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rewrite the same pasted below in structed way Sure — here’s a clean, step-by-step solution with all calculations shown. (Quantities on the graph are in **thousands of tons**; I’ll keep that unit in the algebra and convert to tons at the end.) --- ## Given / inferred from the graph * World price: $$P_W=\$240$$ per ton. * From the graph endpoints: * Demand: $P_D(Q)=420-0.5Q$ (price in \$, $Q$ in **thousands** of tons) * Supply: $P_S(Q)=220+0.5Q$ (same units) Invert to express quantity as functions of price (quantities in thousands): $$ Q_D(P)=\frac{420-P}{0.5}=2(420-P)=840-2P\quad(\text{thousand tons}) $$ $$ Q_S(P)=\frac{P-220}{0.5}=2(P-220)=2P-440\quad(\text{thousand tons}) $$ --- ## Part 1 — Imports under free trade (price = \$240) 1. Compute domestic quantity demanded at $P_W=240$: $$ Q_D(240)=840-2(240)=840-480=360\ \text{(thousand tons)}=360{,}000\ \text{tons} $$ 2. Compute domestic quantity supplied at $P_W=240$: $$ Q_S(240)=2(240)-440=480-440=40\ \text{(thousand tons)}=40{,}000\ \text{tons} $$ 3. Imports = demand − supply: $$ \text{Imports}=360{,}000-40{,}000=320{,}000\ \text{tons} $$ **Answer (Part 1):** **320,000 tons** imported under free trade. --- ## Part 2 — Tariff that reduces imports to exactly 80,000 tons We want imports $=Q_D(P_T)-Q_S(P_T)=80$ (thousand tons). 1. Form the difference: $$ Q_D(P)-Q_S(P)=(840-2P)-(2P-440)=840-2P-2P+440=1280-4P\quad(\text{thousand}) $$ 2. Set equal to 80 (thousand) and solve for $P_T$: $$ 1280-4P_T=80 \implies 4P_T=1200 \implies P_T=300\ (\$ \text{ per ton}) $$ 3. Tariff $$t = P_T - P_W = 300 - 240 = \boxed{\$60\ \text{per ton}}$$. (As a check: at $P_T=300$, $Q_D=840-2(300)=240$ thousand and $Q_S=2(300)-440=160$ thousand, so imports $=240-160=80$ thousand — correct.) **Answer (Part 2):** **\$60 per ton** tariff. --- ## Part 3 — Tariff revenue for the government Tariff revenue $= \text{tariff per ton} \times \text{imports}$: $$ \text{Revenue}=60 \times 80{,}000 = \boxed{\$4{,}800{,}000}. $$ **Answer (Part 3):** **\$4,800,000** in tariff revenue. --- ## Final summary * Imports under free trade: **320,000 tons**. * Tariff needed to cut imports to 80,000 tons: **\$60 per ton**. * Government tariff revenue at that tariff: **\$4,800,000**.

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rewrite the same pasted below in structed way Sure — here’s a clean, step-by-step solution with all calculations shown. (Quantities on the graph are in **thousands of tons**; I’ll keep that unit in the algebra and convert to tons at the end.) --- ## Given / inferred from the graph * World price: $$P_W=\$240$$ per ton. * From the graph endpoints: * Demand: $P_D(Q)=420-0.5Q$ (price in \$, $Q$ in **thousands** of tons) * Supply: $P_S(Q)=220+0.5Q$ (same units) Invert to express quantity as functions of price (quantities in thousands): $$ Q_D(P)=\frac{420-P}{0.5}=2(420-P)=840-2P\quad(\text{thousand tons}) $$ $$ Q_S(P)=\frac{P-220}{0.5}=2(P-220)=2P-440\quad(\text{thousand tons}) $$ --- ## Part 1 — Imports under free trade (price = \$240) 1. Compute domestic quantity demanded at $P_W=240$: $$ Q_D(240)=840-2(240)=840-480=360\ \text{(thousand tons)}=360{,}000\ \text{tons} $$ 2. Compute domestic quantity supplied at $P_W=240$: $$ Q_S(240)=2(240)-440=480-440=40\ \text{(thousand tons)}=40{,}000\ \text{tons} $$ 3. Imports = demand − supply: $$ \text{Imports}=360{,}000-40{,}000=320{,}000\ \text{tons} $$ **Answer (Part 1):** **320,000 tons** imported under free trade. --- ## Part 2 — Tariff that reduces imports to exactly 80,000 tons We want imports $=Q_D(P_T)-Q_S(P_T)=80$ (thousand tons). 1. Form the difference: $$ Q_D(P)-Q_S(P)=(840-2P)-(2P-440)=840-2P-2P+440=1280-4P\quad(\text{thousand}) $$ 2. Set equal to 80 (thousand) and solve for $P_T$: $$ 1280-4P_T=80 \implies 4P_T=1200 \implies P_T=300\ (\$ \text{ per ton}) $$ 3. Tariff $$t = P_T - P_W = 300 - 240 = \boxed{\$60\ \text{per ton}}$$. (As a check: at $P_T=300$, $Q_D=840-2(300)=240$ thousand and $Q_S=2(300)-440=160$ thousand, so imports $=240-160=80$ thousand — correct.) **Answer (Part 2):** **\$60 per ton** tariff. --- ## Part 3 — Tariff revenue for the government Tariff revenue $= \text{tariff per ton} \times \text{imports}$: $$ \text{Revenue}=60 \times 80{,}000 = \boxed{\$4{,}800{,}000}. $$ **Answer (Part 3):** **\$4,800,000** in tariff revenue. --- ## Final summary * Imports under free trade: **320,000 tons**. * Tariff needed to cut imports to 80,000 tons: **\$60 per ton**. * Government tariff revenue at that tariff: **\$4,800,000**.

