Yes, we can minimize the quadratic expression without using derivatives by completing the square.

Given the expression:

$f(x, y) = 4x^2 + 2y^2 + 4xy - 16x - 4y + 23$

We will complete the square step-by-step.

Step 1: Group the $x$ and $y$ terms

$f(x, y) = 4x^2 + 4xy + 2y^2 - 16x - 4y + 23$

Step 2: Rewrite the quadratic terms involving $x$ and $y$

To complete the square for the terms involving both $x$ and $y$, we can combine and rearrange them as follows:

$4x^2 + 4xy + 2y^2 = 2(2x^2 + 2xy + y^2) = 2((2x + y)^2 - y^2 + y^2) = 2((2x + y)^2)$

Step 3: Rewrite the entire function with the completed square term

Rewrite the quadratic expression, including the linear terms:

$f(x, y) = 2((2x + y)^2) - 16x - 4y + 23$

Step 4: Isolate and simplify the linear terms

Next, isolate the linear terms to combine with the square term:

$f(x, y) = 2((2x + y)^2) - 16x - 4y + 23$

Since we have a mixed term $(2x + y)^2$, we need to rewrite the linear terms accordingly. Notice that:

$-16x = -8 \cdot 2x$ $-4y = -4y$

Therefore:

$-16x - 4y = -8(2x + y)$

Step 5: Combine and simplify

Now we need to incorporate these terms back into the expression:

$f(x, y) = 2((2x + y)^2) - 8(2x + y) + 23$

Let $u = 2x + y$. Then the expression becomes:

$f(u) = 2u^2 - 8u + 23$

Step 6: Complete the square for the new variable $u$

Complete the square for the quadratic expression in $u$:

$2u^2 - 8u + 23 = 2(u^2 - 4u) + 23$ $= 2(u^2 - 4u + 4 - 4) + 23$ $= 2((u - 2)^2 - 4) + 23$ $= 2(u - 2)^2 - 8 + 23$ $= 2(u - 2)^2 + 15$

Step 7: Identify the minimum value

The minimum value occurs at the vertex of the completed square term $2(u - 2)^2$, which is zero. Hence, the minimum value of $f(x, y)$ is when $(u - 2)^2 = 0$:

$f(u) = 15$

So the minimum value of $f(x, y)$ is $15$.

Step 8: Determine the values of $x$ and $y$

We know that $u = 2x + y$, and for the minimum value $u - 2 = 0$:

$u = 2$ $2x + y = 2$

Solve this equation for $x$ and $y$. We need another constraint to uniquely determine $x$ and $y$. But since we completed the square for the entire function, the minimum will always occur when $2x + y = 2$. Any $x$ and $y$ that satisfies this equation will minimize the function.

For example, one solution can be:

$x = 1$ $y = 0$

Thus, $x = 1$ and $y = 0$ minimize the expression, but there are infinitely many solutions along the line $2x + y = 2$.

Would you like further details or have any questions?

Questions:

1. How does completing the square help in minimizing quadratic expressions?
2. Can you apply the method of completing the square to higher-dimensional quadratic functions?
3. What is the geometric interpretation of the line $2x + y = 2$?
4. How do we handle quadratic expressions with more than two variables?
5. What are the advantages and disadvantages of completing the square versus using derivatives?
6. How can we verify our solution by substituting different values back into the original function?
7. Are there real-world scenarios where completing the square is particularly useful?
8. How would the solution change if the quadratic terms had different coefficients?

Tip:

When completing the square for functions of multiple variables, carefully group and rearrange terms to isolate mixed terms and simplify the expression systematically.