To solve this problem and find the critical points of the function $f(x, y) = 2x^2 - 4xy + 5y^2 - 18y$, we will follow these steps:

Step 1: Find the partial derivatives

We need to compute the first partial derivatives of $f(x, y)$ with respect to $x$ and $y$, and then set them to zero to find the critical points.

• $f_x(x, y) = \frac{\partial}{\partial x}(2x^2 - 4xy + 5y^2 - 18y)$
• $f_y(x, y) = \frac{\partial}{\partial y}(2x^2 - 4xy + 5y^2 - 18y)$

Step 2: Solve the system of equations

Once the partial derivatives are set to zero, we will solve the resulting system of equations to get the critical points.

Step 3: Classify the critical points

To classify the critical points, we will use the second partial derivatives and the Hessian matrix to calculate the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$. Depending on the value of $D$, we can classify the points as local minima, local maxima, or saddle points.

Let me compute this step-by-step for you.

Step 1: Partial Derivatives

1. $f_x(x, y) = \frac{\partial}{\partial x}(2x^2 - 4xy + 5y^2 - 18y) = 4x - 4y$
2. $f_y(x, y) = \frac{\partial}{\partial y}(2x^2 - 4xy + 5y^2 - 18y) = -4x + 10y - 18$

Step 2: Solve for Critical Points

Set $f_x(x, y) = 0$ and $f_y(x, y) = 0$ to find the critical points:

• $4x - 4y = 0$ → $x = y$
• $-4x + 10y - 18 = 0$

Substitute $x = y$ into the second equation:

• $-4y + 10y - 18 = 0$
• $6y - 18 = 0$ → $y = 3$

So, $y = 3$, and since $x = y$, $x = 3$.

Thus, the critical point is $(3, 3)$.

Step 3: Classify the Critical Point

We now compute the second partial derivatives to classify the critical point:

• $f_{xx} = \frac{\partial}{\partial x}(4x - 4y) = 4$
• $f_{yy} = \frac{\partial}{\partial y}(-4x + 10y - 18) = 10$
• $f_{xy} = \frac{\partial}{\partial y}(4x - 4y) = -4$

Now, calculate the discriminant: $D = f_{xx}f_{yy} - (f_{xy})^2 = (4)(10) - (-4)^2 = 40 - 16 = 24$ Since $D > 0$ and $f_{xx} > 0$, the critical point $(3, 3)$ is a local minimum.

Conclusion:

• Number of critical points: 1
• Classification: Local minimum at $(3, 3)$

Would you like more details or have any questions?

Here are 5 related questions you might find helpful:

1. How do you determine if a critical point is a local maximum or minimum using the Hessian matrix?
2. What is the significance of the discriminant $D$ in classifying critical points?
3. Could the function have saddle points or maxima if the conditions were different?
4. How do we interpret the results geometrically in terms of the surface $f(x, y)$?
5. How would adding higher-order terms in the function affect the classification of critical points?

Tip: Always check both the sign of $D$ and the second partial derivative with respect to $x$ (i.e., $f_{xx}$) when classifying critical points.