Math, asked by mrittika20, 4 months ago

Maximum profit: An barrels 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 + 3500x – 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:





(i) When no barrels are produce what is a profit loss?

(a) 22000 (b) 66000 (c) 11000 (d) 33000

1

(ii) What is the break even point ? (Zero profit point is called break even)

(a) 10 barrels (b) 30 barrels (c) 20 barrels (d) 100 barrels

1

(iii) What is the profit/loss if 175 barrels are produced

(a) Profit 266200 (b) Loss 266200 (c) Profit 240250 (d) Loss 240250

1

(iv) What is the profit/loss if 400 barrels are produced

(a) Profit 266200 (b) Loss 266200 (c) Profit 342000 (d) Loss 342000

1

(v) What is the maximum profit which can manufacturer earn?

(a) Rs 240250 (b) Rs 480500 (c) Rs 680250 (d) Rs 240250​

Answers

Answered by swethassynergy
4

Answer:

(i).When no barrels are produce it is loss of Rs. 66000 and option (b) is correct.

(ii).The break even point  is at 20 barrels and option(c) is correct.

(iii). If 175 barrels are produced the profit is Rs.240250 and option (c) is correct.

(iv) If 400 barrels are produced the loss is Rs.266000 and option (c) is correct.

(v).The maximum profit which can manufacturer earn is Rs.240250 and option (d) is correct.

Step-by-step explanation:

Given :

Up to  300 barrels per day can be produced by a barrel manufacturer.

The profit made  from the sale of these barrels can be modeled by the function

P(x) = - 10 x^{2}  + 3500 x - 66000 where P(x) is the profit in rupees and x  is the number of barrels made and sold.

To Find :

(i).When no barrels are produce, what is a profit or loss.

(ii).What is the break even point ? (Zero profit point is called break even)

(iii) What is the profit/loss if 175 barrels are produced.

(iv).What is the profit/loss if 400 barrels are produced.

(v).What is the maximum profit which can manufacturer earn?

Solution:

(i).When no barrels are produce, what is a profit or loss.

   When no barrels are produce. It means x=0.

   P(x) = - 10 x^{2}  + 3500 x - 66000

     Putting x=0

       P(0)= - 10 ( 0)^{2}  + 3500( 0) - 66000

                = 0 + 0 -66000 = - 66000\ Rs.     [(-ve) sign means Loss].

               It means loss of Rs. 66000.

Thus, when no barrels are produce it is loss of Rs. 66000 and option (b) is correct.

(ii).What is the break even point ? (Zero profit point is called break even)

   At the break even point P(x)=0. Therefore,

   0 = - 10 x^{2}  + 3500 x - 66000\\0=10(-  x^{2}  + 350 x - 6600)\\x^{2}  - 350 x +6600=0\\x^{2}  - 330 x -20 x+6600=0\\x(x-330)-20(x-330)=\\(x-20) (x-330)=0\\x=20\ and \ 330

Thus, the break even point  is at 20 barrels and option(c) is correct.

(iii) What is the profit/loss if 175 barrels are produced.

      175 barrels are produced.

        P(175) = - 10 (175)^{2}  + 3500 (175) - 66000\\

                   =-306250 +612500 -66000\\            = -372250+612500\\=240250 \ Rs.         [(+ve) sign means Profit].                        

Thus, if 175 barrels are produced the profit is Rs.240250 and option (c) is correct.

(iv).What is the profit/loss if 400 barrels are produced.

       400 barrels are produced.

      P(400) = - 10 (400)^{2}  + 3500 (400) - 66000\\

                   =-1600000 +1400000 -66000\\            = -1666000+1400000\\=-266000  \ Rs [(-ve) sign means Loss].

Thus, if 400 barrels are produced the loss is Rs.266000 and option (c) is correct.

(v).What is the maximum profit which can manufacturer earn?    

     P(x) = - 10 x^{2}  + 3500 x - 66000

      Rearranging the given equation we have

      P(x) = - 10 x^{2}  + 3500 x - 66000\\       =-10( x^{2}  - 350 x + 6600)\\       =-10[(x-175)^{2}  - 30625  + 6600]\\        =-  =-10[(x-175)^{2}  - 30625  + 6600]\\          =-10[(x-175)^{2}  - 24025]\\         = -10(x-175)^{2}  +240250

From above equation it is clear that maximum value  of P(x) = - 10 x^{2}  + 3500 x - 66000 is Rs. 240250.    

Thus,the maximum profit which can manufacturer earn is Rs.240250 and option (d) is correct.

         #SPJ3

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