To solve this optimization problem using the method of Lagrange multipliers, follow these steps:

Step 1: Define the given functions

The given function to maximize or minimize is: $f(x, y) = x^2y$ The constraint is: $g(x, y) = x + y - 4 = 0$

Step 2: Set up the Lagrange function

The Lagrange function is: $\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda g(x, y)$ Substituting the given functions: $\mathcal{L}(x, y, \lambda) = x^2y - \lambda (x + y - 4)$

Step 3: Compute partial derivatives

We take the partial derivatives of $\mathcal{L}$ with respect to $x$, $y$, and $\lambda$, and set them equal to zero:

Partial derivatives:

1. $\frac{\partial \mathcal{L}}{\partial x} = 2xy - \lambda = 0$
2. $\frac{\partial \mathcal{L}}{\partial y} = x^2 - \lambda = 0$
3. $\frac{\partial \mathcal{L}}{\partial \lambda} = -(x + y - 4) = 0$

Step 4: Solve the system of equations

From the derivatives, we have the following equations:

1. $2xy = \lambda$
2. $x^2 = \lambda$
3. $x + y = 4$

Substitution:

From Equation (2), $\lambda = x^2$. Substituting into Equation (1): $2xy = x^2$ Simplify: $2y = x$ Thus: $x = 2y$

Use the constraint $x + y = 4$:

Substitute $x = 2y$ into $x + y = 4$: $2y + y = 4$ $3y = 4 \quad \implies \quad y = \frac{4}{3}$ Substitute $y = \frac{4}{3}$ into $x = 2y$: $x = 2 \cdot \frac{4}{3} = \frac{8}{3}$

Step 5: Critical points

The critical points are: $x = \frac{8}{3}, \quad y = \frac{4}{3}$

Would you like me to verify whether this is a maximum or minimum or solve for $f(x, y)$ at this point? Let me know if you'd like more details!

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Here are related questions to deepen your understanding:

1. What is the geometric interpretation of Lagrange multipliers?
2. How do you verify whether a critical point is a maximum or minimum in constrained optimization?
3. What happens if there are multiple constraints in the optimization problem?
4. Can this method be extended to functions with three or more variables?
5. What role does $\lambda$ play in the method of Lagrange multipliers?

Tip: Always ensure the constraint equation is satisfied when using Lagrange multipliers, as it directly impacts the critical points.