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(C) Let the sides of the rectangle be $x$ and $y$. The perimeter $S$ is given by $S = 2(x + y)$.From this,we have $y = \frac{S}{2} - x$.The area $A$ o

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(C) Let the sides of the rectangle be $x$ and $y$. The perimeter $S$ is given by $S = 2(x + y)$.From this,we have $y = \frac{S}{2} - x$.The area $A$ of the rectangle is $A = x 	imes y = x \left( \frac{S}{2} - x ight) = \frac{Sx}{2} - x^2$.To find the maximum area,we differentiate $A$ with respect to $x$:$\frac{dA}{dx} = \frac{S}{2} - 2x$.Setting $\frac{dA}{dx} = 0$,we get $\frac{S}{2} = 2x$,which implies $x = \frac{S}{4}$.Substituting $x = \frac{S}{4}$ into the expression for $y$,we get $y = \frac{S}{2} - \frac{S}{4} = \frac{S}{4}$.Since $x = y = \frac{S}{4}$,the rectangle is a square.Also,$\frac{d^2A}{dx^2} = -2$,which is negative,confirming that the area is maximum at $x = \frac{S}{4}$.

The area of a rectangle will be maximum for a given perimeter when the rectangle is a

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