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- anonymous

Ok so i have a few questions.
When graphing y>x^2 you can only use quadrant one right?
Also i have a word problem that goes like this-Suppose you have 80 ft of fence to enclose a rectangular garden.
The function A = 40x - x2 gives you the area of the garden in square feet
where x is the width in feet. and then it has two parts... a. What width gives you the maximum gardening area? and b. What is the maximum area? Please help because I have no idea what to do. Thanks :)

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- anonymous

- katieb

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- amistre64

unless the problem restricts it; you can use any quadrant that is available to use

- amistre64

2l+2w=80 thats your perimeter right?

- amistre64

and Area = x*y correct?

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- anonymous

yes

- amistre64

A=l*w..i really should keep my variables consistent

- amistre64

with the perimeter we can slve for one of the variables... you want l or w?

- anonymous

sure

- amistre64

either way its the same....and the max area is a square in the end as you will see...hopefully ;)

- amistre64

l=(80-2w)/2 = 40-w
substitute this 'value' into the area formula
A = w(40-w)
A = 40w -w^2

- amistre64

which is what they gave you in the wording; i just went the long way to get to it

- amistre64

now, do you know what the graph of this thing looks like?

- amistre64

its an upside down 'U'

- amistre64

so the highest point gives you the maximum area possible

- amistre64

what do you know about the graph? i can throw out alot of technical names that might mean nothing to you...

- amistre64

the highest point is when w=20.. which is the vertex of the graph
we know that 2w + 2l = 80
2(20) +2l = 80
40 +2l = 80
2l = 40
l = 40/2 = 20
the width is 20, the length is 20 ..its a square :)

- amistre64

the max area is 20(20) = 400 square ....feet? yeah, feet

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