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(C) Let the sides of the rectangle be $x$ and $y$. Given the perimeter $P = 2(x + y) = 100 \, cm$,which simplifies to $x + y = 50$,or $y = 50 - x$. Th

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$25 \, cm$ and $25 \, cm$

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(C) Let the sides of the rectangle be $x$ and $y$. Given the perimeter $P = 2(x + y) = 100 \, cm$,which simplifies to $x + y = 50$,or $y = 50 - x$. The area of the rectangle is $A = xy$. Substituting $y$,we get $A(x) = x(50 - x) = 50x - x^2$. To find the maximum area,we differentiate $A$ with respect to $x$: $\frac{dA}{dx} = 50 - 2x$. Setting $\frac{dA}{dx} = 0$,we get $50 - 2x = 0$,which gives $x = 25 \, cm$. Since $x + y = 50$,we find $y = 50 - 25 = 25 \, cm$. Thus,the sides for maximum area are $25 \, cm$ and $25 \, cm$.

The adjacent sides of a rectangle with a given perimeter of $100 \, cm$ and enclosing the maximum area are:

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