Question

Can I have help for this please1. Model the relationship between the combined area (A) and the two pens and the width of the pens (x). Show a table, a graph, and an equation as part of your model. (Use the grid provided.)

Answer 1

$A=60x-x^2$

Explanation

here we have a rectangle, the perimeter of a recntangle is given by

$\begin{gathered} P=2(l+w) \\ \text{where l is the length} \\ \text{and w is the width} \end{gathered}$

Step 1

Let

$\begin{gathered} \text{Perimeter}=\text{fencing}=120\text{ ft} \\ length=\text{ y} \\ \text{width}=\text{ x} \end{gathered}$

replace

$\begin{gathered} P=2(l+w) \\ 120=2(y+x) \\ \text{divide both sides by 2} \\ \frac{120}{2}=\frac{2(y+x)}{2} \\ 60=y+x \\ \text{subtract x in both sides} \\ 60-x=y+x-x \\ so \\ y=60-x\Rightarrow equation(1) \end{gathered}$

so, the equation that relates x and y is

y=60-x

Step 2

now, lets find an expression for the area

the area of a rectangle is given by

$Area_{rec\tan gle}=length\cdot width$

therefore, the area would be

$\begin{gathered} Area_{rec\tan gle}=length\cdot width \\ A=y\cdot x\Rightarrow equation\text{ (2)} \end{gathered}$

now, replace the y value from equation(1) into equation(2)

$\begin{gathered} A=y\cdot x\Rightarrow equation\text{ (2)} \\ A=(60-x)\cdot x \\ A=60x-x^2\Rightarrow\text{ Model } \end{gathered}$

therefore, the model of equation that shows the relation between the area and the width(x) is

$A=60x-x^2$

Step 3

graph:

a) make a table

i) when x= 0

$\begin{gathered} A=60x-x^2 \\ A=60(0)-(0)^2 \\ A=0-0 \\ A=0 \\ \end{gathered}$

therefore, when the width (x) is 0 the area(A) is zero, obvious but a good example

so, a point of the graph is

(0,0)

ii) when x =1

$\begin{gathered} A=60x-x^2 \\ A=60(1)-(1)^2 \\ A=59 \end{gathered}$

so, the point is (1,59) , when the width is 1, the area is 59

iii) when x = 3

$\begin{gathered} A=60x-x^2 \\ A=60(2)-(2)^2 \\ A=3600-4 \\ A=3594 \\ P3(3,3594) \end{gathered}$

draw a line that passes trought the points

there is a restriction

the area can not be negative, so it must be greater than 0

so

$\begin{gathered} A=60x-x^2 \\ 60x-x^2\ge0 \\ x(60-x)\ge0 \\ \text{divide both sides by x} \\ \frac{x(60-x)}{x}\ge\frac{0}{x} \\ 60-x\ge0 \\ \text{add x in both sides} \\ 60-x+x\ge0+x \\ 60\ge x \end{gathered}$

therefore, the width(x) must be equal or smaller than 60

I hope this helps you