The maximum area of a rectangle that can be f | vedclass

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(B) Let the sides of the rectangle be $x$ and $y$. Given,the perimeter is $20$ units. $2(x + y) = 20 \Rightarrow x + y = 10 \Rightarrow y = 10 - x$. T

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) Let the sides of the rectangle be $x$ and $y$. Given,the perimeter is $20$ units. $2(x + y) = 20 \Rightarrow x + y = 10 \Rightarrow y = 10 - x$. The area $A$ of the rectangle is given by $A = xy$. Substituting $y$,we get $A = x(10 - x) = 10x - x^2$. To find the maximum area,we differentiate $A$ with respect to $x$: $\frac{dA}{dx} = 10 - 2x$. Setting $\frac{dA}{dx} = 0$,we get $10 - 2x = 0$,which implies $x = 5$. Now,we check the second derivative: $\frac{d^2A}{dx^2} = -2$. Since $\frac{d^2A}{dx^2} When $x = 5$,$y = 10 - 5 = 5$. Thus,the maximum area is $A = 5 	imes 5 = 25$ sq units.

The maximum area of a rectangle that can be formed with a fixed perimeter of $20$ units is $ . . . . . . $ sq units.

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