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

The maximum area possible of the rectangular garden in which Simon has 160 meters of fencing to build is 1600 m².=

What is the standard form of quadratic equation?

A quadratic equation is the equation in which the unknown variable is one and the highest power of the unknown variable is two.

The standard form of the quadratic equation is,

$ax^2+bx+c=0$

Here,(a,b, c) is the real numbers and (x) is the variable.

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,

$A(w)=-w(w-80)$

Simplify the eqution,

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

It is a quadratic equation of parabola. Compare it with equation of parabola, we get,

$a=-1\\b=80\\c=0$

The vertex of the parabola is,

$\dfrac{-b}{2a}=\dfrac{-80}{2\times (-1)}\\\dfrac{-b}{2a}=40$

The vertex of the parabola gives the highest value. Thus, the width of the garden is 40 meters. Put this value in the given model,

$A(40)=-40(40-80)\\A(40)=-40\times-40\\A(40)=1600\rm\;m^2$

Thus, the maximum area possible of the rectangular garden in which Simon has 160 meters of fencing to build is 1600 m².

Learn more about the quadratic equation here;

https://brainly.com/question/1214333

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