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

• Wi_N
In summary: 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 =...In summary, the conversation involved finding the maximum and minimum values of a function subject to the constraint of a circle. The participants discussed using partial derivatives and trigonometric changes to solve the problem. The final result was that the maximum value was 17/20 at the point (1/2, √15/4) and there was no global minimum.
Wi_N
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 $$x^2+y^2=4$$
Relevant 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?

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

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

Wi_N said:
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
WWGD said:
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.

Wi_N said:
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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Wi_N said:
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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WWGD said:
no global minimum. Plot the subbed-in curve.
continuous function applied on a compact domain...

StoneTemplePython said:
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.

WWGD said:
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?

Wi_N said:
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 ##

WWGD said:
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 ?

Wi_N said:
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
WWGD said:
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.

Wi_N said:
im sorry but (1/2)^2 + (sqrt15 /4)^2 is not equal to 4.
Clearly @WWGD meant ##\sqrt{15/4}##

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

WWGD said:
##(1/2, \sqrt 15/4)## is on the circle.
Wi_N said:
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
Mark44 said:
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?

Wi_N said:
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}##?

Mark44 said:
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.

Wi_N said:
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##.

Mark44 said:
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.

Wi_N said:
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.

Mark44 said:
Your original function is constrained to the circle x^2 + 4^2 = 4, ##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.

it was constrained to x^2 + y^2 =4, originally.

Wi_N said:
it was constrained to x^2 + y^2 =4, originally.
I mistyped it. I'll change it to what I meant.

Mark44 said:
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.

x^2 +y^2 =4, replace y^2=4-x^2 will just lead to (x^2 +4-x^2)/5 = 4/5. where is the x=0.5.

Wi_N said:
x^2 +y^2 =4, replace y^2=4-x^2 will just lead to (x^2 +4-x^2)/5 = 4/5. where is the x=0.5.
From post #1 ...
##f(x,y)=\frac{x+y^2}{1+x^2+y^2} ##
Replace ##y^2## in this equation by ##4 - x^2##. What do you get?

Wi_N
Mark44 said:
From post #1 ...

Replace ##y^2## in this equation by ##4 - x^2##. What do you get?

edit never mind. thanks.

Correct. The numerator is not just 5, but the function is now much simpler, due to the constraint that its graph must lie on ##x^2 + y^2 = 4##.

Wi_N

## 1. What is the F(x, y) Min Max problem with boundaries?

The F(x, y) Min Max problem with boundaries is a mathematical optimization problem where the goal is to find the minimum or maximum value of a function (F) with two variables (x and y) within a given set of boundaries. This problem is commonly used in economics, engineering, and other fields to find the optimal solution to a problem.

## 2. How do you solve the F(x, y) Min Max problem with boundaries?

To solve the F(x, y) Min Max problem with boundaries, you can use various methods such as the Lagrange multiplier method, the simplex method, or the gradient descent method. These methods involve finding critical points and evaluating them to determine the minimum or maximum value of the function within the given boundaries.

## 3. What are the applications of the F(x, y) Min Max problem with boundaries?

The F(x, y) Min Max problem with boundaries has numerous applications in various fields, including economics, engineering, operations research, and computer science. It can be used to optimize production processes, resource allocation, financial investments, and many other real-world problems.

## 4. What are the limitations of the F(x, y) Min Max problem with boundaries?

One limitation of the F(x, y) Min Max problem with boundaries is that it assumes a continuous and differentiable function, which may not always be the case in real-world problems. Additionally, finding the optimal solution can be computationally intensive, especially for complex functions with multiple variables and constraints.

## 5. How can the F(x, y) Min Max problem with boundaries be extended to higher dimensions?

The F(x, y) Min Max problem with boundaries can be extended to higher dimensions by adding more variables and constraints. This is known as the N-dimensional optimization problem, where N represents the number of variables. The same methods used to solve the 2-dimensional problem can be applied to solve higher-dimensional problems as well.

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