We have the expression:

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

This appears to be a function that includes logarithmic terms and a Lagrange multiplier term, where $\lambda$ is a multiplier and $X$ and $Y$ are variables. The term $(5000 - 4X - 2Y)$ might represent a constraint, possibly indicating optimization with constraints (a common setting in Lagrange multiplier problems).

If you're trying to optimize $L$, such as maximizing or minimizing it, you'd likely take partial derivatives with respect to $X$, $Y$, and $\lambda$, then solve the resulting system of equations.

Partial Derivatives:

1. With respect to $X$: $\frac{\partial L}{\partial X} = \frac{0.8}{X} - 4\lambda$

2. With respect to $Y$: $\frac{\partial L}{\partial Y} = \frac{0.2}{Y} - 2\lambda$

3. With respect to $\lambda$: $\frac{\partial L}{\partial \lambda} = 5000 - 4X - 2Y$

Solving these equations will give the values for $X$, $Y$, and $\lambda$.

• Set the partial derivatives equal to zero to find the critical points: $\frac{0.8}{X} = 4\lambda$ $\frac{0.2}{Y} = 2\lambda$ $5000 - 4X - 2Y = 0$

From here, you can solve for $X$, $Y$, and $\lambda$.

Would you like to proceed with solving for these values? Or do you need a detailed explanation of any particular step?

---

Here are 5 related questions that could expand your understanding:

1. What is the role of the Lagrange multiplier $\lambda$ in optimization problems?
2. How do you solve systems of nonlinear equations?
3. What is the significance of logarithmic functions in optimization problems?
4. How do constraints affect optimization solutions?
5. What are the practical applications of Lagrange multipliers?

Tip: When solving constrained optimization problems, always check if the constraint is active or inactive by evaluating it at the critical points.