P(x) = -5x^2 +125x + 37500 is the total profit function of a company, where x is the production of the company.

Question

1. What will be the production when the profit is maximum?

a. 37500

b. 12.5

c. -12.5

d. –37500

2. What will be the maximum profit?

a. Rs 38,28,125

b. Rs 38281.25

c. Rs 39,000

d. None

3. Check in which interval the profit is strictly increasing .

a. (12.5,∞ )

b. for all real numbers

c. for all positive real numbers

d. (0, 12.5)

4. When the production is 2units what will be the profit of the company?

a. 37500

b. 37,730

c. 37,770

d. None

5. What will be production of the company when the profit is Rs 38250?

a. 15

b. 30

c. 2

d. data is not sufficient to find

Answer 1

1. b) 12.5

2. b) Rs.38281.25

3. d) (0 , 12.5)

4. b) 37,730

5. a) 15

Answer 2

1. (b) 12.5

Given

P(x) = -5x² + 125x + 37500

Differentiating w.r.t. x

Since P''(x) < 0 at x = 12.5

\(\therefore \) x = 12.5 is point of maxima

Thus, P(x) is maximum when x = 12.5

2. (b) Rs 38281.25

Putting x = 12.5 in P(x)

3. (d) (0, 12.5)

Profit is strictly increasing where

P'(x) > 0

-10x + 125 > 0

125 > 10x

10x < 125

x < 12.5

So, profit is strictly increasing for x \(\in\) (0, 12.5).

4. (b) 37,730

Putting x = 2 in P(x)

5. (a) 15

Putting P(x) = 38250 in formula of P(x)

Since x = 15 is in the options.