A wire of length $L$ is bent to form a rectangle. The **maximum area**…

Question

A wire of length $L$ is bent to form a rectangle. The **maximum area** enclosed is:

Accepted 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.)

Answer

$L^2/4$

Answer

$L^2/8$

Answer

$L^2$