Critical Points & the Second Derivative Test

$f_x = 3 - 3x^2$, $f_y = -2y + 4$, $f_{xx} = -6x$, $f_{yy} = -2$, $f_{xy} = 0$.

$3 - 3x^2 = 0 \Rightarrow x = \pm 1$; $\ -2y + 4 = 0 \Rightarrow y = 2$. Points $(1, 2)$ and $(-1, 2)$.

$(1, 2)$: $D = (-6)(-2) - 0 = 12$. $\ (-1, 2)$: $D = (6)(-2) - 0 = -12$.

$(1, 2)$: $D > 0$ and $f_{xx} = -6 < 0$, a local max. $\ (-1, 2)$: $D < 0$, a saddle point.

$f_x = 3x^2 + 3y$, $f_y = 3x - 3y^2$, $f_{xx} = 6x$, $f_{yy} = -6y$, $f_{xy} = 3$.

$f_x = 0 \Rightarrow y = -x^2$. Then $3x - 3x^4 = 0 \Rightarrow 3x(1 - x^3) = 0 \Rightarrow x = 0$ or $x = 1$. Points $(0, 0)$ and $(1, -1)$.

$(0, 0)$: $D = 0 - 9 = -9$. $\ (1, -1)$: $D = (6)(6) - 9 = 27$.

$(0, 0)$: a saddle point. $\ (1, -1)$: $D > 0$ and $f_{xx} = 6 > 0$, a local min.

$f_x = 2xy - 2y$, $f_y = x^2 - 2x + 2y$, $f_{xx} = 2y$, $f_{yy} = 2$, $f_{xy} = 2x - 2$.

$f_x = 2y(x - 1) = 0$. Case $y = 0$: $x^2 - 2x = 0$, so $x = 0$ or $x = 2$. Case $x = 1$: $-1 + 2y = 0$, so $y = \tfrac{1}{2}$. Points $(0, 0)$, $(2, 0)$, $(1, \tfrac{1}{2})$.

$(0, 0)$: $D = 0 - 4 = -4$. $\ (2, 0)$: $D = 0 - 4 = -4$. $\ (1, \tfrac{1}{2})$: $D = (1)(2) - 0 = 2$.

$(0, 0)$ and $(2, 0)$: saddle points. $\ (1, \tfrac{1}{2})$: $D > 0$ and $f_{xx} = 1 > 0$, a local min.

$f_x = 4x + ky$ and $f_y = kx + 16y$ are both $0$ at $(0, 0)$, for every $k$.

$f_{xx} = 4$, $f_{yy} = 16$, $f_{xy} = k$, so $D = 64 - k^2$.

Saddle: $64 - k^2 < 0$.

$k^2 > 64 \Rightarrow |k| > 8$. At $k = \pm 8$, $D = 0$ and $f = 2(x \pm 2y)^2 \ge 0$: a minimum, not a saddle.

$f_x = 2kx + 2y$ and $f_y = 2x - 2y$ are both $0$ at $(0, 0)$.

$f_{xx} = 2k$, $f_{yy} = -2$, $f_{xy} = 2$, so $D = -4k - 4$.

Max: $D = -4k - 4 > 0$ and $f_{xx} = 2k < 0$.

$k < -1$, which already makes $2k < 0$. At $k = -1$, $D = 0$ and the test is silent (there $f = -(x - y)^2$).

$f_x = 4x^3$ and $f_y = 2y$ are both $0$ at $(0, 0)$. There $f_{xx} = 12x^2 = 0$, $f_{yy} = 2$, $f_{xy} = 0$, so $D = 0$.

$f(x,y) - f(0,0) = x^4 + y^2$.

Both terms are $\ge 0$ at every point, and their sum is $> 0$ away from the origin.

Never negative, so $(0, 0)$ is a local (in fact global) minimum.

$f_x = 3x^2$ and $f_y = 2y$ are both $0$ at $(0, 0)$. There $f_{xx} = 6x = 0$, $f_{yy} = 2$, $f_{xy} = 0$, so $D = 0$.

$f(x,y) - f(0,0) = x^3 + y^2$.

On the line $y = 0$ it is $x^3$: positive for $x > 0$, negative for $x < 0$.

Both signs near $(0, 0)$: neither a max nor a min, so a saddle point.

$f = y^2 - 4x^2y + 3x^4$, so $f_x = -8xy + 12x^3$ and $f_y = 2y - 4x^2$, both $0$ at the origin. There $f_{xx} = 0$, $f_{yy} = 2$, $f_{xy} = 0$, so $D = 0$.

$f(x,y) - f(0,0) = (y - x^2)(y - 3x^2)$.

On $y = mx$: $m^2x^2 - 4mx^3 + 3x^4 > 0$ for small $x \ne 0$. On $x = 0$: $y^2 > 0$ for $y \ne 0$. But the curve $y = 2x^2$ lies between $y = x^2$ and $y = 3x^2$. On it, $(x^2)(-x^2) = -x^4 < 0$ for $x \ne 0$.

Positive along every line, negative along $y = 2x^2$: a saddle point.

Base $x \times y$, height $z$. Cost $C = 3xy + 2(2xz + 2yz)$, with $xyz = 6$.

$z = \dfrac{6}{xy}$, so $C = 3xy + \dfrac{24}{x} + \dfrac{24}{y}$ for $x, y > 0$.

$C_x = 3y - \dfrac{24}{x^2} = 0$ and $C_y = 3x - \dfrac{24}{y^2} = 0$, so $x^2y = 8 = xy^2$. Then $x = y$ and $x^3 = 8$: the point $(2, 2)$.

$C_{xx} = \dfrac{48}{x^3} = 6$, $C_{yy} = 6$, $C_{xy} = 3$: $D = 27 > 0$ and $C_{xx} > 0$, a min. $C \to \infty$ at the edges, so it is the cheapest.

Base $2 \times 2$ m, height $z = \dfrac{6}{4} = 1.5$ m, cost $12 + 12 + 12 = 36$ dollars.

Minimize the squared distance $x^2 + y^2 + z^2$ for points on the plane.

$z = 9 - \tfrac{x}{2} - y$, so $g(x,y) = x^2 + y^2 + \left(9 - \tfrac{x}{2} - y\right)^2$.

$g_x = \tfrac{5}{2}x + y - 9 = 0$ and $g_y = x + 4y - 18 = 0$, so $(x, y) = (2, 4)$.

$g_{xx} = \tfrac{5}{2}$, $g_{yy} = 4$, $g_{xy} = 1$: $D = 9 > 0$ and $g_{xx} > 0$, a min. $g \to \infty$ far away, so it is the closest.

$z = 9 - 1 - 4 = 4$: the point $(2, 4, 4)$, at distance $\sqrt{4 + 16 + 16} = 6$.

Maximize $xyz^2$ subject to $x + y + z = 20$.

$z = 20 - x - y$, so $P(x,y) = xy(20 - x - y)^2$ for $x, y > 0$ and $x + y < 20$.

$P_x = y(20 - x - y)(20 - 3x - y)$ and $P_y = x(20 - x - y)(20 - x - 3y)$. Inside the region only the last factors can be $0$: $3x + y = 20$ and $x + 3y = 20$, so $(5, 5)$.

At $(5, 5)$: $P_{xx} = P_{yy} = -150$, $P_{xy} = -50$, so $D = 20000 > 0$ and $P_{xx} < 0$, a max. $P = 0$ on the edges, so it is the largest.

$x = y = 5$, $z = 10$, and $xyz^2 = 2500$.