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


by woodenbox
Tags: derivative test, local max, local min, multivar
woodenbox
woodenbox is offline
#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
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woodenbox
woodenbox is offline
#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
Mark44 is online now
#3
Oct24-10, 08:59 PM
Mentor
P: 20,988
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
woodenbox is offline
#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!


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