Yes, the Lagrange multiplier method can be used to solve this problem. Here's how we approach it:

Problem Restatement:

You want to minimize the objective function:

$f(w_1, w_2, h_1, h_2) = (w_1 + w_2)(h_1 + h_2)$

subject to the constraint:

$g(w_2, h_2) = w_2 h_2 - 4 = 0$

with the conditions $w_1, w_2, h_1, h_2 \geq 0$.

Step 1: Form the Lagrange function

To apply the method of Lagrange multipliers, we define the Lagrange function $\mathcal{L}$ as follows:

$\mathcal{L}(w_1, w_2, h_1, h_2, \lambda) = (w_1 + w_2)(h_1 + h_2) + \lambda (w_2 h_2 - 4)$

where $\lambda$ is the Lagrange multiplier associated with the constraint.

Step 2: Find the partial derivatives

To minimize $\mathcal{L}$, we take the partial derivatives with respect to $w_1, w_2, h_1, h_2$, and $\lambda$, and set them equal to zero.

1. $\frac{\partial \mathcal{L}}{\partial w_1} = (h_1 + h_2) = 0$

   This implies: $h_1 + h_2 = 0 \quad \Rightarrow \quad h_1 = -h_2$

2. $\frac{\partial \mathcal{L}}{\partial w_2} = (h_1 + h_2) + \lambda h_2 = 0$

   Using $h_1 = -h_2$ in this equation, we get: $-h_2 + \lambda h_2 = 0 \quad \Rightarrow \quad \lambda = 1$

3. $\frac{\partial \mathcal{L}}{\partial h_1} = (w_1 + w_2) = 0$

   This implies: $w_1 + w_2 = 0 \quad \Rightarrow \quad w_1 = -w_2$

4. $\frac{\partial \mathcal{L}}{\partial h_2} = (w_1 + w_2) + \lambda w_2 = 0$

   Using $w_1 = -w_2$, this simplifies to: $-w_2 + \lambda w_2 = 0 \quad \Rightarrow \quad \lambda = 1$

5. $\frac{\partial \mathcal{L}}{\partial \lambda} = w_2 h_2 - 4 = 0$

   This is the original constraint, so: $w_2 h_2 = 4$

Step 3: Solve the system of equations

From the first two conditions, we get:

$w_1 = -w_2 \quad \text{and} \quad h_1 = -h_2$

However, since $w_1, w_2, h_1, h_2 \geq 0$, this leads to a contradiction because $w_1$ and $h_1$ cannot be negative.

Thus, this indicates that $w_1$ and $h_1$ must be zero. Therefore, the solution is:

$w_1 = 0, \quad h_1 = 0$

and from the constraint $w_2 h_2 = 4$, we can choose any pair of $w_2$ and $h_2$ such that their product equals 4.

Step 4: Minimize the objective function

Now, substituting $w_1 = 0$ and $h_1 = 0$ into the objective function, we have:

$f(w_1, w_2, h_1, h_2) = w_2 h_2$

But since $w_2 h_2 = 4$, the objective value is minimized when this constraint is satisfied. Hence, the minimum value of the function is 4.

Final Answer:

The solution that minimizes $(w_1 + w_2)(h_1 + h_2)$ subject to the constraint $w_2 h_2 = 4$ is:

$w_1 = 0, \quad h_1 = 0, \quad w_2 h_2 = 4$

Thus, the minimum value of the objective function is 4.

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Would you like more details, or do you have any questions?

Here are 5 related questions for further exploration:

1. How would the solution change if $w_1$ and $h_1$ were not constrained to be non-negative?
2. Can this problem be extended to other constraint types, such as $w_2 h_2 \geq 4$?
3. How does the method of Lagrange multipliers compare to other optimization techniques like gradient descent for this problem?
4. What would happen if you imposed a second constraint, such as $w_2 + h_2 = 10$?
5. Can this problem be solved using quadratic programming instead of Lagrange multipliers?

Tip: Always check whether a constraint's solution leads to contradictions in your system of equations, especially when there are non-negativity conditions involved.