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?

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

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

Therefore the width is 40 meters and the area is

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=

x-coordinate of vertex=

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

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.

hi! i think the answer is 1/2 : )