anonymous
  • anonymous
working with quadratic equations and parabolas in pre-calc. Find two positive real number whose product is a maximum and whose sum is 110.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1333239995618:dw|
radar
  • radar
x=110-y P=y(110-y) substituting for x. P=110y-y^2 \[y ^{2}-110y=-110\]
anonymous
  • anonymous
I followed the P=110y-y^2 part,but I don't understand the next step.

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radar
  • radar
\[y ^{2}-110y+110=0\]\[p ^{'}=2y-110=0\]y=55 x+55=110 x=55
radar
  • radar
Multiplied thru by -1 to obtain a positive coefficient for x^2
anonymous
  • anonymous
\[x + y = 110 \] \[P=xy\] \[x=110-y\] \[P=(110-y)y\] \[P=110-y ^{2}\]
anonymous
  • anonymous
I'm missing something
radar
  • radar
This gives the product P a value of y^2-110y Then differentiate getting 2y -110 For maxima setting that to 0 and solving for y/
anonymous
  • anonymous
Pre calc, no differentiating, graphing a parabola
radar
  • radar
Oh sorry, let me think you could let it stand as P=-x^2+110 which is a formula for a parabola.
radar
  • radar
I left off the y for the 110 typo.
anonymous
  • anonymous
the formula I have for a vertex is \[y=a(x-h)^{2}+k\] where (h, k) is the ordered pair of the vertex
radar
  • radar
I believe this parabolar opens downward as the x^2 has a negative coefficient
anonymous
  • anonymous
that would make sense, and would mean the vertix would be the max, so you already figured it out with calc, the vertex is (55,55) but why?
radar
  • radar
I'm trying to figures this out myself, P=-y^2+110y finding the intercept for P letting y=0 then P=0 that makes sense. Now to get the intercept for x let P=0 0=-y^2 +110y multiply thru by -1 and get 0=y^2-110y y(y-110)=0 y=0, y=110 Now what.
anonymous
  • anonymous
you'd find the vertex of y(110-y), not set to 0
radar
  • radar
Help!!
anonymous
  • anonymous
I'll take it to another resource site and let you know how goes : ), thanks for working on it with me.
anonymous
  • anonymous
x+y=110 p=xy x=(110-x) parabola with equation y=x(110-x), find vertex to get coordinate of maximum y=-x^2+110 we can complete the square from here
anonymous
  • anonymous
y=-(x^2-110)
anonymous
  • anonymous
y=-(x^2-110+55^2-55^2) y=-(x-55)^2+55^2
anonymous
  • anonymous
(h, k) is your vertex, so the max of the parabola is when x=55. this maximum is 55^2
anonymous
  • anonymous
so if x=55 55+y=110 y=55
radar
  • radar
Hey Dockworker thanks, that "completing the square" rings some old bells for me. Thanks for pitching in and showing how those things are solved.
anonymous
  • anonymous
i also forgot the x y=-x^2-110x+55^2-55^2)
anonymous
  • anonymous
yw
anonymous
  • anonymous
damn that was a mess up there, i'll clean it up so when he comes back he can follow. i forgot the x everywhere
anonymous
  • anonymous
x+y=110 p=xy y=110-x parabola with equation y=x(110-x), find vertex to get coordinate of maximum y=-x^2+110x we can complete the square from here y=-(x^2-110x) y=-(x^2-110x+55^2-55^2) y=-(x-55)^2+55^2 (h,k) is your vertex, so the max of the parabola is when x=55. this maximum is 55^2 so if x+y=110 y=110-x=110-55=55
anonymous
  • anonymous
Thanks everyone, I've got it clear now.

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