Suppose k, l, and m grow at constant rates given, respectively, by , , and . What is the growth rate of y = mk1/3 l2/3?

1. Break Down the Problem

We are given a function $y = m k^{1/3} l^{2/3}$ and need to find the growth rate of $y$ based on the growth rates of $m$, $k$, and $l$.

2. Relevant Concepts

To find the growth rate of $y$, we can use the concept of logarithmic differentiation or the total differential. The growth rate of a function based on its variables can often be expressed as:

$\frac{dy}{dt} = \frac{\partial y}{\partial m} \frac{dm}{dt} + \frac{\partial y}{\partial k} \frac{dk}{dt} + \frac{\partial y}{\partial l} \frac{dl}{dt}$

This formula will help us break down the growth of $y$ in terms of the growth rates of $m$, $k$, and $l$.

3. Analysis and Detail

We need to compute the following partial derivatives:

1. $\frac{\partial y}{\partial m} = k^{1/3} l^{2/3}$
2. $\frac{\partial y}{\partial k} = \frac{1}{3} m k^{-2/3} l^{2/3}$
3. $\frac{\partial y}{\partial l} = \frac{2}{3} m k^{1/3} l^{-1/3}$

Now, if we denote the growth rates as follows:

• $\frac{dm}{dt} = g_m$
• $\frac{dk}{dt} = g_k$
• $\frac{dl}{dt} = g_l$

Plugging these into our derivative expression gives us:

$\frac{dy}{dt} = k^{1/3} l^{2/3} g_m + \frac{1}{3} m k^{-2/3} l^{2/3} g_k + \frac{2}{3} m k^{1/3} l^{-1/3} g_l$

4. Verify and Summarize

This total derivative now contains the growth rates of $y$ expressed in terms of $g_m$, $g_k$, and $g_l$. To find the overall growth rate of $y$, we can divide $\frac{dy}{dt}$ by $y$ itself.

The growth rate of $y$ can be expressed in terms of the overall growth rates of $m$, $k$, and $l$:

$\text{Growth Rate of } y = \frac{\frac{dy}{dt}}{y}$

Final Answer

Thus, the overall growth rate of $y$ based on the growth rates $g_m$, $g_k$, and $g_l$ is:

$\text{Growth Rate of } y = \frac{g_m + \frac{1}{3} \frac{m k^{-2/3} l^{2/3}}{k^{1/3} l^{2/3}} g_k + \frac{2}{3} \frac{m k^{1/3} l^{-1/3}}{k^{1/3} l^{2/3}} g_l}{1}$

This can be simplified further based on specific values for $g_m$, $g_k$, and $g_l$.