i have a problem
Give h1, h2, w1, w2, w3 >= 0
h1*w2 = 2
h2*w1 = 4
h2*w3 = 4
find minimum of (h1+h2)*(w1+w2+w3)
Can you solve with linear programming ?

Yes, we can solve this problem using linear programming. The goal is to minimize the expression \((h_1 + h_2) \cdot (w_1 + w_2 + w_3)\), given the constraints:

\[
h_1 \cdot w_2 = 2, \quad h_2 \cdot w_1 = 4, \quad h_2 \cdot w_3 = 4
\]

and

\[
h_1, h_2, w_1, w_2, w_3 \geq 0.
\]

### Reformulating the Problem

To solve this problem with linear programming, we'll introduce some new variables and constraints.

1. **Objective function**: We want to minimize \((h_1 + h_2) \cdot (w_1 + w_2 + w_3)\). Let's denote:
   \[
   z = (h_1 + h_2) \cdot (w_1 + w_2 + w_3).
   \]

2. **Constraints**:
   \[
   h_1 \cdot w_2 = 2, \quad h_2 \cdot w_1 = 4, \quad h_2 \cdot w_3 = 4.
   \]

These are not linear constraints, but we can transform this problem using the fact that we can express either \(h_i\) in terms of \(w_j\), or vice versa. Let's express each \(h_i\) in terms of \(w_i\) using the given equations:

\[
h_1 = \frac{2}{w_2}, \quad h_2 = \frac{4}{w_1}, \quad h_2 = \frac{4}{w_3}.
\]

From the two expressions for \(h_2\):

\[
\frac{4}{w_1} = \frac{4}{w_3} \implies w_1 = w_3.
\]

Thus, \(w_1 = w_3\), and we can rewrite \(h_2\) as:

\[
h_2 = \frac{4}{w_1}.
\]

Now, substitute these into the objective function:

\[
z = \left( \frac{2}{w_2} + \frac{4}{w_1} \right) \cdot (w_1 + w_2 + w_1).
\]

Simplify to:

\[
z = \left( \frac{2}{w_2} + \frac{4}{w_1} \right) \cdot (2w_1 + w_2).
\]

### Converting to a Linear Program

Since the objective function is not linear, we can't directly apply linear programming. Instead, let's find the critical points by minimizing this function using calculus methods.

#### Calculating the Derivatives

We will find the partial derivatives with respect to \(w_1\) and \(w_2\) and set them to zero to find critical points.

Let me calculate the partial derivatives and find the minimum.The solution to the partial derivatives yields \( w_1 = w_2 \). This suggests that for the minimum value, the optimal condition occurs when \( w_1 = w_2 \).

### Substituting \( w_1 = w_2 \) into the Objective Function

Let's substitute \( w_1 = w_2 \) into the expressions for \(h_1\), \(h_2\), and the objective function:

\[
h_1 = \frac{2}{w_2}, \quad h_2 = \frac{4}{w_1} = \frac{4}{w_2}.
\]

Thus,

\[
z = \left( \frac{2}{w_2} + \frac{4}{w_2} \right) \cdot (2w_2 + w_2) = \left( \frac{6}{w_2} \right) \cdot (3w_2) = 18.
\]

### Conclusion

The minimum value of \((h_1 + h_2) \cdot (w_1 + w_2 + w_3)\) is **18**, which occurs when:

\[
w_1 = w_2 = w_3, \quad h_1 = \frac{2}{w_2}, \quad h_2 = \frac{4}{w_2}.
\]

Would you like more details on the steps, or do you have any further questions?

### Related Questions:
1. What is linear programming and when is it used?
2. How do you handle non-linear constraints in optimization problems?
3. Can this problem be solved with different methods, like using Lagrange multipliers?
4. What are the properties of critical points in multivariable functions?
5. How does the choice of variables affect the solution in optimization problems?

**Tip:** Understanding how to convert non-linear constraints into linear ones can help simplify complex problems.

Learn how to minimize the expression (h1 + h2) * (w1 + w2 + w3) using linear programming techniques. Understand the constraints involving h1, h2, w1, w2, and w3 to find the optimal solution. Suitable for advanced high school or college-level students.

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