table 13 - 4\n| quantity sold | price | total revenue | marginal revenue | total cost | marginal cost |…

table 13 - 4\n| quantity sold | price | total revenue | marginal revenue | total cost | marginal cost | profit |\n| ---- | ---- | ---- | ---- | ---- | ---- | ---- |\n| 0 | $10 | $0 |...... | $2 |...... | -$2 |\n| 1 | 9 | 9 | | 8 | | |\n| 2 | 8 | 16 | | 13 | | |\n| 3 | 7 | 21 | | 17 | | |\n| 4 | 6 | 24 | | 20 | | |\n| 5 | 5 | 25 | | 22 | | |\n| 6 | 4 | 24 | | 28 | | |\ntable 13 - 4 lists estimated revenues and costs (per week) for plastic vials (100 vials per box) for the victoria biological supplies company. victoria sells plastic vials to universities and private research laboratories.\nrefer to table 13 - 4. at victorias profit - maximizing output,\n- profit equals $2.\n- total revenue equals $25 and total cost equals $22.\n- total revenue equals $21 and total cost equals $17.\n- total revenue equals $24 and total cost equals $20.
Answer
Explanation:
Step1: Recall profit - maximization condition
Profit is maximized when marginal revenue (MR) equals marginal cost (MC). Calculate the marginal revenue and marginal cost values for each quantity. Marginal revenue for quantity $n$ is $MR_n=TR_n - TR_{n - 1}$ (where $TR$ is total revenue), and marginal cost for quantity $n$ is $MC_n=TC_n - TC_{n - 1}$ (where $TC$ is total cost). For quantity 1: $MR_1=9 - 0=9$, $MC_1=8 - 2 = 6$ For quantity 2: $MR_2=16 - 9 = 7$, $MC_2=13 - 8=5$ For quantity 3: $MR_3=21 - 16 = 5$, $MC_3=17 - 13 = 4$ For quantity 4: $MR_4=24 - 21=3$, $MC_4=20 - 17 = 3$ For quantity 5: $MR_5=25 - 24 = 1$, $MC_5=22 - 20=2$ For quantity 6: $MR_6=24 - 25=- 1$, $MC_6=28 - 22 = 6$ Profit - maximization occurs at quantity 4 where $MR = MC=3$.
Step2: Find total revenue and total cost at profit - maximizing quantity
At quantity $q = 4$, from the table, total revenue $TR = 24$ and total cost $TC = 20$.
Answer:
total revenue equals $24 and total cost equals $20