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.

- anonymous

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- myininaya

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

- 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

there now a is complete

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- anonymous

Okay, that makes sense.

- 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

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

- myininaya

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

- myininaya

what did you get?

- anonymous

l=37.07 or l=22.93

- 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

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

- myininaya

are you sure you got l=37.07 and 22.93

- anonymous

Yeah...

- myininaya

a=1
b=-60
c=850
?

- myininaya

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

- anonymous

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

- myininaya

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

- anonymous

Yeah.

- myininaya

i got the lengths to be appoximately 52.93 and 67.07

- anonymous

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

- 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

let me try again

- myininaya

oops you are right lol

- myininaya

ok so remember from before w=60-l?

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

and greater that 0?

- myininaya

than*

- myininaya

does that make sense?
0

- anonymous

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

- 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

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

Ohhhh okay :D

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