hugy0212 one year ago 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?

1. freckles

Have you number 1?

2. hugy0212

no :(

3. freckles

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

4. freckles

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

5. hugy0212

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

6. freckles

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

7. freckles

but we do know a relationship between x and y

8. hugy0212

oh alright

9. freckles

we know we have 120 feet of fencing

10. freckles

so you should know the perimeter equals?

11. hugy0212

yes

12. freckles

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

13. hugy0212

x+y?

14. 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 )

15. freckles

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

16. freckles

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

17. hugy0212

two sides with x and two sides with y

18. 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

19. freckles

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

20. freckles

which is Perimeter=2L+2W

21. freckles

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

22. freckles

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

23. freckles

what is the area of our rectangle

24. hugy0212

to find area is length times width

25. freckles

in terms of x and y

26. freckles

our length is ? and our width is?

27. hugy0212

so xy?

28. 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

29. freckles

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

30. hugy0212

x=-y+60

31. freckles

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

32. hugy0212

So A= (-y+60)y?

33. freckles

awesome

34. freckles

Now is this a calculus or algebra class?

35. freckles

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

36. hugy0212

Pre calculus

37. freckles

have you differentiated in this class?

38. freckles

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

39. freckles

have you done something like that?

40. hugy0212

no we have not

41. freckles

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

42. 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

43. Directrix