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Finding local min, max, and saddle points in multivariable calculus 
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#1
Oct2410, 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 


#2
Oct2410, 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)). 


#3
Oct2410, 08:59 PM

Mentor
P: 21,297

f_{x} = 0 ==> y = x f_{y} = 2x  2y f_{y} = 0 ==> x = y f_{x} and f_{y} are both zero along the line y = x. 


#4
Oct2410, 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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