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

[woodenbox]

Homework Statement

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

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

Homework 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

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?

 

[woodenbox]

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]

Homework Statement

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

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

Homework 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

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

f_x = 2y - 2x
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.

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]

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