We are asked to find the critical points of the given functions and determine whether they correspond to a local maximum, a local minimum, or a saddle point. We will use Theorem C (second derivative test) to classify the critical points. I will solve problem 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 determine whether they correspond to a local maximum, a local minimum, or a saddle point. We will use Theorem C (second derivative test) to classify the critical points. I will solve problem

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

2. Solution Steps

First, we find the first partial derivatives of $f(x, y)$:

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

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

Next, we find the second partial derivatives:

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

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

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

To find the critical points, we set the first partial derivatives equal to zero and solve for $x$ and $y$:

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

$f_y = 8y = 0 \implies y = 0$

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

Now, we compute the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$:

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

Since $D(2, 0) = 16 > 0$ and $f_{xx}(2, 0) = 2 > 0$, the critical point $(2, 0)$ is a local minimum.

3. Final Answer

The critical point is $(2, 0)$. It is a local minimum.