To solve the given optimization problem using Lagrange multipliers, let's carefully set it up step by step.

Problem Setup

We are given the constraints: $w_1h_2 = 4, \quad w_2h_2 = 4, \quad w_3h_1 = 2,$ and the objective function to minimize: $f(w_1, w_2, w_3, h_1, h_2) = (w_1 + w_2 + w_3)(h_1 + h_2).$

This can be written as the product of two linear terms.

Step 1: Define the objective and constraint functions

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

Now, express the three constraints as functions: $g_1(w_1, h_2) = w_1h_2 - 4 = 0, \quad g_2(w_2, h_2) = w_2h_2 - 4 = 0, \quad g_3(w_3, h_1) = w_3h_1 - 2 = 0.$

Step 2: Set up the Lagrange function

The Lagrange function $\mathcal{L}$ is formed by incorporating the objective function and the constraints using Lagrange multipliers $\lambda_1$, $\lambda_2$, and $\lambda_3$: $\mathcal{L}(w_1, w_2, w_3, h_1, h_2, \lambda_1, \lambda_2, \lambda_3) = (w_1 + w_2 + w_3)(h_1 + h_2) + \lambda_1(w_1h_2 - 4) + \lambda_2(w_2h_2 - 4) + \lambda_3(w_3h_1 - 2).$

Step 3: Compute partial derivatives

We now take the partial derivatives of $\mathcal{L}$ with respect to each variable $w_1, w_2, w_3, h_1, h_2$, and each $\lambda_i$, and set them equal to zero.

1. $\frac{\partial \mathcal{L}}{\partial w_1} = (h_1 + h_2) + \lambda_1 h_2 = 0$
2. $\frac{\partial \mathcal{L}}{\partial w_2} = (h_1 + h_2) + \lambda_2 h_2 = 0$
3. $\frac{\partial \mathcal{L}}{\partial w_3} = (h_1 + h_2) + \lambda_3 h_1 = 0$
4. $\frac{\partial \mathcal{L}}{\partial h_1} = (w_1 + w_2 + w_3) + \lambda_3 w_3 = 0$
5. $\frac{\partial \mathcal{L}}{\partial h_2} = (w_1 + w_2 + w_3) + \lambda_1 w_1 + \lambda_2 w_2 = 0$

For the constraints: $w_1h_2 = 4, \quad w_2h_2 = 4, \quad w_3h_1 = 2.$

Step 4: Solve the system of equations

To minimize the function, we now solve the system of equations formed by these partial derivatives and the constraint equations.

Let me calculate the result step by step.The solutions to the system of equations are:

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

Interpretation:

• The first solution gives negative values for $w_1, w_2, h_2$, which is not valid since $w_1, w_2, w_3, h_1, h_2 \geq 0$.
• The second solution is valid with positive values, meaning: $w_1 = w_2 = w_3, \quad h_1 = \frac{2}{w_3}, \quad h_2 = \frac{4}{w_3}.$

Step 5: Minimize the objective function

Now, substitute these values into the objective function: $f(w_1, w_2, w_3, h_1, h_2) = (w_1 + w_2 + w_3)(h_1 + h_2) = 3w_3 \left( \frac{2}{w_3} + \frac{4}{w_3} \right).$

This simplifies to: $f = 3w_3 \times \frac{6}{w_3} = 18.$

Thus, the minimum value of $(w_1 + w_2 + w_3)(h_1 + h_2)$ is 18.

Would you like further clarification or details?

Here are some related questions to consider:

1. Can you explain the role of the Lagrange multipliers in the solution?
2. How would the solution change if the constraints were different?
3. Could the objective function be minimized using a different method?
4. How can you interpret the geometric meaning of this result?
5. What happens if the non-negativity condition is relaxed?

Tip: Lagrange multipliers help solve constrained optimization by linking gradients of the objective function and the constraints.