Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types

1 answer
Question:

Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types of young animals. the three pens are to have the same area. express the area function for the three pens in terms of x. determine the domain and range for the area function

Answers

The area function is

[tex]A=\frac{135}{2}x-\frac{9}{2}x^2[/tex].

The domain and range of A is [tex](0,15m)[/tex] and [tex](0, 253.125 m^2][/tex].

Step-by-step explanation:

The given length of fencing is [tex]90 m[/tex].

Let the length and width of each pen be [tex]x[/tex] and [tex]y[/tex] respectively as shown in the figure.

As there are 3 pens, so, the total area,

[tex]A= 3 xy \;\cdots (i)[/tex]

From the figure the total length of fencing is [tex]6x+4y[/tex].

Here, for a significant area for the animals, [tex]x0[/tex] as well as [tex]y0[/tex] as [tex]x[/tex] and [tex]y[/tex] are the sides of ben.

From the given value:

[tex]6x+4y=90\;\cdots (ii)[/tex]

[tex]\Rightarrow y=\frac {45}{2}-\frac{3x}{2}[/tex]

Now, from equation (i)

[tex]A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)[/tex]

[tex]\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)[/tex]

This is the required area function in the terms of variable [tex]x[/tex].

For the domain of area function, from equation (ii)

[tex]x=15-\frac{2y}{3}[/tex]

[tex]\Rightarrow x[/tex] [as y>0]

So, the domain of area function is [tex](0,15m)[/tex].

For the range of area function:

As [tex]x \rightarrow 0[/tex] or [tex]y\rightarrow 0[/tex], then [tex]A\rightarrow 0[/tex] [from equation (i)]

[tex]\Rightarrow A0[/tex]

Now, differentiate the area function with respect to [tex]x[/tex] .

[tex]\frac {dA}{dx}=\frac{135}{2}-9x[/tex]

Equate [tex]\frac {dA}{dx}[/tex]  to zero to get the extremum point.

[tex]\frac {dA}{dx}=0[/tex]

[tex]\Rightarrow \frac{135}{2}-9x=0[/tex]

[tex]\Rightarrow x=\frac{15}{2}[/tex]

Check this point by double differentiation

[tex]\frac {d^2A}{dx^2}=-9[/tex]

As,  [tex]\frac {d^2A}{dx^2}[/tex], so, point [tex]x=\frac{15}{2}[/tex] is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

[tex]A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2[/tex]

[tex]\Rightarrow A=253.125 m^2[/tex]

Hence, the range of area function is [tex](0, 253.125 m^2][/tex].


[tex]Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three type[/tex]

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