Question

Eighty metres of fencing are available to enclose a rectangular area. What are the dimensions that enclose the max area. What is max area?

Answer 1

P = the perimeter = 80 m

A = the area

W = the width

L = the length


P = 2 W + 2 L = 2 ( W + L )

80 = 2 ( W + L ) Divide both sides by 2

40 = W + L Subtract L to both sides

40 - L = W + L - L

40 - L = W

W = 40 - L


A = W * L

A = ( 40 - L ) * L

A = 40 L - L ^ 2

A = - L ^ 2 + 40 L


Since the coefficient of x ^ 2 is negative, the parabola opens downward and has a maximum value at the vertex.


The vertex of the parabola is:

- b / 2 a

In this case :

a = - 1

b = 40

- b / 2 a = - 40 / [ 2 * ( - 1) ] = - 40 / - 2 = 20


That's the length of the rectangle of maximum area.

L = 20 m

We need to find the width from the equation :

W = 40 - L

W = 40 - 20

W = 20 m


So we see that the largest area possible is when the rectangle
is chosen to be the square 20 m by 20 m.

Amax = 20 * 20 = 400 m ^ 2

Answer 2

20*20 is not rectangle :D