alleryA wire of length $L$ is bent to form a rectangle. The **maximum area** enclosed is:
A$L^2/4$
B$L^2/8$
C$L^2/16$ (when shape becomes a square of side $L/4$)
D$L^2$
Answer & Solution
Correct answer: C. $L^2/16$ (when shape becomes a square of side $L/4$)
Perimeter $= 2(x + y) = L$ ⇒ $y = L/2 - x$.
Area $A = xy = x(L/2 - x)$.
$dA/dx = L/2 - 2x = 0$ ⇒ $x = L/4$.
Then $y = L/4$ (square!).
Area = $(L/4)^2 = L^2/16$.
(For fixed perimeter, square encloses max area; for fixed area, circle has min perimeter — isoperimetric inequality.)
Keep practising Unit 4-5: Applications of Derivatives & Curve Sketching
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