A wire of length 36 cm is bent to form a rectangle. Find its dimensions, if the area of the rectangle is maximum.
Q.22 A wire of length 36 cm is bent to form a rectangle. Find its dimensions, if the area of the rectangle is maximum.
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Answer: 9 cm by 9 cm (a square), maximum area 81 sq. cm.
Let the length be \(x\) cm and breadth \(y\) cm. Perimeter: \(2(x + y) = 36\), so \(y = 18 - x\).
Area \(A = x(18 - x) = 18x - x^2\). \(\frac{dA}{dx} = 18 - 2x\), \(\frac{d^2A}{dx^2} = -2\).
\(\frac{dA}{dx} = 0 \Rightarrow x = 9\), and \(\frac{d^2A}{dx^2} = -2 \lt 0\), so A is maximum at \(x = 9\). Then \(y = 18 - 9 = 9\).
The rectangle is 9 cm long and 9 cm broad, with maximum area 81 sq. cm.
Doubtly AI
· 28 Oct 2026
AI-verified · Oct 2026
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