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Asked by: mayank
Created at: 2025-09-03 18:52:47

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Certainly! Here’s your solution rewritten in a clear, structured format, with each step labeled and all calculations shown. --- # Step-by-Step Solution ## **I. Information from the Graph** - **World price:** \( P_W = \$240 \) per ton - **Demand equation:** \( P_D(Q) = 420 - 0.5Q \) - **Supply equation:** \( P_S(Q) = 220 + 0.5Q \) - (All quantities \( Q \) are in **thousands of tons**.) ### **Invert to Find Quantities as Functions of Price** - **Demand:** \[ Q_D(P) = \frac{420 - P}{0.5} = 2(420 - P) = 840 - 2P \] - **Supply:** \[ Q_S(P) = \frac{P - 220}{0.5} = 2(P - 220) = 2P - 440 \] - (Units: \( Q \) in thousands of tons) --- ## **II. Imports under Free Trade (\( P_W = \$240 \))** **1. Calculate Domestic Quantity Demanded at \( P_W \):** \[ Q_D(240) = 840 - 2(240) = 840 - 480 = 360 \text{ (thousand tons)} = 360,\!000 \text{ tons} \] **2. Calculate Domestic Quantity Supplied at \( P_W \):** \[ Q_S(240) = 2(240) - 440 = 480 - 440 = 40 \text{ (thousand tons)} = 40,\!000 \text{ tons} \] **3. Find Imports:** \[ \text{Imports} = Q_D - Q_S = 360,\!000 - 40,\!000 = 320,\!000 \text{ tons} \] ### **Final Answer (Part 1):** **Imports under free trade = 320,000 tons** --- ## **III. Tariff to Reduce Imports to 80,000 Tons** We want: \[ \text{Imports} = Q_D(P_T) - Q_S(P_T) = 80 \text{ (thousand tons)} \] **1. Express Imports as a Function of Price:** \[ Q_D(P) - Q_S(P) = (840 - 2P) - (2P - 440) = 840 - 2P - 2P + 440 = 1280 - 4P \] **2. Set Equal to 80 (thousand) and Solve for \( P_T \):** \[ 1280 - 4P_T = 80 \implies 4P_T = 1200 \implies P_T = 300 \] (\$300 per ton) **3. Determine the Tariff:** \[ \text{Tariff} = P_T - P_W = 300 - 240 = \$60 \text{ per ton} \] **4. (Check: At \( P_T = 300 \))** - \( Q_D = 840 - 2(300) = 240 \) thousand - \( Q_S = 2(300) - 440 = 160 \) thousand - \( \text{Imports} = 240 - 160 = 80 \) thousand (correct) ### **Final Answer (Part 2):** **Tariff required = \$60 per ton** --- ## **IV. Tariff Revenue for the Government** **1. Calculate Tariff Revenue:** \[ \text{Revenue} = (\text{Tariff per ton}) \times (\text{Imports}) \] \[ \text{Revenue} = \$60 \times 80,\!000 = \$4,\!800,\!000 \] ### **Final Answer (Part 3):** **Tariff revenue = \$4,800,000** --- ## **V. Summary Table** | Scenario | Quantity (tons) | Tariff (\$/ton) | Revenue (\$) | |----------------------------|:--------------:|:---------------:|:----------------:| | Free trade (no tariff) | 320,000 | 0 | 0 | | With tariff (\$60/ton) | 80,000 | \$60 | \$4,800,000 | **Key Answers:** - **Imports under free trade:** 320,000 tons - **Tariff needed to cut imports to 80,000 tons:** \$60 per ton - **Tariff revenue at that tariff:** \$4,800,000 ---

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