Question

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.

Answer 1

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

Answer 2

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

Answer 3

there now a is complete

Answer 4

Okay, that makes sense.

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

what did you get?

Answer 9

l=37.07 or l=22.93

Answer 10

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

Answer 11

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

Answer 12

are you sure you got l=37.07 and 22.93

Answer 13

Yeah...

Answer 14

a=1
b=-60
c=850
?

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

Yeah.

Answer 19

i got the lengths to be appoximately 52.93 and 67.07

Answer 20

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

Answer 21

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

Answer 22

let me try again

Answer 23

oops you are right lol

Answer 24

ok so remember from before w=60-l?

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

and greater that 0?

Answer 29

than*

Answer 30

does that make sense?
0<l<60

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

Ohhhh okay :D