# Find the length and width of a rectangle that has the given perimeter and a maximum area? Perimeter: 164 meters

Apr 12, 2017

The Reqd. Dims. of the Rectangle for Maximum Area are

$l = 41 m . , \mathmr{and} , w = 82 - l = 41 m .$

#### Explanation:

Let $l \mathmr{and} w$ denote the length and width of the Rectangle.

Its Perimeter is $2 \left(l + w\right)$, which is given to be, $164.$

$\therefore l + w = \frac{164}{2} = 82. . . \left(1\right) .$

Now, Area $A$ of the Rectangle, is, given by, $A = l w .$

$\left(1\right) \Rightarrow A = l \left(82 - l\right) = 82 l - {l}^{2} ,$ which is a function of $l ,$ so let us

write it as $A \left(l\right) = 82 l - {l}^{2.} \ldots \ldots \left(2\right)$

We are reqd. to maximise $A$.

We know that, for ${A}_{\max} , A ' \left(l\right) = 0 , \mathmr{and} , A ' ' \left(l\right) < 0.$

$\left(2\right) \mathmr{and} A ' \left(l\right) = 0 \Rightarrow 82 - 2 l = 0 \Rightarrow l = \frac{82}{2} = 41.$

Further, A""(l)=-2 < 0, AA l, &, in particular, for $l = 41 , \text{ too.}$

Thus, $l = 41 , \text{ gives } {A}_{\max}$.

Hence, the reqd. dims. of the rectangle for maximum area are

$l = 41 m . , \mathmr{and} , w = 82 - l = 41 m .$

Enjoy Maths.!