Question:
A wire of length \(64\) m is to be bent to form a rectangle such that its area is maximum, then its area is.....
A wire of length \(64\) m is to be bent to form a rectangle such that its area is maximum, then its area is.....
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For a fixed perimeter the rectangle of maximum area is a square.
Updated On: Oct 1, 2026
- \(156\) sq.m
- \(324\) sq.m
- \(256\) sq.m
- \(320\) sq.m
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The Correct Option is C
Solution and Explanation
Step 1: Set up
Perimeter \(=64\) m gives \(l+b=32\). Area \(A=l(32-l)\).
Step 2: Maximise
\(\frac{dA}{dl}=32-2l=0\) gives \(l=16\), and \(\frac{d^2A}{dl^2}=-2<0\), so it is a maximum.
Then \(b=16\) and \(A=16\times16=256\) sq. m.
Final Answer:
The maximum area is \(256\) sq. m, option (C). \[ \boxed{\text{(C) }256} \]
Perimeter \(=64\) m gives \(l+b=32\). Area \(A=l(32-l)\).
Step 2: Maximise
\(\frac{dA}{dl}=32-2l=0\) gives \(l=16\), and \(\frac{d^2A}{dl^2}=-2<0\), so it is a maximum.
Then \(b=16\) and \(A=16\times16=256\) sq. m.
Final Answer:
The maximum area is \(256\) sq. m, option (C). \[ \boxed{\text{(C) }256} \]
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