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For each of the following functions, find and classify all critical points. [That is, use the second-derivative test to deduce whether each critical point is a local max, a local min, or a saddle.]

$f(x,y) = (x+y)(1-xy)$

my solution: I first equated gradient of $f$, $\nabla f(a) = 0$

so, $(1-2xy-y^2, 1-2xy-x^2) = (0,0)$ solving two equations I found $y = x,-x$.

and then I found the Hessian matrix

$H = \begin{bmatrix} -2y & -2x-2y \\ -2x-2y & -2x \end{bmatrix}$

Now, I am having trouble in how to put in my critical points and check if they are local max, local min or saddle point.

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  • $\begingroup$ Have you used the second derivative test? $\endgroup$ – Dr. Sonnhard Graubner Jun 29 '18 at 16:54
  • $\begingroup$ I think there is no maximum and no Minimum. $\endgroup$ – Dr. Sonnhard Graubner Jun 29 '18 at 16:55
  • $\begingroup$ @Dr.SonnhardGraubner I know that I have to use second-derivative test. Its written in the question. I just don't know what should the critical points be, should they just be (x,x) and (x,-x) or something else. $\endgroup$ – kronos Jun 29 '18 at 16:56
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With

$f(x, y) = (x + y)(1 - xy) \tag 1$

and

$\nabla f = (f_x, f_y), \tag 2$

we see that

$f_x = 1 - xy + (x + y)(-y) = 1 - xy - xy - y^2 = 1 - 2xy - y^2, \tag 3$

and likewise

$f_y = 1 - xy + (x + y)(-x) = 1 - xy - x^2 - xy = 1 - 2xy - x^2; \tag 4$

at critical points of $f(x, y)$, we have

$\nabla f(x, y) = 0, \tag 5$

whence from (3) and (4), at the critical points,

$1 - 2xy - y^2 = 0 = 1 - 2xy - x^2; \tag 6$

we solve this system by observing that it implies

$y^2 = 1 - 2xy = x^2, \tag 7$

so that

$y = \pm x; \tag 8$

using (8) in (6) we may write an equation for $x$:

$0 = x^2 + 2xy - 1 = x^2 \pm 2x^2 - 1; \tag 9$

$x$ must thus obey

$3x^2 - 1 = 0 \tag{10}$

or

$x^2 + 1 = 0; \tag{11}$

we rule out (11) since $x$ is real; thus

$x = \pm \dfrac{1}{\sqrt 3} = \pm \dfrac{\sqrt 3}{3}; \tag{12}$

again from (6),

$y = \dfrac{1 - x^2}{2x} = \dfrac{\dfrac{2}{3}}{2x} = \dfrac{1}{3x}; \tag{13}$

therefore the critical points are

$(x, y) = \left ( \dfrac{\sqrt 3}{3}, \dfrac{\sqrt 3}{3} \right ), \; (x, y) = \left ( -\dfrac{\sqrt 3}{3}, -\dfrac{\sqrt 3}{3} \right ); \tag{14}$

the Hessian $H_f$ of $f(x, y)$ has been provided for us courtesy of our OP kronos:

$H_f = \begin{bmatrix} -2y & -2x-2y \\ -2x-2y & -2x \end{bmatrix}; \tag{15}$

we thus see that

$\det(H_f) = 4xy - 4(x + y)^2 = 4(xy - (x + y)^2) = -4(x^2 +xy + y^2); \tag{16}$

since at the critical points we have

$x = y = \pm \dfrac{\sqrt 3}{3}, \tag{17}$

it follows that at these points

$\det(H_f) = -4, \tag{18}$

which, as is well-known, implies that each critical point is a saddle.

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You are wrong about the critical points. There are only two of them: $\pm\frac1{\sqrt3}\left(1,1\right)$. In both cases, the Hessian has negative determinant and (since we are working on $\mathbb{R}^2$), the point is a saddle point.

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  • $\begingroup$ Thanks! I knew was doing something wrong. Hopefully, I can figure out how to correctly solve the equations to find the critical points. $\endgroup$ – kronos Jun 29 '18 at 16:59

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