anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
(a) P=2l+2w Perimeter=P=120 length=l width=w 120=2l+2w
myininaya
  • 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
myininaya
  • myininaya
there now a is complete

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anonymous
  • anonymous
Okay, that makes sense.
myininaya
  • myininaya
part b now) If A=850 then 850=60l-l^2 l^2-60l+850=0 now we need to find l here
anonymous
  • anonymous
Oh, so it's a quadratic equation...
myininaya
  • myininaya
right i don't think we can factor it but you use you use the quadratic formula
myininaya
  • myininaya
what did you get?
anonymous
  • anonymous
l=37.07 or l=22.93
myininaya
  • 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
anonymous
  • anonymous
So she can fence in a garden of 850 sq. feet?
myininaya
  • myininaya
are you sure you got l=37.07 and 22.93
anonymous
  • anonymous
Yeah...
myininaya
  • myininaya
a=1 b=-60 c=850 ?
myininaya
  • myininaya
\[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
\[l=(60\pm \sqrt{(-60)^2-4(1)(850)}/2(1)\]
myininaya
  • myininaya
\[\frac{60 \pm \sqrt{60^2-4(1)(850)}}{2}\]
anonymous
  • anonymous
Yeah.
myininaya
  • myininaya
i got the lengths to be appoximately 52.93 and 67.07
anonymous
  • anonymous
It would be \[60\pm14.14/2\]right?
myininaya
  • 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
myininaya
  • myininaya
let me try again
myininaya
  • myininaya
oops you are right lol
myininaya
  • myininaya
ok so remember from before w=60-l?
myininaya
  • myininaya
we want w and l to be positive because negative lengths don't exist
myininaya
  • myininaya
we want w to be positive so we want 60-l>0 so 60>l (l<60)
myininaya
  • myininaya
so do you think it is possible since both of the lengths we got are less than 60?
myininaya
  • myininaya
and greater that 0?
myininaya
  • myininaya
than*
myininaya
  • myininaya
does that make sense? 0
anonymous
  • anonymous
Yeah, it does. Sorry I jsut ate dinner :P
myininaya
  • 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
myininaya
  • 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
anonymous
  • anonymous
Ohhhh okay :D

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