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:​

Answer 1

Given :  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.

To Find : degree of the above polynomial

When no barrels are produce, what is the loss

What is the profit/loss if 175 barrels are produced

What are the factors of the given polynomial

Solution:

p(x) = – 10x² + 3500 x – 66000

Degree  = 2

When no barrels are produce, x = 0

=>  p(0) = 0 + 0 – 66000 = - 66000

-ve sign means Loss

Loss = 66000

175 barrels are produced

=> p(175) = – 10(175)² + 3500 (175) – 66000

= 2,40,250

+Ve sign means profit

Profit 240250

– 10x² + 3500 x – 66000

= - 10(x²  - 350x  + 6600)

= -10(x² - 330x - 20x + 6600)

= -10(x - 330)(x - 20)

$-{10}\left (x - 20\right ) \times \left (x - 330\right )$

(x - 330)(x - 20)  are factors

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Answer 2

Answer:

Given : 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.

To Find : degree of the above polynomial

When no barrels are produce, what is the loss

What is the profit/loss if 175 barrels are produced

What are the factors of the given polynomial

Solution:

p(x) = – 10x² + 3500 x – 66000

Degree = 2

When no barrels are produce, x = 0

=> p(0) = 0 + 0 – 66000 = - 66000

negative sign means Loss

Loss = 66000

175 barrels are produced

=> p(175) = – 10(175)² + 3500 (175) – 66000

= 2,40,250

+Ve sign means profit

Profit 240250

– 10x² + 3500 x – 66000

= - 10(x² - 350x + 6600)

= -10(x² - 330x - 20x + 6600)

= -10(x - 330)(x - 20)