Calculus

Application of Derivatives

Medium

applied

The demand for a certain product is represented by the function

$p = 150 + 10x - x^2$ (in Rs.)

where $x$ is the number of units demanded and $p$ is the price per unit, then the value of marginal revenue, when 10 units are sold is

Rs.50

Rs.100

Rs.150

Rs.200

✅ Correct Option: 1

The demand for a certain product is represented by the function $p = 150 + 10x - x^2$ (in Rs.), where $x$ is the number of units demanded and $p$ is the price per unit.

Marginal Revenue (MR) is the additional revenue earned when one more unit is sold, given by $MR = \dfrac{dR}{dx}$ .

Revenue is given by:

$R = p \times x$

$R = (150 + 10x - x^2) \times x$

$R = 150x + 10x^2 - x^3$

Differentiating revenue with respect to $x$ :

$MR = \dfrac{dR}{dx}$

$MR = 150 + 20x - 3x^2$

When $x = 10$ units:

$MR = 150 + 20(10) - 3(10)^2$

$MR = 150 + 200 - 3(100)$

$MR = 150 + 200 - 300$

$MR = 50$

Therefore, the marginal revenue when 10 units are sold is Rs. 50.

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The demand for a certain product is represented by the function p=150+10x−x2p = 150 + 10x - x^2p=150+10x−x2 (in Rs.) where xxx is the number of units demanded and ppp is the price per unit, then the value of marginal revenue, when 10 units are sold is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The demand for a certain product is represented by the function

$p = 150 + 10x - x^2$ (in Rs.)

where $x$ is the number of units demanded and $p$ is the price per unit, then the value of marginal revenue, when 10 units are sold is