To solve for the marginal productivity functions, we need to calculate the partial derivatives of $P = 100l^{0.76}k^{0.24}$ with respect to $l$ and $k$.

Step 1: Partial derivative with respect to $l$

The formula for $P$ is: $P = 100l^{0.76}k^{0.24}$ Differentiate $P$ with respect to $l$, treating $k$ as a constant: $\frac{\partial P}{\partial l} = \frac{\partial}{\partial l} \left( 100l^{0.76}k^{0.24} \right)$ $\frac{\partial P}{\partial l} = 100 \cdot 0.76 \cdot l^{0.76 - 1} \cdot k^{0.24}$ $\frac{\partial P}{\partial l} = 76l^{-0.24}k^{0.24}$

Step 2: Partial derivative with respect to $k$

Now, differentiate $P$ with respect to $k$, treating $l$ as a constant: $\frac{\partial P}{\partial k} = \frac{\partial}{\partial k} \left( 100l^{0.76}k^{0.24} \right)$ $\frac{\partial P}{\partial k} = 100 \cdot 0.24 \cdot l^{0.76} \cdot k^{0.24 - 1}$ $\frac{\partial P}{\partial k} = 24l^{0.76}k^{-0.76}$

Final Results

The marginal productivity functions are:

1. Marginal productivity with respect to $l$: $\frac{\partial P}{\partial l} = 76l^{-0.24}k^{0.24}$
2. Marginal productivity with respect to $k$: $\frac{\partial P}{\partial k} = 24l^{0.76}k^{-0.76}$

Would you like a detailed step-by-step explanation of any part of the solution? Here are related questions to deepen your understanding:

1. How do the marginal productivity functions change with varying $l$ and $k$?
2. What does the negative exponent in $l^{-0.24}$ and $k^{-0.76}$ imply for the behavior of productivity?
3. Can you explain the economic interpretation of the exponents $0.76$ and $0.24$ in this production function?
4. How would the marginal productivity functions behave if $l$ or $k$ approaches zero?
5. How can this production function be applied to measure returns to scale?

Tip: Always consider the behavior of the partial derivatives when analyzing marginal effects in production economics, especially for values close to zero or infinity.