You have presented the following expression:

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

This appears to be part of an optimization problem where $\lambda$ represents a Lagrange multiplier, and $X$ and $Y$ are variables involved in the constraint function $5000 - 4X - 2Y = 0$.

To find the optimal values of $X$ and $Y$, we can follow these steps:

1. Set up the Lagrange function: $\mathcal{L}(X, Y, \lambda) = 0.8 \log X + 0.2 \log Y + \lambda (5000 - 4X - 2Y)$

2. Take the partial derivatives of $\mathcal{L}$ with respect to $X$, $Y$, and $\lambda$, and set them equal to zero:

   • Partial derivative with respect to $X$: $\frac{\partial \mathcal{L}}{\partial X} = \frac{0.8}{X} - 4\lambda = 0$

   • Partial derivative with respect to $Y$: $\frac{\partial \mathcal{L}}{\partial Y} = \frac{0.2}{Y} - 2\lambda = 0$

   • Partial derivative with respect to $\lambda$ (this gives the constraint): $\frac{\partial \mathcal{L}}{\partial \lambda} = 5000 - 4X - 2Y = 0$

3. Solve the system of equations:

   From the first equation: $\frac{0.8}{X} = 4\lambda \quad \Rightarrow \quad \lambda = \frac{0.8}{4X}$

   From the second equation: $\frac{0.2}{Y} = 2\lambda \quad \Rightarrow \quad \lambda = \frac{0.2}{2Y} = \frac{0.1}{Y}$

   Now, set the expressions for $\lambda$ equal to each other: $\frac{0.8}{4X} = \frac{0.1}{Y} \quad \Rightarrow \quad \frac{0.2}{X} = \frac{0.1}{Y}$

   This simplifies to: $2Y = X \quad \Rightarrow \quad X = 2Y$

4. Substitute into the constraint: From the constraint $5000 - 4X - 2Y = 0$, substitute $X = 2Y$: $5000 - 4(2Y) - 2Y = 0 \quad \Rightarrow \quad 5000 - 8Y - 2Y = 0$ $5000 = 10Y \quad \Rightarrow \quad Y = 500$ Substituting $Y = 500$ into $X = 2Y$: $X = 2(500) = 1000$

Thus, the optimal values are $X = 1000$ and $Y = 500$.

Would you like more details or have any questions? Here are some related problems:

1. How would the results change if the coefficients in the constraint were different?
2. What is the role of the Lagrange multiplier in this optimization?
3. How would this problem look in a different logarithmic base, such as natural logarithms?
4. Could this system be solved graphically? What insights could that provide?
5. What would happen if $X$ and $Y$ had boundary conditions, like non-negativity constraints?

Tip: Always ensure that the values of variables $X$ and $Y$ satisfy the constraint function after solving the system.