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

by woodenbox
Tags: derivative test, local max, local min, multivar
 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
 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)).
Mentor
P: 19,675
 Quote by woodenbox 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 by woodenbox 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.

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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