give answer in 2 step with explanation at the end of each step and final answer at the end: question steps for3. Suppose that R = R(q1. 2). where g; and g; are themselves functions of k and I.
Write

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:  question steps for3. Suppose that R = R(q1. 2). where g; and g; are themselves functions of k and I.
Write down the chain rule formulae for 9R/dk and OR/0l. Evaluate OR/0k and OR/l as functions
of k and I when

Raa) =a. alkl)=3k+4 and gk0) =5k+6L
4. (a) Caleulate dy/dz at the point (zy) = (1,1) when y is the function of x defined implicitly by
the equation 2? + 2zy + 3y* = 6.
(b) Using an extension of the method described in the notes and the book, find dg/dk and dq/dl at
the point where (k,I) = (1,1) when g, k and I are related by the equation ¢’k + k*I + gk*l = 3.
5. Show that the Cobb-Douglas production function

q(k,0) = AKI" (A and a are constants with 0 < a < 1)

is homogeneous of degree 1.
Show that the marginal product of labour, u is positive and decreases as I increases [for fixed &].
Verify Euler's theorem and explain its significance in terms of the distribution of the total product
amongst capital and labour.

VIPSolutions · Accepted Answer

Let's solve **Question 3** step-by-step as you requested:

---

### **Step 1: Write Down the Chain Rule Formulae**

Suppose $ R = R(q_1, q_2) $ where $ q_1 = q_1(k, l) $ and $ q_2 = q_2(k, l) $.

By the multivariable chain rule:
$$
\frac{\partial R}{\partial k} = \frac{\partial R}{\partial q_1} \frac{\partial q_1}{\partial k} + \frac{\partial R}{\partial q_2} \frac{\partial q_2}{\partial k}
$$
$$
\frac{\partial R}{\partial l} = \frac{\partial R}{\partial q_1} \frac{\partial q_1}{\partial l} + \frac{\partial R}{\partial q_2} \frac{\partial q_2}{\partial l}
$$

**Explanation:**  
The chain rule allows us to compute the derivatives of $ R $ with respect to $ k $ and $ l $ by considering how $ R $ depends on $ q_1 $ and $ q_2 $, which in turn depend on $ k $ and $ l $.

---

### **Step 2: Find the Partial Derivatives**

Given:
- $ R(q_1, q_2) = q_1^2 q_2 $
- $ q_1(k, l) = 3k + 4l $
- $ q_2(k, l) = 5k + 6l $

First, compute the required derivatives:
- $ \frac{\partial R}{\partial q_1} = 2q_1 q_2 $
- $ \frac{\partial R}{\partial q_2} = q_1^2 $
- $ \frac{\partial q_1}{\partial k} = 3 $, $ \frac{\partial q_1}{\partial l} = 4 $
- $ \frac{\partial q_2}{\partial k} = 5 $, $ \frac{\partial q_2}{\partial l} = 6 $

Now plug into the chain rule:

#### For $ \frac{\partial R}{\partial k} $:
$$
\frac{\partial R}{\partial k} = \frac{\partial R}{\partial q_1} \frac{\partial q_1}{\partial k} + \frac{\partial R}{\partial q_2} \frac{\partial q_2}{\partial k}
= [2q_1 q_2](3) + [q_1^2](5)
= 6q_1 q_2 + 5q_1^2
$$

#### For $ \frac{\partial R}{\partial l} $:
$$
\frac{\partial R}{\partial l} = \frac{\partial R}{\partial q_1} \frac{\partial q_1}{\partial l} + \frac{\partial R}{\partial q_2} \frac{\partial q_2}{\partial l}
= [2q_1 q_2](4) + [q_1^2](6)
= 8q_1 q_2 + 6q_1^2
$$

**Explanation:**  
We substituted all the partial derivatives into the chain rule expressions to get the final forms of $ \frac{\partial R}{\partial k} $ and $ \frac{\partial R}{\partial l} $ as functions of $ q_1 $ and $ q_2 $, which themselves are functions of $ k $ and $ l $.

---

### **Final Answers**

$$
\boxed{
\frac{\partial R}{\partial k} = 6q_1 q_2 + 5q_1^2
}
$$
$$
\boxed{
\frac{\partial R}{\partial l} = 8q_1 q_2 + 6q_1^2
}
$$

where $ q_1 = 3k + 4l $ and $ q_2 = 5k + 6l $.

VIPSolutions · Answer

Let's solve **Question 4** step-by-step as you requested:

---

### **Step 1: Calculate $\frac{dy}{dz}$**

We start with the implicit equation:
$$
2z^2 + 2zy + 3y^2 = 6
$$

To find $\frac{dy}{dz}$, we will differentiate both sides with respect to $z$:
$$
\frac{d}{dz}(2z^2) + \frac{d}{dz}(2zy) + \frac{d}{dz}(3y^2) = \frac{d}{dz}(6)
$$

Calculating each term:
- $\frac{d}{dz}(2z^2) = 4z$
- Using the product rule on $2zy$: 
  $$
  2 \left( z \frac{dy}{dz} + y \right)
  $$
- For $3y^2$, using the chain rule:
  $$
  6y \frac{dy}{dz}
  $$

Putting it all together:
$$
4z + 2 \left( z \frac{dy}{dz} + y \right) + 6y \frac{dy}{dz} = 0
$$

Rearranging gives:
$$
(2z + 6y) \frac{dy}{dz} = -4z - 2y
$$
$$
\frac{dy}{dz} = \frac{-4z - 2y}{2z + 6y}
$$

**Explanation:**  
We applied implicit differentiation to find how $y$ changes with respect to $z$, leading to a formula for $\frac{dy}{dz}$.

---

### **Step 2: Evaluate at the Point (1, 1)**

Now we substitute $z = 1$ and $y = 1$ into our derivative:
$$
\frac{dy}{dz} = \frac{-4(1) - 2(1)}{2(1) + 6(1)} = \frac{-4 - 2}{2 + 6} = \frac{-6}{8} = -\frac{3}{4}
$$

**Final Answer for $\frac{dy}{dz}$**:
$$
\boxed{-\frac{3}{4}}
$$

---

### **Step 3: Find $dg/dk$ and $dq/dl$**

Given the equation:
$$
\phi(k, l, g) = k + k^2 l + g(k, l) = 3
$$

Differentiating implicitly with respect to $k$:
$$
1 + 2kl + \frac{dg}{dk} = 0 \implies \frac{dg}{dk} = -1 - 2kl
$$

Differentiating implicitly with respect to $l$:
$$
k + k^2 + \frac{dg}{dl} = 0 \implies \frac{dg}{dl} = -k - k^2
$$

**Explanation:**  
We used implicit differentiation to find how $g$ changes with respect to $k$ and $l$.

---

### **Step 4: Evaluate at the Point (1, 1)**

Substituting $k = 1$ and $l = 1$:
- For $\frac{dg}{dk}$:
$$
\frac{dg}{dk} = -1 - 2(1)(1) = -1 - 2 = -3
$$

- For $\frac{dg}{dl}$:
$$
\frac{dg}{dl} = -1 - 1^2 = -1 - 1 = -2
$$

**Final Answers for $dg/dk$ and $dg/dl$**:
$$
\boxed{\frac{dg}{dk} = -3}
$$
$$
\boxed{\frac{dg}{dl} = -2}
$$

---

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