We need to find the global maximum and minimum values of the given functions $f(x, y)$ on the set $S$, and indicate where each occurs. Problem 11: $f(x, y) = 3x + 4y$, $S = \{(x, y): 0 \le x \le 1, -1 \le y \le 1\}$ Problem 12: $f(x, y) = x^2 + y^2$, $S = \{(x, y): -1 \le x \le 3, -1 \le y \le 4\}$ Problem 13: $f(x, y) = x^2 - y^2 + 1$, $S = \{(x, y): x^2 + y^2 \le 1\}$ Problem 14: $f(x, y) = x^2 - 6x + y^2 - 8y + 7$, $S = \{(x, y): x^2 + y^2 \le 1\}$

1. Problem Description

We need to find the global maximum and minimum values of the given functions $f(x, y)$ on the set $S$, and indicate where each occurs.

Problem 11: $f(x, y) = 3x + 4y$, $S = \{(x, y): 0 \le x \le 1, -1 \le y \le 1\}$

Problem 12: $f(x, y) = x^2 + y^2$, $S = \{(x, y): -1 \le x \le 3, -1 \le y \le 4\}$

Problem 13: $f(x, y) = x^2 - y^2 + 1$, $S = \{(x, y): x^2 + y^2 \le 1\}$

Problem 14: $f(x, y) = x^2 - 6x + y^2 - 8y + 7$, $S = \{(x, y): x^2 + y^2 \le 1\}$

2. Solution Steps

Problem 11: $f(x, y) = 3x + 4y$, $S = \{(x, y): 0 \le x \le 1, -1 \le y \le 1\}$

Since $f(x, y)$ is linear, the extrema occur at the vertices of the region $S$. The vertices are $(0, -1)$, $(0, 1)$, $(1, -1)$, $(1, 1)$.

$f(0, -1) = 3(0) + 4(-1) = -4$

$f(0, 1) = 3(0) + 4(1) = 4$

$f(1, -1) = 3(1) + 4(-1) = -1$

$f(1, 1) = 3(1) + 4(1) = 7$

Minimum value is $-4$ at $(0, -1)$ and maximum value is $7$ at $(1, 1)$.

Problem 12: $f(x, y) = x^2 + y^2$, $S = \{(x, y): -1 \le x \le 3, -1 \le y \le 4\}$

$f(x, y)$ represents the squared distance from the origin.

The minimum value occurs at $(0, 0)$ if it is in the region $S$. Since $-1 \le 0 \le 3$ and $-1 \le 0 \le 4$, $(0, 0)$ is in $S$. So the minimum value is $f(0, 0) = 0^2 + 0^2 = 0$.

For the maximum value, we examine the corners of the rectangular region $S$: $(-1, -1)$, $(-1, 4)$, $(3, -1)$, $(3, 4)$.

$f(-1, -1) = (-1)^2 + (-1)^2 = 1 + 1 = 2$

$f(-1, 4) = (-1)^2 + 4^2 = 1 + 16 = 17$

$f(3, -1) = 3^2 + (-1)^2 = 9 + 1 = 10$

$f(3, 4) = 3^2 + 4^2 = 9 + 16 = 25$

The maximum value is $25$ at $(3, 4)$.

Problem 13: $f(x, y) = x^2 - y^2 + 1$, $S = \{(x, y): x^2 + y^2 \le 1\}$

We can use Lagrange multipliers.

$g(x, y) = x^2 + y^2 - 1 = 0$

$\nabla f = (2x, -2y)$

$\nabla g = (2x, 2y)$

$2x = \lambda 2x$

$-2y = \lambda 2y$

$x^2 + y^2 = 1$

Case 1: $x = 0$. Then $y^2 = 1$, so $y = \pm 1$.

$f(0, 1) = 0^2 - 1^2 + 1 = 0$

$f(0, -1) = 0^2 - (-1)^2 + 1 = 0$

Case 2: $y = 0$. Then $x^2 = 1$, so $x = \pm 1$.

$f(1, 0) = 1^2 - 0^2 + 1 = 2$

$f(-1, 0) = (-1)^2 - 0^2 + 1 = 2$

Case 3: $x \ne 0$ and $y \ne 0$. Then $\lambda = 1$ and $\lambda = -1$, which is a contradiction.

Consider the interior points: $\frac{\partial f}{\partial x} = 2x = 0$ and $\frac{\partial f}{\partial y} = -2y = 0$.

