Provide the complete calculated solution in a proper structured way from start to end and write the solution in the following format:

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1. Complete Solution:

Solve the problem step-by-ste

Question

Provide the complete calculated solution in a proper structured way from start to end and write the solution in the following format:


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1. Complete Solution:

Solve the problem step-by-step, showing all calculations clearly without skipping any steps.

Include formulas, substitution of values, and simplification in proper order.

Write units wherever applicable, ensuring all given data is first converted to standard SI units (if needed).

If the question has multiple subparts, label them clearly as (a), (b), (c), etc. and solve each part separately using the same structure.

After solving each subpart, briefly verify the answer to check its correctness.



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2. Explanation of Each and Every Part:

After each step, briefly explain why that step was taken or which rule, concept, or formula was applied.

For multiple subparts, provide explanations individually for each part.



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3. Final Conclusion Statement:

Clearly state the final answer for each subpart with correct units.

If there are multiple subparts, provide a separate conclusion for each and then an overall summary.

Make the final answer stand out by using bullet points or nu

VIPSolutions · Accepted Answer

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## 1. Complete Solution

### 1(a) **Find the total welfare using the market equilibrium condition**

#### **Step 1: Set up the equilibrium condition**

At equilibrium:  
Supply = Demand  
So, set the price equations equal:

$$
\ln(Q^3_S - 3) = 22e^{-\arctan(Q_D)}
$$

Since at equilibrium, $ Q_S = Q_D = Q^* $:

$$
\ln((Q^*)^3 - 3) = 22e^{-\arctan(Q^*)}
$$

---

#### **Step 2: Solve for $ Q^* $ using Newton's Method**

Let $ f(Q) = \ln(Q^3 - 3) - 22e^{-\arctan(Q)} = 0 $.

- Compute $ f(Q) $ and $ f'(Q) $:

$$
f(Q) = \ln(Q^3 - 3) - 22e^{-\arctan(Q)}
$$

$$
f'(Q) = \frac{3Q^2}{Q^3 - 3} + 22 e^{-\arctan(Q)} \cdot \frac{1}{1+Q^2}
$$

- Use Newton's iteration:

$$
Q_{n+1} = Q_n - \frac{f(Q_n)}{f'(Q_n)}
$$

Let’s start with an initial guess, say $ Q_0 = 2 $:

- $ Q_0 = 2 $:  
  $ Q^3 - 3 = 8 - 3 = 5 $  
  $ \ln(5) \approx 1.609 $  
  $ \arctan(2) \approx 1.107 $  
  $ e^{-1.107} \approx 0.331 $  
  $ 22 \times 0.331 \approx 7.282 $  
  $ f(2) = 1.609 - 7.282 = -5.673 $

- $ f'(2) = \frac{3 \times 4}{5} + 22 \times 0.331 \times \frac{1}{5} $  
  $ = 2.4 + 22 \times 0.0662 \approx 2.4 + 1.456 = 3.856 $

- Newton update:  
  $ Q_1 = 2 - \frac{-5.673}{3.856} \approx 2 + 1.472 = 3.472 $

- $ Q_1 = 3.472 $:  
  $ Q^3 - 3 \approx 41.85 - 3 = 38.85 $  
  $ \ln(38.85) \approx 3.661 $  
  $ \arctan(3.472) \approx 1.292 $  
  $ e^{-1.292} \approx 0.274 $  
  $ 22 \times 0.274 \approx 6.028 $  
  $ f(3.472) = 3.661 - 6.028 = -2.367 $

- $ f'(3.472) = \frac{3 \times (3.472)^2}{38.85} + 22 \times 0.274 \times \frac{1}{1+(3.472)^2} $  
  $ = \frac{3 \times 12.06}{38.85} + 22 \times 0.274 / 13.06 $  
  $ = \frac{36.18}{38.85} + 0.461 \approx 0.931 + 0.461 = 1.392 $

- Newton update:  
  $ Q_2 = 3.472 - \frac{-2.367}{1.392} \approx 3.472 + 1.701 = 5.173 $

- (Continue iterating or approximate from here; let's try one more step.)

