Certainly! The problem is to minimize the objective function $(w_1 + w_2 + w_3)(h_1 + h_2)$, subject to the constraints:

1. $w_1 h_2 = 4$
2. $w_2 h_2 = 4$
3. $w_3 h_1 = 2$

We will use the method of Lagrange multipliers to solve this. Here's the step-by-step approach.

1. Define the Objective Function

The objective function is: $f(w_1, w_2, w_3, h_1, h_2) = (w_1 + w_2 + w_3)(h_1 + h_2)$

2. Define the Constraints

The constraints can be written as: $g_1(w_1, h_2) = w_1 h_2 - 4 = 0$ $g_2(w_2, h_2) = w_2 h_2 - 4 = 0$ $g_3(w_3, h_1) = w_3 h_1 - 2 = 0$

We introduce Lagrange multipliers $\lambda_1, \lambda_2, \lambda_3$ for each of these constraints.

3. Define the Lagrange Function

The Lagrange function is: $\mathcal{L}(w_1, w_2, w_3, h_1, h_2, \lambda_1, \lambda_2, \lambda_3) = (w_1 + w_2 + w_3)(h_1 + h_2) + \lambda_1 (w_1 h_2 - 4) + \lambda_2 (w_2 h_2 - 4) + \lambda_3 (w_3 h_1 - 2)$

4. Compute the Partial Derivatives

We now take partial derivatives of $\mathcal{L}$ with respect to all the variables $w_1, w_2, w_3, h_1, h_2$, and the Lagrange multipliers.

Partial derivatives with respect to $w_1, w_2, w_3$:

$\frac{\partial \mathcal{L}}{\partial w_1} = h_1 + h_2 + \lambda_1 h_2 = 0$ $\frac{\partial \mathcal{L}}{\partial w_2} = h_1 + h_2 + \lambda_2 h_2 = 0$ $\frac{\partial \mathcal{L}}{\partial w_3} = h_1 + h_2 + \lambda_3 h_1 = 0$

Partial derivatives with respect to $h_1, h_2$:

$\frac{\partial \mathcal{L}}{\partial h_1} = w_1 + w_2 + w_3 + \lambda_3 w_3 = 0$ $\frac{\partial \mathcal{L}}{\partial h_2} = w_1 + w_2 + w_3 + \lambda_1 w_1 + \lambda_2 w_2 = 0$

Partial derivatives with respect to $\lambda_1, \lambda_2, \lambda_3$:

These give us back the constraints: $\frac{\partial \mathcal{L}}{\partial \lambda_1} = w_1 h_2 - 4 = 0$ $\frac{\partial \mathcal{L}}{\partial \lambda_2} = w_2 h_2 - 4 = 0$ $\frac{\partial \mathcal{L}}{\partial \lambda_3} = w_3 h_1 - 2 = 0$

5. Solve the System of Equations

Now we have a system of equations from the partial derivatives. We solve these equations to find the optimal values of $w_1, w_2, w_3, h_1, h_2$.

Let me solve these for you.The system of equations has two possible solutions:

1. $w_1 = -w_3, w_2 = -w_3, w_3 = w_3, h_1 = \frac{2}{w_3}, h_2 = -\frac{4}{w_3}$
2. $w_1 = w_3, w_2 = w_3, w_3 = w_3, h_1 = \frac{2}{w_3}, h_2 = \frac{4}{w_3}$

Since $w_1, w_2, w_3, h_1, h_2 \geq 0$ are required by the problem, the first solution is not valid due to negative terms.

Thus, the valid solution is: $w_1 = w_2 = w_3 = w_3, \quad h_1 = \frac{2}{w_3}, \quad h_2 = \frac{4}{w_3}$

Minimizing the Objective Function

Now that we have the values of $w_1, w_2, w_3, h_1, h_2$, the objective function becomes:

= 3w_3 \left( \frac{2}{w_3} + \frac{4}{w_3} \right) = 3w_3 \cdot \frac{6}{w_3} = 18$$ Thus, the minimum value of the objective function is $$18$$. Would you like to discuss any specific details or have further questions on this solution? Here are 5 follow-up questions: 1. Can Lagrange multipliers handle inequalities directly? 2. How would this method change with more than 3 constraints? 3. What happens if one of the constraints becomes nonlinear? 4. How does this solution change if some variables are strictly positive? 5. How do Lagrange multipliers relate to gradient-based optimization? **Tip:** Lagrange multipliers are useful for constrained optimization problems, but make sure to always check the validity of solutions based on the constraints!