# Finding maxima and minima of a mathematical function

Find the local maximum and minimum values and the saddle points of f(x,y) = 2xy + x-y.

Â© BrainMass Inc. brainmass.com December 24, 2021, 4:57 pm ad1c9bdddfhttps://brainmass.com/math/calculus-and-analysis/finding-maxima-and-minima-of-a-mathematical-function-18497

## SOLUTION This solution is **FREE** courtesy of BrainMass!

A function of two variables has a local maximum at (a,b)

<br>if f(x,y) =< f(a,b) when (x,y) is near (a,b)

<br>

<br>The number f(a,b) is called a local maximum value.

<br>

<br>If f(x,y) >= f(a,b) when (x,y) is near (a,b) then f(a,b) is a local minimum value.

<br>

<br>

<br>our function is,

<br>

<br>f(x,y) = 2xy + x -y

<br>

<br>let me use fx = partial derivative with respect to x and fy for y

<br>

<br>fx = 2 y + 1 -----(1)

<br>

<br>fy = 2x - 1 -----(2)

<br>

<br>to find the crical point, solve the equations,

<br>

<br>2y+ 1 = 0 and 2x-1 = 0

<br>

<br>2y+1=0 ====> y = -1/2

<br>

<br>2x-1=0 ====> x = 1/2

<br>

<br>Therefore, the critical point is (1/2, -1/2)

<br>

<br>Now to the second derivative test

<br>

<br>D = D(a,b) = fxx (a,b) * fyy(a,b)- [fxy (a,b)]^2

<br>(Please see the attached figure)

<br>

<br>fxx(a,b) = 0 (differentiate eq 1 with respect to x again)

<br>fyy(a,b) = 0

<br>

<br>fxy = 2 (differentiate eq 1 with respect to y)

<br>

<br>at(1/2, -1/2)

<br>

<br>D = 0 - 2 = -2

<br>

<br>Now refer to the figure, (1/2, -1/2) is a saddle point

https://brainmass.com/math/calculus-and-analysis/finding-maxima-and-minima-of-a-mathematical-function-18497