- $ Q_2 = 5.173 $:  
  $ Q^3 - 3 \approx 138.5 - 3 = 135.5 $  
  $ \ln(135.5) \approx 4.910 $  
  $ \arctan(5.173) \approx 1.384 $  
  $ e^{-1.384} \approx 0.251 $  
  $ 22 \times 0.251 \approx 5.522 $  
  $ f(5.173) = 4.910 - 5.522 = -0.612 $

- $ f'(5.173) = \frac{3 \times 26.78}{135.5} + 22 \times 0.251 / (1+26.78) $  
  $ = \frac{80.33}{135.5} + 22 \times 0.251 / 27.78 $  
  $ = 0.593 + 0.199 = 0.792 $

- Newton update:  
  $ Q_3 = 5.173 - \frac{-0.612}{0.792} \approx 5.173 + 0.773 = 5.946 $

- Now check at $ Q = 5.946 $:

$ Q^3 - 3 \approx 210.1 - 3 = 207.1 $  
  $ \ln(207.1) \approx 5.333 $  
  $ \arctan(5.946) \approx 1.404 $  
  $ e^{-1.404} \approx 0.246 $  
  $ 22 \times 0.246 \approx 5.412 $  
  $ f(5.946) = 5.333 - 5.412 = -0.079 $

- $ f'(5.946) = \frac{3 \times 35.36}{207.1} + 22 \times 0.246 / (1+35.36) $  
  $ = \frac{106.08}{207.1} + 22 \times 0.246 / 36.36 $  
  $ = 0.513 + 0.149 = 0.662 $

- Newton update:  
  $ Q_4 = 5.946 - \frac{-0.079}{0.662} \approx 5.946 + 0.119 = 6.065 $

- At $ Q = 6.065 $:

$ Q^3 - 3 \approx 223.1 - 3 = 220.1 $  
  $ \ln(220.1) \approx 5.393 $  
  $ \arctan(6.065) \approx 1.406 $  
  $ e^{-1.406} \approx 0.245 $  
  $ 22 \times 0.245 \approx 5.390 $  
  $ f(6.065) = 5.393 - 5.390 = 0.003 $ (very close to zero)

So, the equilibrium quantity is $ Q^* \approx 6.06 $.

---

#### **Step 3: Find the equilibrium price**

Plug $ Q^* $ into either supply or demand:

$$
P^* = \ln((6.065)^3 - 3) = \ln(220.1) \approx 5.393
$$

---

#### **Step 4: Calculate Total Welfare**

- **Consumer surplus (CS):**
$$
CS = \int_{Q=0}^{Q^*} P_D(Q) \, dQ - P^* Q^*
$$

- **Producer surplus (PS):**
$$
PS = P^* Q^* - \int_{Q=0}^{Q^*} P_S(Q) \, dQ
$$

- **Total Welfare (W):**
$$
W = CS + PS = \int_{Q=0}^{Q^*} [P_D(Q) - P_S(Q)] \, dQ
$$

Where:
- $ P_D(Q) = 22e^{-\arctan(Q)} $
- $ P_S(Q) = \ln(Q^3 - 3) $

So:
$$
W = \int_{Q=0}^{Q^*} [22e^{-\arctan(Q)} - \ln(Q^3 - 3)] \, dQ
$$

- The lower limit for the supply integral must start from $ Q = \sqrt[3]{3} \approx 1.442 $ because $ Q^3 - 3 > 0 $.

- Therefore:
$$
W = \int_{Q=1.442}^{6.065} [22e^{-\arctan(Q)} - \ln(Q^3 - 3)] \, dQ
$$

- This integral can be evaluated numerically, but let's estimate:

Let’s use the midpoint approximation for a rough estimate:

- $ Q_m = \frac{1.442 + 6.065}{2} \approx 3.753 $
- $ \arctan(3.753) \approx 1.310 $
- $ e^{-1.310} \approx 0.270 $
- $ 22 \times 0.270 \approx 5.94 $
- $ (3.753)^3 - 3 \approx 52.8 - 3 = 49.8 $
- $ \ln(49.8) \approx 3.91 $
- Difference: $ 5.94 - 3.91 = 2.03 $
- Interval length: $ 6.065 - 1.442 = 4.623 $

So, approximate welfare:  
$ W \approx 2.03 \times 4.623 \approx 9.39 $

---

#### **Step 5: Verification**

Check that $ Q^* $ and $ P^* $ satisfy both equations.

- $ P^* = 5.393 $
- $ P_S(Q^*) = \ln((6.065)^3 - 3) \approx 5.393 $
- $ P_D(Q^*) = 22e^{-\arctan(6.065)} \approx 22 \times 0.245 \approx 5.39 $

Both match.

---

### 1(b) **Find equilibrium price using welfare optimization**

The welfare maximization problem is:

\[
\max_{Q

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