A metal wire 108 m long is bent to form a re | vedclass

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(B) Let the sides of the rectangle be $x$ and $y$.Given the perimeter is $108 \ m$,we have $2x + 2y = 108$,which simplifies to $x + y = 54$,or $y = 54

Vedclass · Accepted Answer

$27 \ m, 27 \ m$

Vedclass · Answer

(B) Let the sides of the rectangle be $x$ and $y$.Given the perimeter is $108 \ m$,we have $2x + 2y = 108$,which simplifies to $x + y = 54$,or $y = 54 - x$.The area of the rectangle is $A = x 	imes y$.Substituting $y$,we get $A = x(54 - x) = 54x - x^2$.To find the maximum area,we differentiate $A$ with respect to $x$: $\frac{dA}{dx} = 54 - 2x$.Setting $\frac{dA}{dx} = 0$,we get $54 - 2x = 0$,which implies $x = 27$.Since $\frac{d^2A}{dx^2} = -2 Then $y = 54 - 27 = 27$.Thus,the dimensions are $27 \ m$ and $27 \ m$.

$A$ metal wire $108 \ m$ long is bent to form a rectangle. If the area of the rectangle is maximum,then its dimensions are

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