A farmer has 120 feet of fencing available to build a rectangular pen for her pygmy goats. She wants to give them as much room as possible to run.
1.Draw a diagram to represent this problem.
2.Write an expression in terms of a single variable that would represent the area of a rectangle in this family.
3.Find the dimensions of the rectangle with maximum area.
4.What is another name for this kind of rectangle?

- hugy0212

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- freckles

Have you number 1?

- hugy0212

no :(

- freckles

Hmmm... you should know what a rectangle looks like...

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## More answers

- freckles

|dw:1443890804767:dw|
opposite sides are equal
|dw:1443890824575:dw|

- hugy0212

I do not need to apply any numbers to the sides?

- freckles

what don't know the numbers....I called the numbers x and y.

- freckles

but we do know a relationship between x and y

- hugy0212

oh alright

- freckles

we know we have 120 feet of fencing

- freckles

so you should know the perimeter equals?

- hugy0212

yes

- freckles

so can you state an equation relating x and y (think perimeter)

- hugy0212

x+y?

- freckles

x+y is an expression
I'm looking for an equation that relates x and y (the hint I gave for this equation was to think perimeter )

- freckles

you have 120 feet amount of fencing
in terms of x and y you have how much amount of fencing ?

- freckles

how many sides have measurement x?
how many sides have measurement y?

- hugy0212

two sides with x and two sides with y

- freckles

right
x+x+y+y is how much fencing we have
or simplifying
2x+2y is how much fencing we have
but we were also given that we had 120 feet of fencing
so 2x+2y=120

- freckles

Also you guys probably know the formula for finding the perimeter of a rectangle ...

- freckles

which is Perimeter=2L+2W

- freckles

Anyways to make 2x+2y=120 prettier I would divide both sides by 2
giving x+y=60

- freckles

anyways we have started doing number 2 sorta of
except we need to mention something about the area of this rectangle

- freckles

what is the area of our rectangle

- hugy0212

to find area is length times width

- freckles

in terms of x and y

- freckles

our length is ?
and our width is?

- hugy0212

so xy?

- freckles

yes we have Area=xy
I'm just going to use A for area
A=xy
now number 2 also says write in terms of one variable
this is where that perimeter equation we came up with will come in use

- freckles

remember we had x+y=60
solve for either x or y
(your choice; no need to solve for both though)

- hugy0212

x=-y+60

- freckles

right good
so replace the x in A=xy with x=-y+60
and you will have completed number 2

- hugy0212

So A= (-y+60)y?

- freckles

awesome

- freckles

Now is this a calculus or algebra class?

- freckles

if it is calculus we differentiate or you could use algebraic ways because this is a parabola

- hugy0212

Pre calculus

- freckles

have you differentiated in this class?

- freckles

example: like the derivative of 6x^2 is 12x\

- freckles

have you done something like that?

- hugy0212

no we have not

- freckles

ok then this is parabola
so you just need to find the vertex of the parabola to find the max

- freckles

\[f(x)=ax^2+bx+c \\ f(x)=ax^2+\frac{a}{a}bx+c \\ \text{ I multiply second term by } \frac{a}{a} \\ \text{ so it would be easier for you to see how I'm going to factor out } a \\ \text{ from the first two terms } \\ f(x)=a(x^2+\frac{1}{a}bx)+c \\ \text{ so } \frac{1}{a} \cdot b =\frac{b}{a} \\ f(x)=a(x^2+\frac{b}{a}x)+c \\ \text{ now I'm going to leave a space inside the ( ) } \\ \text{ this space will be for completing the square } \\ \text{ I will also mention whatever we add in } \\ \text{ we must subtract out } \\ f(x)=a(x^2+\frac{b}{a}x+?)+c-a? \]
\[\text{ anyways recall } x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \\ \text{ so that means our } k=\frac{b}{a} \\ \text{ so } (x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(x+\frac{b}{2a})^2 \\ \text{ so going back to } f(x) \\ f(x)=a(x^2+\frac{b}{a}x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2 \\ f(x)=a(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2\]
You should recall this is the vertex form of a quadratic.
Where there vertex is:
\[(-\frac{b}{2a},c-a(\frac{b}{2a})^2)\]
you try finding the vertex form (or vertex) of your parabola

- Directrix

At the link are two versions of a solution. They may be helpful.
http://openstudy.com/study#/updates/55035308e4b01982d66287df
@hugy0212

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