The problem asks to find all critical points of the given functions, and determine whether they correspond to local maximum, local minimum or saddle point. We'll solve the first two questions here. 1. $f(x, y) = x^2 + 4y^2 - 4x$

1. Problem Description

The problem asks to find all critical points of the given functions, and determine whether they correspond to local maximum, local minimum or saddle point. We'll solve the first two questions here.

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

Step 1: Find the first partial derivatives.

$f_x = \frac{\partial f}{\partial x} = 2x - 4$

$f_y = \frac{\partial f}{\partial y} = 8y$

Step 2: Find the critical points by setting the first partial derivatives equal to zero.

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

$8y = 0 \implies y = 0$

Thus, the critical point is $(2, 0)$.

Step 3: Find the second partial derivatives.

$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2$

$f_{yy} = \frac{\partial^2 f}{\partial y^2} = 8$

$f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0$

Step 4: Calculate the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$.

$D = (2)(8) - (0)^2 = 16 > 0$

Step 5: Determine the nature of the critical point.

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

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

Step 1: Find the first partial derivatives.

$f_x = \frac{\partial f}{\partial x} = 2x - 2$

$f_y = \frac{\partial f}{\partial y} = 8y + 8$

Step 2: Find the critical points by setting the first partial derivatives equal to zero.

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

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

Thus, the critical point is $(1, -1)$.

Step 3: Find the second partial derivatives.

$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2$

$f_{yy} = \frac{\partial^2 f}{\partial y^2} = 8$

$f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0$

Step 4: Calculate the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$.

$D = (2)(8) - (0)^2 = 16 > 0$

Step 5: Determine the nature of the critical point.

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

3. Final Answer

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

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