What is the maximum area of a rectangle that | vedclass

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(B) Let the lengths of the adjacent sides of the rectangle be $x \ cm$ and $y \ cm$. The perimeter of the rectangle is given by $p = 2(x + y)$,which i

Vedclass · Accepted Answer

$\frac{p^2}{16} \ cm^2$

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(B) Let the lengths of the adjacent sides of the rectangle be $x \ cm$ and $y \ cm$. The perimeter of the rectangle is given by $p = 2(x + y)$,which implies $y = \frac{p}{2} - x$. The area of the rectangle is $A = x y$. Substituting $y$,we get $A = x(\frac{p}{2} - x) = \frac{px}{2} - x^2$. To find the maximum area,we differentiate $A$ with respect to $x$ and set it to zero: $\frac{dA}{dx} = \frac{p}{2} - 2x = 0$. This gives $x = \frac{p}{4} \ cm$. Consequently,$y = \frac{p}{2} - \frac{p}{4} = \frac{p}{4} \ cm$. Thus,the maximum area is $A = \frac{p}{4} 	imes \frac{p}{4} = \frac{p^2}{16} \ cm^2$. Therefore,option $B$ is correct.

What is the maximum area of a rectangle that can be formed with a fixed perimeter $p \ cm$?

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