# F(x, y) Min Max problem with boundaries

## Homework Statement:

Find max or min value of function $$f(x,y)=\frac{x+y^2}{1+x^2+y^2}$$ which is defined by $$x^2+y^2=4$$

## Homework Equations:

$$f(x,y)=\frac{x+y^2}{1+x^2+y^2}$$
First I took the partial derivatives (f'x=0 and f'y=0) and got x=-1 or 1 and y=0 but these values are outside our definition right?

I did a parameter change to trigs and got result f(0,2) is max and f(-2,0) is min. is this correct?

edit: maybe its f(1, 3) is max?

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WWGD
Gold Member
The statement seems unclear. Do you mean your $f(x,y)$ is subject to the constraint that $x,y$ lie on the EDIT: circle $C(0,2)$? If so, you can simplify your $f$ and sub in $4$ for $x^2+y^2$ in the denominator.

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Ok so i get 4/5 as max value and -2/5 as min value. is this correct?

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WWGD
Gold Member
If you simplify with the sub in then $f_x=1, f_y=2y$. Have you tried that?

If you simplify with the sub in then $f_x=1, f_y=2y$. Have you tried that?
interesting. thanks.

WWGD
Gold Member
interesting. thanks.
Sorry, forgot to include that in the resulting expression after the sub in , you can also turn $x+ y^2$ into a function of 1 variable ( seemingly easier in terms of $x$) , which simplifies things even further.

• Wi_N
Sorry, forgot to include that in the resulting expression after the sub in , you can also turn $x+ y^2$ into a function of 1 variable ( seemingly easier in terms of $x$) , which simplifies things even further.
since the circle itself is only defined is there any point in taking derivatives? since the max min have to be on the circle.

WWGD
Gold Member
since the circle itself is only defined is there any point in taking derivatives? since the max min have to be on the circle.
If you can figure it out, no need to do so. But how would you prove it's the maximum? Edit: never mind. Once you do the sub ins the max will be pretty clear. Edit2: No, you're right, the equation is defined on the circle so that the max will also be.

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WWGD
Gold Member
Ok so i get 4/5 as max value and -2/5 as min value. is this correct?
Max is 17/4 at x=1/2 , no global minimum. Plot the subbed-in curve. EDIT: As pointed out, there is a min. within the circle. The curve is a parabola, which we can intersect with the circle.

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StoneTemplePython
Gold Member
no global minimum. Plot the subbed-in curve.
continuous function applied on a compact domain...

WWGD
Gold Member
continuous function applied on a compact domain...
Ah, my bad, I was considering the global min of 4+x -x^2 , not the one restricted to the circle. We get a parabola in the global one before restricting. Edited. I thought subbing -in the equation of the circle on the objective function would take care of it. Clearly not.

Max is 17/4 at x=1/2 , no global minimum. Plot the subbed-in curve. EDIT: As pointed out, there is a min. within the circle. The curve is a parabola, which we can intersect with the circle.
how can x=1/2 when its not part of the domain?

WWGD
Gold Member
how can x=1/2 when its not part of the domain?
Try the point in the circle with coordinate $x =1/2$ Should be $\sqrt {15} /2$

Try the point in the circle with coordinate $x =1/2$ Should be $\sqrt {15} /2$
how does inserting x=1/2 into the function yield 17/4 when our boundries are x^2 +y^2=4 ?????

WWGD
Gold Member
how does inserting x=1/2 into the function yield 17/4 when our boundries are x^2 +y^2=4 ?????
Edit: You're right ,my bad. Objective function is $f(x,y)= \frac {x+ y^2}{5}$ for points$(x,y)$on the circle. $(1/2, \sqrt 15/4)$ is on the circle. So max is 17/20, not 17/4, sorry.

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• Wi_N
Edit: You're right ,my bad. Objective function is $f(x,y)= \frac {x+ y^2}{5}$ for points$(x,y)$on the circle. $(1/2, \sqrt 15/4)$ is on the circle. So max is 17/20, not 17/4, sorry.
im sorry but (1/2)^2 + (sqrt15 /4)^2 is not equal to 4.

haruspex
Homework Helper
Gold Member
2018 Award
im sorry but (1/2)^2 + (sqrt15 /4)^2 is not equal to 4.
Clearly @WWGD meant $\sqrt{15/4}$

• WWGD and Wi_N
Try the point in the circle with coordinate $x =1/2$ Should be $\sqrt {15} /2$
how did you get that point?

Mark44
Mentor
$(1/2, \sqrt 15/4)$ is on the circle.
how did you get that point?
He got it by misunderstanding how LaTeX works. If you write $\sqrt 15/4$, it renders as $\sqrt 15/4$, but if you write it as $\sqrt{15/4}$, with braces around the quantity in the radical, the expression renders as $\sqrt{15/4}$. The latter form is what I believe he intended.

• WWGD
He got it by misunderstanding how LaTeX works. If you write $\sqrt 15/4$, it renders as $\sqrt 15/4$, but if you write it as $\sqrt{15/4}$, with braces around the quantity in the radical, the expression renders as $\sqrt{15/4}$. The latter form is what I believe he intended.
i meant how did he find it on the circle which method did he use? derivative. trig change? how?

Mark44
Mentor
i meant how did he find it on the circle which method did he use? derivative. trig change? how?
Are you asking how he got the x value of 1/2? Or are you asking how he got the y value of $\sqrt{15/4}$?

Are you asking how he got the x value of 1/2? Or are you asking how he got the y value of $\sqrt{15/4}$?
well both. since they are related.

Mark44
Mentor
well both.
I'll let @WWGD explain how he came up with x = 1/2. As for the y value, just substitute x = 1/2 into the equation $x^2 + y^2 = 4$.

I'll let @WWGD explain how he came up with x = 1/2. As for the y value, just substitute x = 1/2 into the equation $x^2 + y^2 = 4$.
well yes, i understood that part. but where did x=0.5 come from. cause you don't get it with partial derivates and you don't get it from trig change either.

Mark44
Mentor
well yes, i understood that part. but where did x=0.5 come from. cause you don't get it with partial derivates and you don't get it from trig change either.
Edited to fix a typo:
Your original function is constrained to the circle $x^2 + y^2 = 4$, so what do you get if you replace $y^2$ in your original function with $4 - x^2$? If you do that, you get a critical point for x = 1/2.