Question

A rectangular area is formed having a perimeter of 40 cm. Determine the length and breadth of the rectangle if it is to enclose the maximum possible area​

Answer 1

Let x and y be the dimensions of the rectangle. If the perimeter is 40, we have

$2(x+y)=40 \iff x+y=20$

We can expression one variable in terms of the others as

$x+y=20 \iff x=20-y$

Since the area is the product of the dimensions, we have

$xy=(20-y)y=-y^2+20y$

This is a parabola facing down, so it's vertex is the maximum:

$f(y)=-y^2+20y \implies f'(y)=-2y+20$

So, the maximum is

$f'(y)=0 \iff -2y+20=0 \iff 2y=20 \iff y=10$

And since we know that $x+y=20$, we have $x=10$ as well.

This is actually a well known theorem: out of all the rectangles with given perimeter, the one with the greatest area is the square.

Answer 2

We use Derivative test for maximization of any quantity.

Both length and breadth of the rectangle is 10 cm.

If length and breadth of rectangle is L and B respectively.

Perimeter of rectangle = $2(L + B)$

Since, perimeter is 40 cm.

                 $2(L + B)=40\\\\L + B=20\\\\L=20-B$

Area of rectangle is,$A=L*B$

Substituting value of L in above equation.

We get,    $A=(20-B)B=20B-B^{2}$

For maximum possible area, differentiate Area with respect to breadth B and then equate with zero..

         $\frac{dA}{dB} =20-2B=0\\\\B=\frac{20}{2}=10cm$

Substituting value of B in equation $L+B=20$

We get, $L=10cm$

Thus, Both length and breadth of the rectangle is 10 cm.

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