Question

The demand for a certain product is represented by the equation $p=500+25x-\frac{x^2}{3}$ in rupees where $x$ is the number of units and $p$ is the price $3$ per unit. Find:
(i) Marginal revenue function.
(ii) The marginal revenue when $10$ units are sold.

Answer 1

i ) To find Marginal Revenue function

Demand for a certain Product is represented by the Equation

$p=500+25x-\frac{x^2}{3}$

Where $x$ is the number units and $p$ is the price per unit

Marginal Revenue function is the derivative of the revenue function

So , Revenue Function is

$R=x.p$

$R=x.(500+25x-\frac{x^2}{3})$

$R=(500x+25x^2-\frac{x^3}{3})$

Now , Marginal Revenue function can be Calculated as

$=\frac{dR}{dx}$

$=\frac{d}{dx}(500x+25x^2-\frac{x^3}{3})$

$=(500+50x-\frac{3x^2}{3})$

$=(500+50x-x^2)$

Hence , Marginal Revenue function$=500+50x-x^2$ units

ii) To find Marginal Revenue when 10 units are sold

Marginal Revenue function $=500+50x-x^2$

At $x=10units$

$=500+50\left(10\right)-5^2$

$=1000-25$

$=975$ units