Question

Simon has 160160160 meters of fencing to build a rectangular garden. The garden's area (in square meters) as a function of the garden's width xxx (in meters) is modeled by A(x)=-x(x-80)A(x)=−x(x−80)A, left parenthesis, x, right parenthesis, equals, minus, x, left parenthesis, x, minus, 80, right parenthesis What width will produce the maximum garden area?

Answer 1

The width for maximum area will be 40 metres.

Given,Simon has 160 metres of fencing to build a rectangular garden.

The garden's area (in square meters) as a function of the garden's width x(in meters) is modeled by,

$A(x)=-x(x-80)$.

We know that perimeter of rectangle will be,

$P=2(L+B)$

Here p is 160,

So,$160=2(L+B)$

$\L+B= 80$

Now we have the sum of length and width off the rectangular garden is 80.

Since,

$A(x)=-x(x-80)$

So, $A(x)=x(80-x)$

We know that the area of rectangle will be the product of length and , here in question width is $x$ so the length will be$(80-x)$.

Now we have to calculate the width for which the area will be maximum.

The area will be maximum when the first derivative of area function will becomes zero.

So,

$\frac{\mathrm{d} }{\mathrm{d} x} A(x)=\frac{\mathrm{d} }{\mathrm{d} x}(-x)(x-80)$

$\frac{\mathrm{d} }{\mathrm{d} x}A(x)=\frac{\mathrm{d} }{\mathrm{d} x}(-x^{2} +80x)$

$\frac{\mathrm{d} }{\mathrm{d} x}A(x)=-2x+80$

For maximum area ,

$\frac{\mathrm{d} }{\mathrm{d} x}A(x)=0$

Hence,

$-2x+80=0\\x=40$

Hence the width for maximum area will be 40 metres.

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Answer 2

Answer:

The maximum area is 1600 sq meters.

Step-by-step explanation:

The garden's area is modeled by a quadratic function, whose graph is a parabola.

The maximum area is reached at the vertex.

So in order to find the maximum area, we need to find the vertex's y-coordinate.

We will start by finding the vertex's x-coordinate, and then plug that into A(x).

The vertex's x-coordinate is the average of the two zeros, so let's find those first.

A(x)=0    -x(x-80)=0

  ↓                  ↓

-x=0 or x-80=0

x=0 or x=80

Now let's take the zeros' average:

$\frac{(0)+(80)}{2}=\frac{80}{2}=40$

The vertex's x-coordinate is 40. Now lets find A(40):

A(40)= -(40)(40-80)

        = -(40)(-40)

        = 1600

in conclusion, the maximum area is 1600 square meters.