Application of Derivatives Question - CBSE

Question

A wire of length $28	ext{ m}$ 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 square and the circle is minimum?

Accepted Answer

Total area $A(x) = \dfrac{x^2}{16} + \dfrac{(28-x)^2}{4\pi}$. [1.0 Mark]
$\dfrac{dA}{dx} = \dfrac{x}{8} - \dfrac{28-x}{2\pi} = 0 \Rightarrow x(\pi + 4) = 112 \Rightarrow x = \dfrac{112}{\pi + 4}$. [2.0 Marks]
Verify $\dfrac{d^2 A}{dx^2} > 0 \Rightarrow$ Minimum. [0.5 Mark]
Lengths of pieces: Square piece $= \dfrac{112}{\pi + 4} ext{ m}$, Circle piece $= \dfrac{28\pi}{\pi + 4} ext{ m}$. [1.5 Marks]

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🎯 Official CBSE Marking Scheme:
Setting up combined area function: 1.0 Mark
Differentiating and solving $x = 112/(\pi+4)$: 2.0 Marks
Applying Second Derivative Test: 0.5 Mark
Evaluating lengths of both pieces: 1.5 Marks

Question:

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