We are asked to find the critical points of several functions and classify them as local maxima, local minima, or saddle points, using Theorem C. Let's solve question 4, 5, 6 and 7.

1. Problem Description

We are asked to find the critical points of several functions and classify them as local maxima, local minima, or saddle points, using Theorem C. Let's solve question 4, 5, 6 and

7.

2. Solution Steps

Question 4: $f(x, y) = xy^2 - 6x^2 - 3y^2$

First, find the partial derivatives:

$f_x = y^2 - 12x$

$f_y = 2xy - 6y$

Set the partial derivatives equal to zero and solve for $x$ and $y$:

$y^2 - 12x = 0$

$2xy - 6y = 0$ => $2y(x-3) = 0$

From the second equation, either $y = 0$ or $x = 3$.

If $y = 0$, then from the first equation, $0^2 - 12x = 0$, so $x = 0$. Thus, $(0, 0)$ is a critical point.

If $x = 3$, then from the first equation, $y^2 - 12(3) = 0$, so $y^2 = 36$, which gives $y = \pm 6$. Thus, $(3, 6)$ and $(3, -6)$ are critical points.

Now find the second partial derivatives:

$f_{xx} = -12$

$f_{yy} = 2x - 6$

$f_{xy} = 2y$

Calculate $D = f_{xx}f_{yy} - f_{xy}^2 = (-12)(2x - 6) - (2y)^2 = -24x + 72 - 4y^2$.

At $(0, 0)$, $D = 72 - 0 = 72 > 0$. Since $f_{xx}(0, 0) = -12 < 0$, $(0, 0)$ is a local maximum.

At $(3, 6)$, $D = -24(3) + 72 - 4(36) = -72 + 72 - 144 = -144 < 0$. Thus, $(3, 6)$ is a saddle point.

At $(3, -6)$, $D = -24(3) + 72 - 4(-6)^2 = -72 + 72 - 144 = -144 < 0$. Thus, $(3, -6)$ is a saddle point.

Question 5: $f(x, y) = xy$

$f_x = y$

$f_y = x$

Setting them equal to zero:

$y = 0$

$x = 0$

Thus, $(0, 0)$ is the only critical point.

$f_{xx} = 0$

$f_{yy} = 0$

$f_{xy} = 1$

$D = f_{xx}f_{yy} - f_{xy}^2 = (0)(0) - (1)^2 = -1 < 0$. Thus, $(0, 0)$ is a saddle point.

Question 6: $f(x, y) = x^3 + y^3 - 6xy$

$f_x = 3x^2 - 6y$

$f_y = 3y^2 - 6x$

Setting them equal to zero:

$3x^2 - 6y = 0$ => $x^2 = 2y$

$3y^2 - 6x = 0$ => $y^2 = 2x$

Substitute $y = x^2/2$ into $y^2 = 2x$:

$(x^2/2)^2 = 2x$

$x^4/4 = 2x$

$x^4 = 8x$

$x^4 - 8x = 0$

$x(x^3 - 8) = 0$

$x = 0$ or $x^3 = 8$ => $x = 2$

If $x = 0$, then $y = x^2/2 = 0$.

If $x = 2$, then $y = x^2/2 = 4/2 = 2$.

The critical points are $(0, 0)$ and $(2, 2)$.

$f_{xx} = 6x$

$f_{yy} = 6y$

$f_{xy} = -6$

$D = f_{xx}f_{yy} - f_{xy}^2 = (6x)(6y) - (-6)^2 = 36xy - 36$.

At $(0, 0)$, $D = 36(0)(0) - 36 = -36 < 0$. Thus, $(0, 0)$ is a saddle point.

At $(2, 2)$, $D = 36(2)(2) - 36 = 144 - 36 = 108 > 0$. Since $f_{xx}(2, 2) = 6(2) = 12 > 0$, $(2, 2)$ is a local minimum.

Question 7: $f(x, y) = xy + \frac{2}{x} + \frac{4}{y}$

$f_x = y - \frac{2}{x^2}$

$f_y = x - \frac{4}{y^2}$

Setting them equal to zero:

$y - \frac{2}{x^2} = 0$ => $y = \frac{2}{x^2}$

$x - \frac{4}{y^2} = 0$ => $x = \frac{4}{y^2}$

Substitute $y = 2/x^2$ into $x = 4/y^2$:

$x = \frac{4}{(2/x^2)^2} = \frac{4}{4/x^4} = x^4$

$x = x^4$

$x^4 - x = 0$

$x(x^3 - 1) = 0$

Since $x$ and $y$ cannot be zero because of the $\frac{2}{x}$ and $\frac{4}{y}$ terms, we must have $x^3 = 1$, so $x = 1$.

If $x = 1$, then $y = \frac{2}{1^2} = 2$.

The critical point is $(1, 2)$.

$f_{xx} = \frac{4}{x^3}$

$f_{yy} = \frac{8}{y^3}$

$f_{xy} = 1$

$D = f_{xx}f_{yy} - f_{xy}^2 = \frac{4}{x^3} \cdot \frac{8}{y^3} - 1 = \frac{32}{x^3 y^3} - 1$.

At $(1, 2)$, $D = \frac{32}{(1)^3 (2)^3} - 1 = \frac{32}{8} - 1 = 4 - 1 = 3 > 0$. Since $f_{xx}(1, 2) = \frac{4}{(1)^3} = 4 > 0$, $(1, 2)$ is a local minimum.

3. Final Answer

Question 4: Critical points are $(0, 0)$, $(3, 6)$, and $(3, -6)$. $(0, 0)$ is a local maximum, and $(3, 6)$ and $(3, -6)$ are saddle points.

Question 5: Critical point is $(0, 0)$. $(0, 0)$ is a saddle point.

Question 6: Critical points are $(0, 0)$ and $(2, 2)$. $(0, 0)$ is a saddle point, and $(2, 2)$ is a local minimum.

Question 7: Critical point is $(1, 2)$. $(1, 2)$ is a local minimum.