**ECO401 assignment 2 solution 2021**

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**ECO401 Assignment No.2 2021**

**Question No.1 **

** **

**Solution:**

**Profit is maximum at MR=MC**

MR= dTR/dQ

dTR50/4Q= d150AQ – d4Q2/dQ = MR

MR= 150 — 8Q

MC= dTC/dQ

dTC/dQ= d40/4Q + dl0Q/4Q + d3Q^{2}/4Q = MC

MC= 10 + 6Q

MR=MC

150-8Q=10+6

150-10=6Q+8Q

14Q= 140

Q = 140/14

Q= 10

**The output is 10 units at the profit maximization level**

**Question No.2**

**Solution:**

** **

π=TR-TC

π= (150Q – 4Q2) – (40-F 10Q + 3Q^{2})

π=150Q-4Q2-40-10Q-3Q^{2}

π= -7Q2 +140Q – 40

π= -7(10)2+ 140(10)- 40

π= -700 + 1400 – 40

π= 660

**At profit maximization level profit is 660rs**

**Question No.3**

**Solution:**

** **

AR = TR/Q

TR/Q = AR = 150Q/Q – 4Q^{2}/Q

AR=150-4Q

Slope of AR = dAR/dQ

dAR/dQ = Slope of AR = d150/dQ – d4Q/dQ

Slope of AR = -4

MR= 150 – 8Q

Slope of MR = dMR/dQ

dMR/4Q = Slope of MR = d150/dQ – d8Q/dQ

Slope of MR = -8

Slope of MR is -8 and Slope of AR is -4

**So, the Slope of MR is twice the slope of AR prove.**

**ECO401 assignment 2 solution**- eco401 assignment 2 solution 2020
- eco401 assignment 2 solution 2021

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