So $(0, 0)$ is a critical point. $f(0, 0) = 0^2 - 0^2 + 1 = 1$.

The maximum value is $2$ at $(1, 0)$ and $(-1, 0)$. The minimum value is $0$ at $(0, 1)$ and $(0, -1)$.

Problem 14: $f(x, y) = x^2 - 6x + y^2 - 8y + 7$, $S = \{(x, y): x^2 + y^2 \le 1\}$

Complete the square:

$f(x, y) = (x^2 - 6x + 9) + (y^2 - 8y + 16) + 7 - 9 - 16 = (x - 3)^2 + (y - 4)^2 - 18$

Let $x = r \cos \theta$ and $y = r \sin \theta$, where $r^2 \le 1$, so $0 \le r \le 1$.

$f(r \cos \theta, r \sin \theta) = (r \cos \theta - 3)^2 + (r \sin \theta - 4)^2 - 18 = r^2 \cos^2 \theta - 6r \cos \theta + 9 + r^2 \sin^2 \theta - 8r \sin \theta + 16 - 18 = r^2 - 6r \cos \theta - 8r \sin \theta + 7$

$f(x, y)$ represents the squared distance from $(3, 4)$ minus

1

8. The distance from $(3, 4)$ to the origin is $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$. Since $5 > 1$, $(3, 4)$ is not in $S$.

The point in $S$ closest to $(3, 4)$ is the intersection of the line from $(0, 0)$ to $(3, 4)$ with the boundary $x^2 + y^2 = 1$.

The line is $y = \frac{4}{3}x$.

$x^2 + (\frac{4}{3}x)^2 = 1$

$x^2 + \frac{16}{9}x^2 = 1$

$\frac{25}{9}x^2 = 1$

$x^2 = \frac{9}{25}$

$x = \pm \frac{3}{5}$. Since we want the point closest to $(3, 4)$, $x = \frac{3}{5}$.

$y = \frac{4}{3} (\frac{3}{5}) = \frac{4}{5}$.

The point closest is $(\frac{3}{5}, \frac{4}{5})$. $f(\frac{3}{5}, \frac{4}{5}) = (\frac{3}{5} - 3)^2 + (\frac{4}{5} - 4)^2 - 18 = (\frac{3-15}{5})^2 + (\frac{4-20}{5})^2 - 18 = (\frac{-12}{5})^2 + (\frac{-16}{5})^2 - 18 = \frac{144}{25} + \frac{256}{25} - 18 = \frac{400}{25} - 18 = 16 - 18 = -2$

The point in S farthest from $(3, 4)$ is $(-3/5, -4/5)$.

$f(-3/5, -4/5) = (-3/5 - 3)^2 + (-4/5 - 4)^2 - 18 = (-18/5)^2 + (-24/5)^2 - 18 = 324/25 + 576/25 - 18 = 900/25 - 18 = 36 - 18 = 18$

Check the center $(0, 0)$. $f(0, 0) = (0 - 3)^2 + (0 - 4)^2 - 18 = 9 + 16 - 18 = 25 - 18 = 7$

Minimum:

Let $u = x - 3$ and $v = y - 4$.

$f(x, y) = u^2 + v^2 - 18$.

$x = u + 3$ and $y = v + 4$.

$x^2 + y^2 \le 1$, so $(u + 3)^2 + (v + 4)^2 \le 1$

$f(x, y) = x^2 - 6x + y^2 - 8y + 7$.

$\frac{\partial f}{\partial x} = 2x - 6 = 0$, so $x = 3$.

$\frac{\partial f}{\partial y} = 2y - 8 = 0$, so $y = 4$.

$(3, 4)$ is not in the region $S$.

Minimum value is $-2$ at $(3/5, 4/5)$

Maximum value is $18$ at $(-3/5, -4/5)$.

3. Final Answer

Problem 11:

Maximum: $7$ at $(1, 1)$

Minimum: $-4$ at $(0, -1)$

Problem 12:

Maximum: $25$ at $(3, 4)$

Minimum: $0$ at $(0, 0)$

Problem 13:

Maximum: $2$ at $(1, 0)$ and $(-1, 0)$

Minimum: $0$ at $(0, 1)$ and $(0, -1)$

Problem 14:

Maximum: $18$ at $(-3/5, -4/5)$

Minimum: $-2$ at $(3/5, 4/5)$