How to evaluate maximum of a function with 2 variables?

maximum of a function?

Hi,
How do I evaluate the max of a function of two variables
ex: f = ax + by

maximum of a function?

At first, you have to find a math book about function with more than one variable. ANALYSES MATH BOOK.
Here is just a small introduction about this topic. You know, this is a chapter in my math book.
Let's take an example for you to take the overview: remember that we can not solve the others like this. They have cases.
We have a function with 2 variables: f(x,y)=2x^4 + y^4 - x^2 - 2y^2
Then:
df/dx = 8x^3 - 2x = 0
df/dy = 4y^3 - 4y = 0
we have 9 points to determine
P1(0,0)
P2(0,1)
P3(0,-1)
P4(1/2,0)
P5(-1/2,0)
P6(1/2,1)
p7(1/2,-1)
P8(-1/2,1)
P9(-1/2,-1)
Besides
d^2f/dx^2 = 24x^2 - 2 = A
d^2f/dxdy = 0 = B
d^2f/dy^2 = 12^2 - 4 = C
So at P1, f is maximum with close meaning (just maximum to all the points near P1)
If you want to know the maximum of f beyond RxRxR, you'll have to do more.
at P6,P7,P8,P9 f is minimum with close meaning
at P2, P3, P4, P5 we don't have any because B^2 - AC > 0

Re: maximum of a function?

The same way you do with 1 variable but consider a vector of variables instead.
For 2 variables x y
\[\vec{x}= (x,y)\]
The maximum point \[\vec{x}^{*}\] must satisfy
(a) (Gradient) \[\nabla f(\vec{x}^{*})=0\]
(b) (Hessian) \[\nabla ^{2} f(\vec{x}^{*})\] negative definite

Re: maximum of a function?

Hey guys thanks for your insights but its much simpler than that I suppose, my specific problem is
f = x + (y-x)*p where 0<=p<=1, say x and y are limited to some finite value K, then will this function max be under K? (lets assume the values to be positive)

Re: maximum of a function?

f = x + (y-x)*p where 0<=p<=1, say x and y are limited to some finite value K

Click to expand...

This is linear f= (1-p)x + py. f is on the line joining x and y.
If x<=K and y<=K,
max value of f must be f = (1-p)*max(x) + p*max = K

Re: maximum of a function?

you can go in to way , first go and get the jacobian matrix of the funtion and see it in all the points or the short way uuse the second derive proof , in this like you do when you have a single variable funcion you get its second derive but in this case is the second partial derive, the critical points , and replace it in a formula
a*b-c

a=δxy/δxx
b=δxy/δyy
c=δxy/δxy
then the result you get if ≤0, chair point , ≥0 maximun if is 0 it say nothing