Questions: The output of an economic system Q, subject to two inputs, such as labor L and capital K, is often modeled by the Cobb-Douglas production function Q=c L^a K^b. When a+b=1, the case is called constant returns to scale. Suppose Q=1451, a=8/9, b=1/9, and c=72. Complete parts (a) and (b). a. Find the rate of change of capital with respect to labor, dK/dL. First substitute the given values of Q, a, b, and c into the production function.

Solution Steps

To find the rate of change of capital with respect to labor, $\frac{dK}{dL}$, we need to use implicit differentiation on the Cobb-Douglas production function $Q = c L^a K^b$. Given the values $Q = 1451$, $a = \frac{8}{9}$, $b = \frac{1}{9}$, and $c = 72$, we can substitute these into the function and then differentiate implicitly with respect to $L$.

1. Substitute the given values into the Cobb-Douglas production function.
2. Differentiate implicitly with respect to $L$.
3. Solve for $\frac{dK}{dL}$.

The Cobb-Douglas production function is given by: $$Q = c L^a K^b$$

We are given: $$Q = 1451$$ $$a = \frac{8}{9}$$ $$b = \frac{1}{9}$$ $$c = 72$$

Substituting these values into the production function, we get: $$1451 = 72 L^{\frac{8}{9}} K^{\frac{1}{9}}$$

To find the rate of change of capital with respect to labor, we first need to express $K$ in terms of $L$. Rearrange the equation to solve for $K$: $$1451 = 72 L^{\frac{8}{9}} K^{\frac{1}{9}}$$ $$\frac{1451}{72} = L^{\frac{8}{9}} K^{\frac{1}{9}}$$ $$\left( \frac{1451}{72} \right)^9 = \left( L^{\frac{8}{9}} K^{\frac{1}{9}} \right)^9$$ $$\left( \frac{1451}{72} \right)^9 = L^8 K$$ $$K = \frac{\left( \frac{1451}{72} \right)^9}{L^8}$$

Now, we differentiate $K$ with respect to $L$: $$K = \frac{\left( \frac{1451}{72} \right)^9}{L^8}$$ Let $C = \left( \frac{1451}{72} \right)^9$, then: $$K = \frac{C}{L^8}$$

Differentiate both sides with respect to $L$: $$\frac{dK}{dL} = \frac{d}{dL} \left( \frac{C}{L^8} \right)$$ $$\frac{dK}{dL} = C \cdot \frac{d}{dL} \left( L^{-8} \right)$$ $$\frac{dK}{dL} = C \cdot (-8) L^{-9}$$ $$\frac{dK}{dL} = -8C L^{-9}$$

Substitute back $C = \left( \frac{1451}{72} \right)^9$: $$\frac{dK}{dL} = -8 \left( \frac{1451}{72} \right)^9 L^{-9}$$

Final Answer