- anonymous

Maximize Q = xy2, where x and y are positive numbers, such that x + y2 = 4.

- schrodinger

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- anonymous

does y2 means y squared?

- anonymous

yes

- 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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## More answers

- anonymous

Do you know how to maximize 4x-x^2?

- anonymous

I plug test points into 4x-x^2?

- anonymous

the critical number is 4?

- anonymous

^^ yeah exactly.

- anonymous

so the max is 3?

- 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

oooh. i dont know

- anonymous

hey wait!!

- anonymous

I made a mistake :(

- anonymous

lol oh no

- anonymous

the critical point is not x=4

- anonymous

I didn't make it, you did and I didn't see it :P

- anonymous

Don't worry.. It's so easy. Do you know how to find the derivative of 4x-x^2?

- anonymous

4- 2x

- anonymous

Right!! Now solve for x when 4-2x=0.

- anonymous

2!

- anonymous

haha so hard

- anonymous

LOL

- anonymous

Now 2 is your critical point, and at x=2 Q is a maximum value.

- anonymous

So now just plug x=2 in Q=4x-x^2, and you're done!!

- anonymous

yay thanks

- anonymous

What is the maximum value?

- anonymous

4

- anonymous

:) I hope you can do any similar problem next time!

- anonymous

thanks for your help!!

- anonymous

no problem.

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