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To solve this problem using the method of Lagrange multipliers, we need to maximize or minimize the function \( f(x, y) = x^2 + 4xy \) subject to the constraint \( g(x, y) = x + y - 145 = 0 \).

### Step 1: Set Up the Lagrange Function
Define the Lagrange function:
\[
\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda \cdot g(x, y)
\]
Substitute \( f(x, y) = x^2 + 4xy \) and \( g(x, y) = x + y - 145 \):
\[
\mathcal{L}(x, y, \lambda) = x^2 + 4xy - \lambda (x + y - 145)
\]

### Step 2: Take the Partial Derivatives
We’ll take partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), and \( \lambda \) and set them to zero.

1. **Partial derivative with respect to \( x \):**
   \[
   \frac{\partial \mathcal{L}}{\partial x} = 2x + 4y - \lambda = 0
   \]

2. **Partial derivative with respect to \( y \):**
   \[
   \frac{\partial \mathcal{L}}{\partial y} = 4x - \lambda = 0
   \]

3. **Partial derivative with respect to \( \lambda \):**
   \[
   \frac{\partial \mathcal{L}}{\partial \lambda} = -(x + y - 145) = 0
   \]
   which simplifies to the constraint:
   \[
   x + y = 145
   \]

### Step 3: Solve the System of Equations
We now have the following system of equations:
1. \( 2x + 4y - \lambda = 0 \)
2. \( 4x - \lambda = 0 \)
3. \( x + y = 145 \)

From equation (2), we find:
\[
\lambda = 4x
\]
Substitute \( \lambda = 4x \) into equation (1):
\[
2x + 4y - 4x = 0
\]
This simplifies to:
\[
-2x + 4y = 0 \quad \Rightarrow \quad 2y = x
\]

Now substitute \( x = 2y \) into the constraint \( x + y = 145 \):
\[
2y + y = 145 \quad \Rightarrow \quad 3y = 145 \quad \Rightarrow \quad y = \frac{145}{3}
\]
Then substitute \( y = \frac{145}{3} \) back into \( x = 2y \):
\[
x = 2 \cdot \frac{145}{3} = \frac{290}{3}
\]

### Step 4: Calculate \( f(x, y) \)
Now that we have \( x = \frac{290}{3} \) and \( y = \frac{145}{3} \), substitute these values into \( f(x, y) = x^2 + 4xy \):
\[
f\left(\frac{290}{3}, \frac{145}{3}\right) = \left(\frac{290}{3}\right)^2 + 4 \cdot \frac{290}{3} \cdot \frac{145}{3}
\]
Calculate each term separately:

1. \( \left(\frac{290}{3}\right)^2 = \frac{84100}{9} \)
2. \( 4 \cdot \frac{290}{3} \cdot \frac{145}{3} = \frac{168200}{9} \)

So,
\[
f\left(\frac{290}{3}, \frac{145}{3}\right) = \frac{84100}{9} + \frac{168200}{9} = \frac{252300}{9} = 28033.\overline{3}
\]

### Final Answer
The optimal value of \( f(x, y) \) is:
\[
\boxed{28033.\overline{3}}
\]

Would you like more details or have any questions?

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Here are some related questions that might help deepen your understanding:
1. How would you handle a different constraint equation, like \( x^2 + y^2 = 145 \)?
2. What is the significance of the Lagrange multiplier \( \lambda \) in optimization problems?
3. Can the Lagrange multiplier method find saddle points?
4. How would you interpret the solution if the constraint was a strict inequality?
5. What happens if there are no feasible points that satisfy both \( f(x, y) \) and \( g(x, y) \)?

**Tip:** In problems involving Lagrange multipliers, always remember to double-check if the constraint leads to real and feasible solutions.

Use Lagrange multipliers to find the maximum and minimum values of f(x, y) = x^2 + 4xy subject to the constraint x + y = 145. If such values exist, enter the exact answer. If there is no global maximum or global minimum, enter NA.

Solve for the maximum and minimum values of the function f(x, y) = x^2 + 4xy using Lagrange multipliers, constrained by x + y = 145. This problem involves calculus concepts suitable for undergraduate students.

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