$$\displaystyle A = \frac{x^2}{4\pi}+\frac{(10-x)^2}{16}\geq \frac{x+10-x}{4\pi+16} = \frac{10}{4\pi+16}$$

and equality hold when $$\displaystyle \frac{x}{4\pi} = \frac{10-x}{16}\Rightarrow x=\frac{10\pi}{\pi+4}$$

So length of wire to form a circle $\displaystyle x = \frac{10\pi}{\pi+4}$ and

length of wire to form a square $\displaystyle (10-x) = 10-\frac{10\pi}{\pi+4} = \frac{40}{\pi+4}$

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