We are asked to find the critical points of the given functions and classify them as local maxima, local minima, or saddle points. We will use Theorem C, which involves calculating the second partial derivatives and the determinant $D$. The functions are: 1. $f(x, y) = x^2 + 4y^2 - 4x$

1. Problem Description

We are asked to find the critical points of the given functions and classify them as local maxima, local minima, or saddle points. We will use Theorem C, which involves calculating the second partial derivatives and the determinant $D$.

The functions are:

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

2. $f(x, y) = x^2 + 4y^2 - 2x + 8y - 1$

2. Solution Steps

Problem 1: $f(x, y) = x^2 + 4y^2 - 4x$

a. Find the first partial derivatives:

$f_x = 2x - 4$

$f_y = 8y$

b. Set the first partial derivatives equal to zero to find critical points:

$2x - 4 = 0 \implies x = 2$

$8y = 0 \implies y = 0$

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

c. Find the second partial derivatives:

$f_{xx} = 2$

$f_{yy} = 8$

$f_{xy} = 0$

d. Calculate the determinant $D = f_{xx}f_{yy} - (f_{xy})^2$:

$D(x, y) = (2)(8) - (0)^2 = 16$

e. Evaluate $D$ at the critical point $(2, 0)$:

$D(2, 0) = 16 > 0$

Since $f_{xx}(2, 0) = 2 > 0$ and $D(2, 0) > 0$, the critical point $(2, 0)$ is a local minimum.

Problem 2: $f(x, y) = x^2 + 4y^2 - 2x + 8y - 1$

a. Find the first partial derivatives:

$f_x = 2x - 2$

$f_y = 8y + 8$

b. Set the first partial derivatives equal to zero to find critical points:

$2x - 2 = 0 \implies x = 1$

$8y + 8 = 0 \implies y = -1$

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

c. Find the second partial derivatives:

$f_{xx} = 2$

$f_{yy} = 8$

$f_{xy} = 0$

d. Calculate the determinant $D = f_{xx}f_{yy} - (f_{xy})^2$:

$D(x, y) = (2)(8) - (0)^2 = 16$

e. Evaluate $D$ at the critical point $(1, -1)$:

$D(1, -1) = 16 > 0$

Since $f_{xx}(1, -1) = 2 > 0$ and $D(1, -1) > 0$, the critical point $(1, -1)$ is a local minimum.

3. Final Answer

Problem 1: The critical point is $(2, 0)$, which is a local minimum.

Problem 2: The critical point is $(1, -1)$, which is a local minimum.