Question

Simon has 160 meters of fencing to build a rectangular garden. the garden's area (in square meters) as a function of the garden's width www (in meters) is modeled by a(w)=-w(w-80) what is the maximum area possible?

Answer 1

It is given in the question , that Simon has 160 meters of fencing to build a rectangular garden. the garden's area (in square meters) as a function of the garden's width w (in meters) is modeled by

$A(w) = -w(w-80) \\ A(w) = -w^2 + 80w$

Which represents parabola and the parabola is maximum at its vertex, that is

$w = -\frac{b}{2a} = - \frac{80}{2} = 40 meters$

Therefore the width is 40 meters and the area is

$A(40) = -40(40-80) = -40*-40 = 1600 meter ^2$

Answer 2

Simon has 160 meters of fencing to build a rectangular garden. the garden's area (in square meters) as a function of the garden's width www (in meters) is modeled by a(w)=-w(w-80) what is the maximum area possible?

Solution:

Let width of the rectangular garden=w

Perimeter of the rectangular garden=2(length+width)

160=2(length+w)

Divide by 2 on both sides

80=length+w

So, Length= 80-w

So, Area of rectangular garden= Length* Width

Area, A(w)=(80-w)(w)

Area, A(w)=-w²+ 80 w

Area is a quadratic equation. And, quadratic equation makes a parabola.

For the maximum area, We need to find the vertex of the parabola.

The formula for x-coordinate of the vertex=$\frac{-b}{2a}$

x-coordinate of vertex=$\frac{-(80)}{2(-1)}$

x-coordinate of vertex=40

So, Width= 40 meters

Length=80-w=80-40=40 meters

Maximum area possible=Length* Width=40*40=1600 square meters