Register to reply

Finding local min, max, and saddle points in multivariable calculus

Share this thread:
woodenbox
#1
Oct24-10, 06:55 PM
P: 9
1. The problem statement, all variables and given/known data

Find the local maximum and minimum values and saddle point(s) of the function.

f(x,y) = 1 + 2xy - x^2 - y^2

2. Relevant equations

The Second Derivative Test: let D = D(a,b) = fxx(a,b)*fyy(a,b) - [fxy(a,b)]^2
if D > 0 and fxx(a,b) > 0, then f(a,b) is a local minimum
if D > 0 and fxx(a,b) < 0, then f(a,b) is a local maximum
if D < 0 then f(a,b) is a saddle point
if D = 0 then the test is inconclusive

3. The attempt at a solution

I tried to use the Second Derivative Test to find the local mins, maxes, and saddle points but it's inconclusive, and I don't know how else to find them. My textbook says the answer is "f has a local maximum value of 1 at all points of the form (x, x)"

This is my work for the Second Derivative Test:

fx = 2y - 2x = 0 --> 2y = 2x --> y = x

fy = 2x - 2y = 0 --> 2x - 2(x) = 0 --> 0 = 0

so i guess there are critical points at every value where y = x... which matches the textbook's answer.
and then:

fxx = -2

fyy = -2

fxy = 2

so D = fxx * fyy - (fxy)^2 = (-2)*(-2) - 2^2 = 4 - 4 = 0 so the test is inconclusive

Is there a different way to find the local mins, maxes, and saddle points?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
woodenbox
#2
Oct24-10, 08:57 PM
P: 9
got the answer, for anyone else who looks this up.

Rewriting the function as f(x, y) = (x - y)^2 + 1,
we see that the minimum value must be 1 (since 0 is the smallest value of a square),
and this is attained whenever y = x (i.e., points of the form (x, x)).
Mark44
#3
Oct24-10, 08:59 PM
Mentor
P: 21,312
Quote Quote by woodenbox View Post
1. The problem statement, all variables and given/known data

Find the local maximum and minimum values and saddle point(s) of the function.

f(x,y) = 1 + 2xy - x^2 - y^2

2. Relevant equations

The Second Derivative Test: let D = D(a,b) = fxx(a,b)*fyy(a,b) - [fxy(a,b)]^2
if D > 0 and fxx(a,b) > 0, then f(a,b) is a local minimum
if D > 0 and fxx(a,b) < 0, then f(a,b) is a local maximum
if D < 0 then f(a,b) is a saddle point
if D = 0 then the test is inconclusive

3. The attempt at a solution

I tried to use the Second Derivative Test to find the local mins, maxes, and saddle points but it's inconclusive, and I don't know how else to find them. My textbook says the answer is "f has a local maximum value of 1 at all points of the form (x, x)"

This is my work for the Second Derivative Test:

fx = 2y - 2x = 0 --> 2y = 2x --> y = x

fy = 2x - 2y = 0 --> 2x - 2(x) = 0 --> 0 = 0
fx = 2y - 2x
fx = 0 ==> y = x

fy = 2x - 2y
fy = 0 ==> x = y

fx and fy are both zero along the line y = x.
Quote Quote by woodenbox View Post

so i guess there are critical points at every value where y = x... which matches the textbook's answer.
and then:

fxx = -2

fyy = -2

fxy = 2

so D = fxx * fyy - (fxy)^2 = (-2)*(-2) - 2^2 = 4 - 4 = 0 so the test is inconclusive

Is there a different way to find the local mins, maxes, and saddle points?
The function can be written as f(x, y) = 1 - (x2 - 2xy + y2), and the right side can be written in factored form, which should give you some ideas for finding the global maxima and minima.

woodenbox
#4
Oct24-10, 09:06 PM
P: 9
Finding local min, max, and saddle points in multivariable calculus

that does make a lot of sense... i never thought to take a more direct approach. thank you for your help!


Register to reply

Related Discussions
Finding Rel. Max, Min, and Saddle Points on Levelset Calculus & Beyond Homework 1
Local min/max/saddle points of 3d graphs Calculus & Beyond Homework 1
Multivariable Calculus: finding relative extrema/saddle points Calculus & Beyond Homework 2
Finding the critical points of a multivariable function and determining local extrema Calculus & Beyond Homework 1
Finding Critical Points and Local Extrema of a Multivariable Function Calculus & Beyond Homework 4