anonymous
  • 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 :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
unless the problem restricts it; you can use any quadrant that is available to use
amistre64
  • amistre64
2l+2w=80 thats your perimeter right?
amistre64
  • amistre64
and Area = x*y correct?

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anonymous
  • anonymous
yes
amistre64
  • amistre64
A=l*w..i really should keep my variables consistent
amistre64
  • amistre64
with the perimeter we can slve for one of the variables... you want l or w?
anonymous
  • anonymous
sure
amistre64
  • amistre64
either way its the same....and the max area is a square in the end as you will see...hopefully ;)
amistre64
  • amistre64
l=(80-2w)/2 = 40-w substitute this 'value' into the area formula A = w(40-w) A = 40w -w^2
amistre64
  • amistre64
which is what they gave you in the wording; i just went the long way to get to it
amistre64
  • amistre64
now, do you know what the graph of this thing looks like?
amistre64
  • amistre64
its an upside down 'U'
amistre64
  • amistre64
so the highest point gives you the maximum area possible
amistre64
  • amistre64
what do you know about the graph? i can throw out alot of technical names that might mean nothing to you...
amistre64
  • 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
  • amistre64
the max area is 20(20) = 400 square ....feet? yeah, feet

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