Question

11. A wire 16m long has to be formed into a rectangle. What dimensions should the rectangle have to maximize the area?
[x -4, y=4]​

Answer 1

Step-by-step explanation:

This picture show the complete amswer

Answer 2

The dimensions for which rectangle has maximum area is $x=4$ and $y=4$.

Given:

A wire 16m long has to be formed into a rectangle.

To Find:

The dimensions should the rectangle have to maximize the area.

Solution:

Let x be the length and y be the length of the rectangle.

Also, let $2x+2y=16$, then

$x+y=8$

Write the equation for y.

$y=8-x$

The area of a rectangle is given by $xy$.

Substitute the expression for y into the formula of area.

$x(8-x)=8x-x^2$

To maximize the area $8x-x^2$ will be differentiated twice.

Find the first derivative of $8x-x^2$.

$\frac{d(8x-x^2)}{dx} =8-2x$

Equate $8-2x$ to 0 and find x.

$8-2x=0\\2x=8\\x=4$

Find $\frac{d(8-2x)}{dx}$.

$\frac{d(8-2x)}{dx}=-2$

Since -2<0 so this is the point of maxima. So, substitute $x=4$ in $y=8-x$ and find y.

$y=8-4\\=4$

Thus, the dimensions for which rectangle has maximum area is $x=4$ and $y=4$.