Question

Determine the maximized area of a rectangle that has a perimeter equal to 56m by creating and solving a quadratic equation. What is the length and width?

Answer 1

Answer:

Area of rectangle = $196\,m^2$

Length of rectangle = 14 m

Width of rectangle = 14 m

Step-by-step explanation:

Given:

Perimeter of rectangle is 56 m

To find: the maximized area of a rectangle and the length and width

Solution:

A function $y=f(x)$ has a point of maxima at $x=x_0$ if $f''(x_0)$

Let x, y denotes length and width of the rectangle.

Perimeter of rectangle = 2( length + width )

$=2(x+y)$

Also, perimeter of rectangle is equal to 56 m.

So,

$56=2(x+y)\\x+y=28\\y=28-x$

Let A denotes area of rectangle.

A = length × width

$A=xy\\=x(28-x)\\=28x-x^2$

Differentiate with respect to x

$\frac{dA}{dx}=28-2x$

Put $\frac{dA}{dx}=0$

$28-2x=0\\2x=28\\x=14$

Also,

$\frac{d^2A}{dx^2}=-2$

At x = 14, $\frac{d^2A}{dx^2}=-2$

So, x = 14 is a point of maxima

So,

$y=28-x=28-14=14$

Area of rectangle:

$A=xy=14(14)=196\,m^2$

Length of rectangle = 14 m

Width of rectangle = 14 m