Step 1: Define the variables. Let the number of plates sold be x, and the price per plate be P. Initially, P = 160 and x = 300.
Step 2: Relationship between price and number of plates sold. For every Rs. 10 increase in price, 10 fewer plates are sold. Let y be the number of Rs. 10 increments in price above Rs. 160. Then:
P = 160 + 10y
The number of plates sold decreases by 10 for each increment in price, so:
x = 300 − 10y
Step 3: Profit function. The cost per plate is Rs. 120, so the profit per plate is:
Profit per plate = P − 120 = (160 + 10y) − 120 = 40 + 10y
Thus, the total profit is:
Total profit = (40 + 10y)(300 − 10y)
Step 4: Maximize the profit. Expand the profit function:
Profit = (40 + 10y)(300 − 10y) = 12000 + 400y − 120y − 100y2 = 12000 + 280y − 100y2
To maximize the profit, take the derivative with respect to y and set it equal to 0:
$\frac{d}{dy}$(12000 + 280y − 100y2) = 280 − 200y
Set the derivative equal to 0:
280 − 200y = 0 => y = 1.4
Since y must be an integer, round y = 1.
Step 5: Calculate the maximum profit. For y = 1, the price per plate is:
P = 160 + 10(1) = 170
The number of plates sold is:
x = 300 − 10(1) = 290
Thus, the total profit is:
Profit = (170 − 120)(290) = 50 × 290 = 14,500
Answer: Rs. 41,400
A shopkeeper buys an item for Rs 2000 and marks it up by 50% to set the marked price. He then offers a 20% discount on the marked price. What is the profit earned by the shopkeeper?
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |