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A Wire of Length 28 M is to Be Cut into Two Pieces. One of the Pieces is to Be Made into a Square and the Other into a Circle. What Should Be the Lengths - CBSE (Arts) Class 12 - Mathematics

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Question

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?

Solution

\[\text { Suppose the wire, which is to be made into a square and a circle, is cut into two pieces of length x m and y m, respectively . Then, } \]

\[x + y = 28 . . . \left( 1 \right)\]

\[\text { Perimeter of square }, 4\left( side \right) = x\]

\[ \Rightarrow \text { Side } = \frac{x}{4}\]

\[ \Rightarrow \text { Area of square } = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16}\]

\[\text { Circumference of circle }, 2\pi r = y\]

\[ \Rightarrow r = \frac{y}{2\pi}\]

\[\text { Area of circle =} \pi r^2 = \pi \left( \frac{y}{2\pi} \right)^2 = \frac{y^2}{4\pi}\]

\[\text { Now, }\]

\[z = \text { Area of square + Area of circle }\]

\[ \Rightarrow z = \frac{x^2}{16} + \frac{y^2}{4\pi}\]

\[ \Rightarrow z = \frac{x^2}{16} + \frac{\left( 28 - x \right)^2}{4\pi}\]

\[ \Rightarrow \frac{dz}{dx} = \frac{2x}{16} - \frac{2\left( 28 - x \right)}{4\pi}\]

\[\text { For maximum or minimum values of z, we must have }\]

\[\frac{dz}{dx} = 0\]

\[ \Rightarrow \frac{2x}{16} - \frac{2\left( 28 - x \right)}{4\pi} = 0 .............\left[ \text { From eq }. \left( 1 \right) \right]\]

\[ \Rightarrow \frac{x}{4} = \frac{\left( 28 - x \right)}{\pi}\]

\[ \Rightarrow \frac{x\pi}{4} + x = 28\]

\[ \Rightarrow x\left( \frac{\pi}{4} + 1 \right) = 28\]

\[ \Rightarrow x = \frac{28}{\left( \frac{\pi}{4} + 1 \right)}\]

\[ \Rightarrow x = \frac{112}{\pi + 4}\]

\[ \Rightarrow y = 28 - \frac{112}{\pi + 4} ............\left[ \text { From eq } . \left( 1 \right) \right]\]

\[ \Rightarrow y = \frac{28\pi}{\pi + 4}\]

\[ \frac{d^2 z}{d x^2} = \frac{1}{8} + \frac{1}{2\pi} > 0\]

\[\text { Thus, z is minimum when x } = \frac{112}{\pi + 4} \text { and }y = \frac{28\pi}{\pi + 4} . \]

\[\text { Hence, the length of the two pieces of wire are } \frac{112}{\pi + 4} m \text { and } \frac{28\pi}{\pi + 4} \text { m respectively }.\]

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Solution A Wire of Length 28 M is to Be Cut into Two Pieces. One of the Pieces is to Be Made into a Square and the Other into a Circle. What Should Be the Lengths Concept: Graph of Maxima and Minima.
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