JEE Mains · Maths · STD 12 - 6. Application of derivatives
A wire of length \(20 m\) is to be cut into two pieces. A piece of length \(\ell_1\) is bent to make a square of area \(A_1\) and the other piece of length \(\ell_2\) is made into a circle of area \(A _2\). If \(2 A _1+3 A _2\) is minimum then \(\left(\pi \ell_1\right): \ell_2\) is equal to:
- A \(6: 1\)
- B \(3: 1\)
- C \(1: 6\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(A) \(6: 1\)
Step-by-step Solution
Detailed explanation
\(\ell_1+\ell_2=20 \Rightarrow \frac{ d \ell_2}{ d \ell_1}=-1\) \(A _1=\left(\frac{\ell_1}{4}\right)^2 \text { and } A _2=\pi\left(\frac{\ell_2}{2 \pi}\right)^2\) Let \(S =2 A _1+3 A _2=\frac{\ell_1^2}{8}+\frac{3 \ell_2^2}{4 \pi}\)…
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