The maximum area of a rectangle with a perime | vedclass

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Question

(A) Let the length of the rectangle be $x \ cm$ and the width be $y \ cm$.Given perimeter $P = 2(x + y) = 176 \ cm$.Therefore,$x + y = 88$,which impli

Vedclass · Accepted Answer

$1936$

Vedclass · Answer

(A) Let the length of the rectangle be $x \ cm$ and the width be $y \ cm$.Given perimeter $P = 2(x + y) = 176 \ cm$.Therefore,$x + y = 88$,which implies $y = 88 - x$.The area $A$ of the rectangle is given by $A = x 	imes y = x(88 - x) = 88x - x^2$.To find the maximum area,we differentiate $A$ with respect to $x$ and set it to zero:$\frac{dA}{dx} = 88 - 2x = 0$.Solving for $x$,we get $2x = 88$,so $x = 44 \ cm$.Since $x = 44$,then $y = 88 - 44 = 44 \ cm$.Thus,the rectangle is a square with side $44 \ cm$.The maximum area is $A = 44 	imes 44 = 1936 \ sq. \ cm$.

The maximum area of a rectangle with a perimeter of $176 \ cm$ is .......... $sq. \ cm$.

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