Question

CASE STUDY-4) – MAXIMUM PROFIT
A barrel manufacturer can produce up to 300 barrels per day. The profit made from the sale of these barrels can be modelled by the function p(x) = – 10x2 + 3500 x – 66000, where p(x) is the profit in rupees and x is the number of barrels made and sold. Based on this model answer the following questions:
When no barrels are produced, what is the profit /loss?
i) 22000 (ii) 11000 ((iii) −66000
iv) 33000
What is the breakeven point? (Zero profit point is called
breakeven)
(i) 10 barrels (ii) 20 barrels iii)30 barrels iv) 100 barrels
What is the profit/loss if 175 barrels are produced?
i) Profit 266200 (ii) Profit 240250 (iii) Loss 266200 (iv) Loss 240250
What is the profit/loss if 400 barrels are produced?
(i) Profit 266200 (ii) Profit 342000 (iii) Loss 266200 (iv) Loss 342000
Whether profit or loss would result, if 10 barrels are produced?
(i) Profit ii)loss
(iii) No profit no loss (iv) cannot find​

Answer 1

Answer:

1. i) b ii c iii. c iv. b v. a

Step-by-step explanation: