The table below represents the cost function of a poultry farm. The price of a crate of egg is $21. Use the information contained in the table to answer the questions that follow.
| Quantity of eggs (in crates) | Total cost (in $) |
| 0 | 50 |
| 1 | 55 |
| 2 | 62 |
| 3 | 75 |
| 4 | 96 |
| 5 | 125 |
| 6 | 162 |
| 7 | 203 |
| 8 | 248 |
(a) What Is the fixed cost of the farm? (2 marks]
(b)(i) Calculate the marginal cost at each level of output. [9 marks]
(ii) What is the profit maximizing output of the farm? [3 marks]
(c) Draw the demand curve for the farm. [6 marks].
Explanation
(a)
| Quantity of eggs (in crates) | Total cost (in $) |
| 0 | 50 |
| 1 | 55 |
| 2 | 62 |
| 3 | 75 |
| 4 | 96 |
| 5 | 125 |
| 6 | 162 |
| 7 | 203 |
| 8 | 248 |
$50 (When output is zero, total cost is $50. This implioes that fixed cost is $5)
(b) (i) MC\(_0\) = 50
MC\(_1\) = \(\frac{55-50}{1-0}\) = \(\frac{5}{1}\) = 5 OR MC\(_1\) = \(\frac{55-50}\) = 5
MC\(_2\) = \(\frac{62-55}{2-1}\) = \(\frac{7}{1}\) = 7
MC\(_3\) = \(\frac{75-62}{3-2}\) = \(\frac{13}{1}\) = 13
MC\(_4\) = \(\frac{96-75}{4-3}\) = \(\frac{21}{1}\) = 21
MC\(_5\) = \(\frac{125-96}{5-4}\) = \(\frac{29}{1}\) = 29
MC\(_6\) = \(\frac{162-125}{6-5}\) = \(\frac{37}{1}\) = 37
MC\(_7\) = \(\frac{203-162}{7-6}\) = \(\frac{41}{1}\) = 41
MC\(_8\) = \(\frac{248-203}{8-7}\) = \(\frac{45}{1}\) = 45
(ii) The profit maximizing output is 4 crates of eggs. This is when MC = P
(c) D=AR (AR = $21)
