We are asked to find the critical points of the following three functions and classify them as local maxima, local minima, or saddle points: 8. $f(x, y) = e^{-(x^2 + y^2 - 4y)}$ 9. $f(x, y) = \cos x + \cos y + \cos(x + y)$ with $0 < x < \frac{\pi}{2}$ and $0 < y < \frac{\pi}{2}$ 10. $f(x, y) = x^2 + a^2 - 2ax \cos y$ with $-\pi < y < \pi$

1. Problem Description

We are asked to find the critical points of the following three functions and classify them as local maxima, local minima, or saddle points:

8. $f(x, y) = e^{-(x^2 + y^2 - 4y)}$

9. $f(x, y) = \cos x + \cos y + \cos(x + y)$ with $0 < x < \frac{\pi}{2}$ and $0 < y < \frac{\pi}{2}$

1

0. $f(x, y) = x^2 + a^2 - 2ax \cos y$ with $-\pi < y < \pi$

2. Solution Steps

8. $f(x, y) = e^{-(x^2 + y^2 - 4y)}$

$f_x = e^{-(x^2 + y^2 - 4y)} (-2x) = -2x e^{-(x^2 + y^2 - 4y)}$

$f_y = e^{-(x^2 + y^2 - 4y)} (-2y + 4) = (4 - 2y) e^{-(x^2 + y^2 - 4y)}$

To find the critical points, we set $f_x = 0$ and $f_y = 0$.

$-2x e^{-(x^2 + y^2 - 4y)} = 0$, so $x = 0$.

$(4 - 2y) e^{-(x^2 + y^2 - 4y)} = 0$, so $4 - 2y = 0$, which implies $y = 2$.

The critical point is $(0, 2)$.

$f_{xx} = -2 e^{-(x^2 + y^2 - 4y)} + (-2x)^2 e^{-(x^2 + y^2 - 4y)} = (-2 + 4x^2) e^{-(x^2 + y^2 - 4y)}$

$f_{yy} = -2 e^{-(x^2 + y^2 - 4y)} + (4 - 2y)^2 e^{-(x^2 + y^2 - 4y)} = (-2 + (4 - 2y)^2) e^{-(x^2 + y^2 - 4y)}$

$f_{xy} = (-2x)(4 - 2y) e^{-(x^2 + y^2 - 4y)}$

At $(0, 2)$:

$f_{xx}(0, 2) = -2 e^{-(0 + 4 - 8)} = -2 e^4$

$f_{yy}(0, 2) = -2 e^{-(0 + 4 - 8)} = -2 e^4$

$f_{xy}(0, 2) = 0$

$D = f_{xx} f_{yy} - f_{xy}^2 = (-2 e^4) (-2 e^4) - 0^2 = 4 e^8 > 0$.

Since $f_{xx}(0, 2) = -2 e^4 < 0$, the critical point $(0, 2)$ is a local maximum.

9. $f(x, y) = \cos x + \cos y + \cos(x + y)$ with $0 < x < \frac{\pi}{2}$ and $0 < y < \frac{\pi}{2}$

$f_x = -\sin x - \sin(x + y)$

$f_y = -\sin y - \sin(x + y)$

Setting $f_x = 0$ and $f_y = 0$, we have

$\sin x = -\sin(x + y)$ and $\sin y = -\sin(x + y)$.

Thus, $\sin x = \sin y$, and since $0 < x < \frac{\pi}{2}$ and $0 < y < \frac{\pi}{2}$, we have $x = y$.

So $\sin x = -\sin(2x) = -2 \sin x \cos x$.

$\sin x + 2 \sin x \cos x = 0$, $\sin x (1 + 2 \cos x) = 0$.

Since $\sin x \ne 0$, $1 + 2 \cos x = 0$, so $\cos x = -\frac{1}{2}$. Thus, $x = \frac{2 \pi}{3}$.

Since we require $0 < x < \frac{\pi}{2}$, there is no critical point.

