MHT CET · Maths · Application of Derivatives
A metal wire 108 meters long is bent to form a rectangle. If the area of the rectangle
is maximum, then its dimensions are
- A \(28 \mathrm{~m}, 28 \mathrm{~m}\)
- B 27 m, \(27 \mathrm{~m}\)
- C \(25 \mathrm{~m}, 25 \mathrm{~m}\)
- D \(26 \mathrm{~m}, 26 \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) 27 m, \(27 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Let sides of rectangle be \(x\) and \(y\)
Thus \(2 x+2 y=108 \Rightarrow x+y=54 \Rightarrow y=54-x\)
Now Area \(=A=x y\)
\(\therefore \quad=x(54-x)=54 x-x^{2}\)
Differentiating w.r.t. \(\mathrm{x}\), we get
\(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=54 \times 1-2 \mathrm{x}\) and \(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=-2 < 0\)
When \(\frac{d A}{d x}=0\), we get \(54-2 x=0 \Rightarrow x=27 \Rightarrow y=54-x=27\)
Thus \(2 x+2 y=108 \Rightarrow x+y=54 \Rightarrow y=54-x\)
Now Area \(=A=x y\)
\(\therefore \quad=x(54-x)=54 x-x^{2}\)
Differentiating w.r.t. \(\mathrm{x}\), we get
\(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=54 \times 1-2 \mathrm{x}\) and \(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=-2 < 0\)
When \(\frac{d A}{d x}=0\), we get \(54-2 x=0 \Rightarrow x=27 \Rightarrow y=54-x=27\)
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