obtaining global minimum

**Sambit** · January 25th 2011, 10:08 PM

Consider the function of 2 variables $F(x,y)=21x-12x^2-2y^2+x^3+xy^2$
Find the points of local minimum.
And show that F has no global minimum.

For the 1st part, I have obtained x=2, and correspondingly the value of y. But how to show the 2nd part? Am I to show F takes $-\infty$ as minimum value?
Don't know whether the answer of 1st part is needed to solve the 2nd part.

Help needed....

**TKHunny** · January 26th 2011, 06:16 PM

How did you obtain x = 2? First off, there are two critical points with x = 2. Can you find both?

What is your plan for the other two critical points (1,0) and (7,0)?

You do know you need both first partial derivatives and all three second partial derivatives, right?

**Sambit** · January 26th 2011, 06:58 PM

I am getting x=2, corresponding values of y is $+\sqrt{15} , -\sqrt{15}$ . This is obtained by solving $f_y(x,y)=0$ and $f_x(x,y)=0$ . How are you getting (1,0) and (7,0)?

**TKHunny** · January 26th 2011, 07:33 PM

$f_{y}(x,y) = 0$ has two solutions. You got x = 2. Please get the other one.

**Krahl** · January 26th 2011, 08:08 PM

You have 2 variables.
$F_x(x,y)$ demonstrates how the functions value changes with changes in x. so if this equals 0 it means that the rate of change of F w.r.t x is 0.
similarly $F_y(x,y)$ w.r.t y.

$F_x(x,y)=0$ is pretty hard to solve for x. so why not first solve $F_y(x,y)=0$ to find the y value where the rate of change is 0 (y=0) and then apply that to $F_x(x,y)$ .

**Sambit** · January 26th 2011, 08:10 PM

oops...i ignored that. It gives the solutions (1,0) and (7,0). (1,0) gives maximum value and (7,0) gives the minimum one. These are local optimum points I think. how to find the global ones then? Or rather, how to show that global minimum does not exist?

@ Krahl: That is what I have done.

**Krahl** · January 26th 2011, 08:36 PM

the value of F at each of the extrema is 10 and -98, but F(0,10) = -200 and F(1000,0) > 10.
so the function has no global extrema. So F goes to -Infinity as you said.
also (1,0) is a local maxima since $F_x(0.9,0) > 0$ and $F_x(1.1,0) < 0$ i.e. considering the functions behaviour close to x = 1.

**Sambit** · January 26th 2011, 08:48 PM

So what is the general way to show that a function has no global optima? Is it sufficient to show that for some value of the arguments, the function attains a value which is more than local maximum or less than local minimum?

**Krahl** · January 26th 2011, 08:56 PM

Originally Posted by Wikipedia

In mathematics, maxima and minima, known collectively as extrema (singular: extremum), are the largest value (maximum) or smallest value (minimum), that a function takes in a point either within a given neighborhood (local extremum) or on the function domain in its entirety (global or absolute extremum).

So yes you are right. If you can show that there is a point where the function has a lower(higher) value than any of the local minima(maxima) then there is no global minimum(maximum).

**Sambit** · January 26th 2011, 08:58 PM

Thanks for the clarification. This is what I wanted to know.

**Krahl** · January 26th 2011, 09:05 PM

Welcome Sambit

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