If from a wire of length 36 m a rectangle of | vedclass

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(B) Let the sides of the rectangle be $a$ and $b$. The perimeter of the rectangle is given by $2(a + b) = 36 \ m$,which implies $a + b = 18$,or $b = 1

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$9, 9$

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(B) Let the sides of the rectangle be $a$ and $b$. The perimeter of the rectangle is given by $2(a + b) = 36 \ m$,which implies $a + b = 18$,or $b = 18 - a$.The area $A$ of the rectangle is given by $A = a 	imes b = a(18 - a) = 18a - a^2$.To find the maximum area,we differentiate $A$ with respect to $a$:$\frac{dA}{da} = 18 - 2a$.Setting the derivative to zero for critical points:$18 - 2a = 0 \implies a = 9$.Substituting $a = 9$ back into the equation for $b$:$b = 18 - 9 = 9$.Since the second derivative $\frac{d^2A}{da^2} = -2 < 0$,the area is maximum when $a = 9$ and $b = 9$. Thus,the adjacent sides are $9 \ m$ and $9 \ m$.

If from a wire of length $36 \ m$ a rectangle of greatest area is made,then its two adjacent sides in metres are

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