Question on min/max question (1 Viewer)

piethepker · Nov 8, 2015

A 3m piece of wire is cut into two pieces and bent around to form a square and a circle. Find the two lengths, correct to 2 decimals places, that will make the total area of the square and circle a minimum.

Thanks.

InteGrand · Nov 8, 2015

piethepker said:
A 3m piece of wire is cut into two pieces and bent around to form a square and a circle. Find the two lengths, correct to 2 decimals places, that will make the total area of the square and circle a minimum.

Thanks.

$Let the length of wire (in metres) for the circle be $C\geq 0$ and that for the square be $S \geq0$. Then we must have $C+S =3$ (since we're using a total of 3 metres). We want to find the values of $C$ and $S$ that will minimise the total area.$

$Now, since the perimeter of the square is $S$, its side length is $\frac{S}{4}$, so its area is $A_\text{square}=\frac{S^2}{16}$.$

$Since the circle's circumference is $C$, its radius is $\frac{C}{2\pi}$ (since $C=2\pi r $), so its area is $A_\text{circle}=\pi\cdot \left( \frac{C}{2\pi} \right)^2 = \frac{C^2}{4\pi}$.$

$So the total area is $A=\frac{S^2}{16} + \frac{C^2}{4\pi}$. But $C=3-S$, since $C+S=3$. Hence $A=\frac{S^2}{16} + \frac{(3-S)^2}{4\pi}$. You should be able to find the value of $S$ (call it $S^*$) that makes $A$ minimum from here, since $A$ is now a function of a single variable, and then the corresponding value of $C$ (says $C^*$) is given by $C^* = 3-S^*$.$

Question on min/max question (1 Viewer)

piethepker

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InteGrand

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