## anonymous 5 years ago A gardener has 120 feet of fencing to fence in a rectangular flower garden. a) Find a function that models the area of the garden she can fence. b) Can she fence a garden with an area of 850 square feet? c) Find the dimensions of the largest area she can fence.

1. myininaya

(a) P=2l+2w Perimeter=P=120 length=l width=w 120=2l+2w

2. myininaya

oops Area=A and since 120=2l+2w 60=l+w so w=60-l so A=lw=l(60-l)=60l-l^2

3. myininaya

there now a is complete

4. anonymous

Okay, that makes sense.

5. myininaya

part b now) If A=850 then 850=60l-l^2 l^2-60l+850=0 now we need to find l here

6. anonymous

Oh, so it's a quadratic equation...

7. myininaya

right i don't think we can factor it but you use you use the quadratic formula

8. myininaya

what did you get?

9. anonymous

l=37.07 or l=22.93

10. myininaya

ok so if l=37.07 then w=60-37.07 which is possible since the w is positive and if l=22.93, then is also possible since w is positive if we have this length

11. anonymous

So she can fence in a garden of 850 sq. feet?

12. myininaya

are you sure you got l=37.07 and 22.93

13. anonymous

Yeah...

14. myininaya

a=1 b=-60 c=850 ?

15. myininaya

$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

16. anonymous

$l=(60\pm \sqrt{(-60)^2-4(1)(850)}/2(1)$

17. myininaya

$\frac{60 \pm \sqrt{60^2-4(1)(850)}}{2}$

18. anonymous

Yeah.

19. myininaya

i got the lengths to be appoximately 52.93 and 67.07

20. anonymous

It would be $60\pm14.14/2$right?

21. myininaya

she can fence the garden as long as the length is less that 60 we have 52.93 so it is possible to fence an area of 850 square feet

22. myininaya

let me try again

23. myininaya

oops you are right lol

24. myininaya

ok so remember from before w=60-l?

25. myininaya

we want w and l to be positive because negative lengths don't exist

26. myininaya

we want w to be positive so we want 60-l>0 so 60>l (l<60)

27. myininaya

so do you think it is possible since both of the lengths we got are less than 60?

28. myininaya

and greater that 0?

29. myininaya

than*

30. myininaya

does that make sense? 0<l<60

31. anonymous

Yeah, it does. Sorry I jsut ate dinner :P

32. myininaya

ok find the vertex of A and you will find the maximum area put the A (which is a parabola) in vertex is form

33. myininaya

A=60l-l^2 A=-l^2+60l A=-(l^2-60l) A=-(l^2-60l+(60/2)^2)+(60/2)^2 A=-(l^2-60l+30^2)+30^2 A=-(l-30)^2+900 the vertex is (30,900) so max length is 30 and max area is 900

34. anonymous

Ohhhh okay :D