To solve for the variables in the given equation:

\[
L = 0.8 \log X + 0.2 \log Y + \lambda (5000 - 4X - 2Y)
\]

We can approach this in multiple ways depending on the goal, such as maximizing or minimizing \( L \) with respect to \( X \), \( Y \), and \( \lambda \), or solving for a specific variable. This equation involves a Lagrange multiplier \( \lambda \), which suggests we are dealing with an optimization problem subject to the constraint \( 5000 - 4X - 2Y = 0 \).

### Steps for finding the maximum or minimum:

1. **Formulate the Lagrange equations**:
   To solve this, we differentiate \( L \) with respect to \( X \), \( Y \), and \( \lambda \), and set the derivatives equal to zero (first-order necessary conditions for optimality). 

   - Differentiate \( L \) with respect to \( X \):
     \[
     \frac{\partial L}{\partial X} = \frac{0.8}{X \ln 10} - 4\lambda = 0
     \]
     
   - Differentiate \( L \) with respect to \( Y \):
     \[
     \frac{\partial L}{\partial Y} = \frac{0.2}{Y \ln 10} - 2\lambda = 0
     \]

   - Differentiate \( L \) with respect to \( \lambda \):
     \[
     \frac{\partial L}{\partial \lambda} = 5000 - 4X - 2Y = 0
     \]

2. **Solve the system of equations**:
   - From the first equation: 
     \[
     \lambda = \frac{0.8}{4X \ln 10}
     \]
   - From the second equation:
     \[
     \lambda = \frac{0.2}{2Y \ln 10}
     \]
   - Set the two expressions for \( \lambda \) equal:
     \[
     \frac{0.8}{4X \ln 10} = \frac{0.2}{2Y \ln 10}
     \]
     Simplify:
     \[
     \frac{0.8}{4X} = \frac{0.2}{2Y} \quad \Rightarrow \quad \frac{1}{X} = \frac{1}{Y} \quad \Rightarrow \quad X = Y
     \]

3. **Substitute \( X = Y \) into the constraint**:
   Use the constraint equation:
   \[
   5000 - 4X - 2Y = 0
   \]
   Substituting \( X = Y \):
   \[
   5000 - 4X - 2X = 0 \quad \Rightarrow \quad 5000 = 6X \quad \Rightarrow \quad X = \frac{5000}{6} \approx 833.33
   \]
   Since \( X = Y \), \( Y = 833.33 \).

Thus, the solution is:
\[
X = Y = 833.33
\]

### Summary:
The optimal values for \( X \) and \( Y \) that satisfy the equation are approximately \( 833.33 \).

---

Would you like further details or have any questions on this? Here are some related questions you might find helpful:

1. How does the Lagrange multiplier method work in optimization problems?
2. Can we solve for \( X \) and \( Y \) if different constraints are applied?
3. What happens if we change the logarithmic coefficients (0.8 and 0.2)?
4. How would the solution change if \( \lambda \) had a specific value?
5. How does the log function affect the behavior of the solution?

**Tip**: Always check the constraint equation after solving, as it helps confirm if the derived values satisfy the original problem condition.

L = 0.8logX + 0.2logY + λ(5000 - 4X - 2Y)

Learn how to solve an optimization problem using logarithms and Lagrange multipliers. The problem involves maximizing or minimizing L = 0.8logX + 0.2logY + λ(5000 - 4X - 2Y). Discover key steps, including forming Lagrange equations, differentiating with respect to X, Y, and λ, and solving for X = Y = 833.33. Suitable for university students studying advanced calculus and optimization.

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