However, the image shows $0 < x < \frac{\pi}{2}$ and $0 < y < \frac{\pi}{2}$. There appears to be a mistake, then.

If $x = y$ still holds, then $\sin x = -\sin(2x)$ becomes $\sin(x) + \sin(2x) = 0$.

So $\sin(x) (1 + 2 \cos(x)) = 0$, which gives $\sin(x) = 0$ or $\cos(x) = -\frac{1}{2}$.

If $\sin(x) = 0$, then $x=0$ which is outside the range.

If $\cos(x) = -\frac{1}{2}$, then $x = \frac{2\pi}{3}$, which is also outside the range. So no critical points exist in $(0, \frac{\pi}{2}) \times (0, \frac{\pi}{2})$.

1

0. $f(x, y) = x^2 + a^2 - 2ax \cos y$ with $-\pi < y < \pi$

$f_x = 2x - 2a \cos y$

$f_y = 2ax \sin y$

Setting $f_x = 0$ and $f_y = 0$,

$2x - 2a \cos y = 0$, so $x = a \cos y$.

$2ax \sin y = 0$.

If $a = 0$, then $x = 0$ and $f(x, y) = x^2$. Thus $f(0,y) = 0$, any $y$ can work.

If $a \neq 0$, then either $x = 0$ or $\sin y = 0$.

If $x = 0$, then $0 = a \cos y$. Since $a \neq 0$, $\cos y = 0$, so $y = \pm \frac{\pi}{2}$.

If $\sin y = 0$, then $y = 0$ or $y = \pi$ or $y = -\pi$. (Since $-\pi < y < \pi$, $y=\pi$ and $y=-\pi$ represent the same value).

When $y = 0$, $x = a \cos 0 = a$.

When $y = \pi$, $x = a \cos \pi = -a$.

So we have the critical points $(0, \frac{\pi}{2})$, $(0, -\frac{\pi}{2})$, $(a, 0)$, $(-a, \pi)$.

$f_{xx} = 2$

$f_{yy} = -2ax \cos y$

$f_{xy} = 2a \sin y$

At $(0, \frac{\pi}{2})$, $f_{xx} = 2$, $f_{yy} = 0$, $f_{xy} = 2a$.

$D = (2)(0) - (2a)^2 = -4a^2 < 0$. Saddle point.

At $(0, -\frac{\pi}{2})$, $f_{xx} = 2$, $f_{yy} = 0$, $f_{xy} = -2a$.

$D = (2)(0) - (-2a)^2 = -4a^2 < 0$. Saddle point.

At $(a, 0)$, $f_{xx} = 2$, $f_{yy} = -2a^2$, $f_{xy} = 0$.

$D = (2)(-2a^2) - 0^2 = -4a^2$. Since $a \ne 0$, then $-4a^2 < 0$. Saddle point.

At $(-a, \pi)$, $f_{xx} = 2$, $f_{yy} = 2a^2$, $f_{xy} = 0$.

$D = (2)(2a^2) - 0^2 = 4a^2 > 0$. Since $f_{xx} = 2 > 0$, this is a local minimum.

3. Final Answer

8. Critical point: $(0, 2)$. Local maximum.

9. No critical points in the given domain.

1

0. Critical points: $(0, \frac{\pi}{2})$, $(0, -\frac{\pi}{2})$, $(a, 0)$, $(-a, \pi)$.

$(0, \frac{\pi}{2})$ and $(0, -\frac{\pi}{2})$ are saddle points.

$(a, 0)$ is a saddle point.

$(-a, \pi)$ is a local minimum.

If $a = 0$, then $f(x, y) = x^2$, thus $f(0,y) = 0$. $f_x = 2x = 0$ at $x=0$. $f_{xx}=2$. So a local minimum at x=0, for any $y$.

For part 10 if $a = 0$ we have critical point $(0,y)$ for any $y$. It is a local minimum. If $a \neq 0$, the answer above is good.