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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.

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  1. myininaya
    • 5 years ago
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    (a) P=2l+2w Perimeter=P=120 length=l width=w 120=2l+2w

  2. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    there now a is complete

  4. anonymous
    • 5 years ago
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    Okay, that makes sense.

  5. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    Oh, so it's a quadratic equation...

  7. myininaya
    • 5 years ago
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    right i don't think we can factor it but you use you use the quadratic formula

  8. myininaya
    • 5 years ago
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    what did you get?

  9. anonymous
    • 5 years ago
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    l=37.07 or l=22.93

  10. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    So she can fence in a garden of 850 sq. feet?

  12. myininaya
    • 5 years ago
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    are you sure you got l=37.07 and 22.93

  13. anonymous
    • 5 years ago
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    Yeah...

  14. myininaya
    • 5 years ago
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    a=1 b=-60 c=850 ?

  15. myininaya
    • 5 years ago
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    \[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  16. anonymous
    • 5 years ago
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    \[l=(60\pm \sqrt{(-60)^2-4(1)(850)}/2(1)\]

  17. myininaya
    • 5 years ago
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    \[\frac{60 \pm \sqrt{60^2-4(1)(850)}}{2}\]

  18. anonymous
    • 5 years ago
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    Yeah.

  19. myininaya
    • 5 years ago
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    i got the lengths to be appoximately 52.93 and 67.07

  20. anonymous
    • 5 years ago
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    It would be \[60\pm14.14/2\]right?

  21. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    let me try again

  23. myininaya
    • 5 years ago
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    oops you are right lol

  24. myininaya
    • 5 years ago
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    ok so remember from before w=60-l?

  25. myininaya
    • 5 years ago
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    we want w and l to be positive because negative lengths don't exist

  26. myininaya
    • 5 years ago
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    we want w to be positive so we want 60-l>0 so 60>l (l<60)

  27. myininaya
    • 5 years ago
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    so do you think it is possible since both of the lengths we got are less than 60?

  28. myininaya
    • 5 years ago
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    and greater that 0?

  29. myininaya
    • 5 years ago
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    than*

  30. myininaya
    • 5 years ago
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    does that make sense? 0<l<60

  31. anonymous
    • 5 years ago
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    Yeah, it does. Sorry I jsut ate dinner :P

  32. myininaya
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Ohhhh okay :D

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