A company is selling a certain product. The demand function of the product is linear. The company can sell 1000 units when the price is ₹8 per unit and when the price is ₹4 per unit, it can sell 2000 units. The total revenue function (R(x)) is given by :

$\frac{-x^2}{250}+12$

The correct answer is Option (2) → $\frac{-x^2}{250}+12$

$R(x)=p(x).x$

Since the demand function is linear

$p(x)=mx+c$

Here, m ⇒ slope, c ⇒ intercept

Points of demand function (given)

$x=1000,p=8$

$x=2000,p=4$

$M=\frac{p_2-p_1}{x_2-x_1}=\frac{4-8}{2000-1000}=\frac{-4}{1000}=\frac{-1}{250}$

$P(x)=\frac{-1}{250}x+c$

Using point (1000, 8)

$8=\frac{-1}{250}(1000)+c$

$8=-4+c$

$c=12$

$∴P(x)=\frac{-1}{250}x+12$

$R(x)=\left(\frac{-1}{250}x+12\right)x$

$=\frac{-1}{250}x^2+12x$