Find x and y that minimize the expression of 4x
2 + 2y
2 + 4xy − 16x − 4y + 23.

To find the values of \( x \) and \( y \) that minimize the expression \( 4x^2 + 2y^2 + 4xy - 16x - 4y + 23 \), we will use calculus, specifically finding the critical points by setting the partial derivatives equal to zero.

Given the expression:

\[ f(x, y) = 4x^2 + 2y^2 + 4xy - 16x - 4y + 23 \]

First, we need to find the partial derivatives with respect to \( x \) and \( y \).

1. **Partial derivative with respect to \( x \):**

\[ \frac{\partial f}{\partial x} = 8x + 4y - 16 \]

2. **Partial derivative with respect to \( y \):**

\[ \frac{\partial f}{\partial y} = 4y + 4x - 4 \]

Next, we set these partial derivatives equal to zero to find the critical points.

\[ 8x + 4y - 16 = 0 \]
\[ 4x + 4y - 4 = 0 \]

We solve this system of linear equations to find \( x \) and \( y \).

From the second equation, we can simplify:

\[ 4x + 4y - 4 = 0 \]
\[ x + y = 1 \]

Now substitute \( y = 1 - x \) into the first equation:

\[ 8x + 4(1 - x) - 16 = 0 \]
\[ 8x + 4 - 4x - 16 = 0 \]
\[ 4x - 12 = 0 \]
\[ 4x = 12 \]
\[ x = 3 \]

Substitute \( x = 3 \) back into \( x + y = 1 \):

\[ 3 + y = 1 \]
\[ y = 1 - 3 \]
\[ y = -2 \]

So, the critical point is \( (3, -2) \).

Next, we need to check if this critical point is a minimum. We use the second partial derivatives to find the Hessian matrix.

The second partial derivatives are:

\[ \frac{\partial^2 f}{\partial x^2} = 8 \]
\[ \frac{\partial^2 f}{\partial y^2} = 4 \]
\[ \frac{\partial^2 f}{\partial x \partial y} = 4 \]

The Hessian matrix \( H \) is:

\[ H = \begin{bmatrix}
8 & 4 \\
4 & 4 
\end{bmatrix} \]

The determinant of the Hessian matrix is:

\[ \det(H) = (8 \cdot 4) - (4 \cdot 4) = 32 - 16 = 16 \]

Since \( \det(H) > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), the function has a local minimum at \( (3, -2) \).

Therefore, the values of \( x \) and \( y \) that minimize the expression are:

\[ x = 3 \]
\[ y = -2 \]

Would you like further details or have any questions?

### Questions:

1. How do we determine if a critical point is a minimum, maximum, or saddle point using the Hessian matrix?
2. What is the role of the determinant of the Hessian matrix in optimization problems?
3. Can we use another method to find the minimum of this quadratic function?
4. How does the nature of the quadratic terms influence the shape of the function?
5. What happens if the Hessian matrix determinant is zero?
6. How can we verify our solution by substituting the values back into the original function?
7. Are there any real-world applications of finding minima in quadratic expressions?
8. What are the implications if \( \frac{\partial^2 f}{\partial x^2} \) were negative?

### Tip:

When dealing with optimization problems involving quadratic functions, always check the second partial derivatives and the Hessian matrix to confirm the nature of the critical points.

Learn how to find the values of x and y that minimize the quadratic expression 4x^2 + 2y^2 + 4xy - 16x - 4y + 23 using calculus. Explore the process of finding critical points, checking for minima using the Hessian matrix, and verifying solutions. Suitable for advanced undergraduate students and beyond.

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