Question

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 :)

Answer 1

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

Answer 2

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

Answer 3

and Area = x*y correct?

Answer 4

yes

Answer 5

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

Answer 6

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

Answer 7

sure

Answer 8

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

Answer 9

l=(80-2w)/2 = 40-w

substitute this 'value' into the area formula

A = w(40-w)

A = 40w -w^2

Answer 10

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

Answer 11

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

Answer 12

its an upside down 'U'

Answer 13

so the highest point gives you the maximum area possible

Answer 14

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

Answer 15

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 :)

Answer 16

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