A wire of length 64 text{ m} is to be bent to | vedclass

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(C) Let the length and breadth of the rectangle be $l$ and $b$ respectively.The perimeter of the rectangle is $2(l + b) = 64 	ext{ m}$, so $l + b = 3

Vedclass · Accepted Answer

$256$

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(C) Let the length and breadth of the rectangle be $l$ and $b$ respectively.The perimeter of the rectangle is $2(l + b) = 64 	ext{ m}$, so $l + b = 32 	ext{ m}$, which implies $b = 32 - l$.The area $A$ is given by $A = l 	imes b = l(32 - l) = 32l - l^2$.To find the maximum area, we differentiate $A$ with respect to $l$: $\frac{dA}{dl} = 32 - 2l$.Setting $\frac{dA}{dl} = 0$, we get $32 - 2l = 0$, so $l = 16 	ext{ m}$.Then $b = 32 - 16 = 16 	ext{ m}$.The maximum area is $A = 16 	imes 16 = 256 	ext{ m}^2$.

$A$ wire of length $64 \text{ m}$ is to be bent to form a rectangle such that its area is maximum. What is the maximum area (in $\text{ m}^2$)?

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