A company is selling a certain product. The demand function for the product is linear. The company can sell 2000 units when the price is ₹8 per unit and it can sell 3000 units when the price is ₹4 per unit. Determine the total revenue function.

$R(x)=16x-\frac{x^2}{250}$

The correct answer is Option (2) → $R(x)=16x-\frac{x^2}{250}$

As the demand function for the product is linear (given), let

$p = a + bx$   ...(1)

where p is the price per unit and x units is the demand of the product, a and b are constants.

Given $x = 2000$ when $p = 8$ and $x = 3000$ when $p = 4$.

Substituting these values in (1), we get

$8 = a + 2000b$   ...(2)

$4 = a + 3000 b$   ...(3)

Subtracting (2) from (3), we get

$1000b=-4⇒b=-\frac{1}{250}$

Substituting this value of b in (2), we get

$8 = a + 2000(-\frac{1}{250})$

$⇒ 8=a-8⇒ a=16$.

Substituting these values of a and b in (1), we get

$p=16-\frac{1}{250}x$.

The demand function is $p=16-\frac{1}{250}x$.

Total revenue by selling x units of the product

$=px=\left(16-\frac{1}{250}x\right)x=16x-\frac{x^2}{250}$

Hence, the revenue function is $R(x) =16x-\frac{x^2}{250}$