anonymous
  • anonymous
Maximize Q = xy2, where x and y are positive numbers, such that x + y2 = 4.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
does y2 means y squared?
anonymous
  • anonymous
yes
anonymous
  • anonymous
If so then the solution of the problem is: You want to maximize Q, that means you need to find the derivative and critical points of Q. The problem here is that Q is in terms of x and y. You should make it in term of only one of them. So we are going to use the second equation to do so. \[x+y^2=4 \implies y^2=4-x\] substitute for the value of y^2 into Q, you get: \[Q=xy^2=x(4-x)=4x-x^2\]

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anonymous
  • anonymous
Do you know how to maximize 4x-x^2?
anonymous
  • anonymous
I plug test points into 4x-x^2?
anonymous
  • anonymous
the critical number is 4?
anonymous
  • anonymous
^^ yeah exactly.
anonymous
  • anonymous
so the max is 3?
anonymous
  • anonymous
you solved for x when the derivative is zero, and you got tgat x=4. this is your only critical point, so it's either maximum or minimum. Which one do you think it is?
anonymous
  • anonymous
oooh. i dont know
anonymous
  • anonymous
hey wait!!
anonymous
  • anonymous
I made a mistake :(
anonymous
  • anonymous
lol oh no
anonymous
  • anonymous
the critical point is not x=4
anonymous
  • anonymous
I didn't make it, you did and I didn't see it :P
anonymous
  • anonymous
Don't worry.. It's so easy. Do you know how to find the derivative of 4x-x^2?
anonymous
  • anonymous
4- 2x
anonymous
  • anonymous
Right!! Now solve for x when 4-2x=0.
anonymous
  • anonymous
2!
anonymous
  • anonymous
haha so hard
anonymous
  • anonymous
LOL
anonymous
  • anonymous
Now 2 is your critical point, and at x=2 Q is a maximum value.
anonymous
  • anonymous
So now just plug x=2 in Q=4x-x^2, and you're done!!
anonymous
  • anonymous
yay thanks
anonymous
  • anonymous
What is the maximum value?
anonymous
  • anonymous
4
anonymous
  • anonymous
:) I hope you can do any similar problem next time!
anonymous
  • anonymous
thanks for your help!!
anonymous
  • anonymous
no problem.

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