Calculus

Application of Derivatives

Medium

applied

The demand function (in Rs.) for a product is given by $P = 20 - 0.25x$ , where P is the price per unit and x is the number of units sold, then the price of one unit, when the revenue is maximized, is:

Rs.20

Rs.10

Rs.30

Rs.40

✅ Correct Option: 2

The demand function for the product is $P = 20 - 0.25x$ , where $P$ is the price per unit and $x$ is the number of units sold.

Revenue is the product of price and quantity.

Revenue $R = P \times x$

$R = (20 - 0.25x) \times x$

$R = 20x - 0.25x^2$

To find the maximum revenue, differentiate $R$ with respect to $x$ :

$\dfrac{dR}{dx} = 20 - 0.5x$

For maximum revenue, set the derivative equal to zero:

$20 - 0.5x = 0$

$0.5x = 20$

$x = 40$

Substitute $x = 40$ into the demand function to find the price:

$P = 20 - 0.25x$

$P = 20 - 0.25(40)$

$P = 20 - 10$

$P = 10$

Therefore, the price per unit when revenue is maximized is Rs. 10.

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Choose the correct answer from the options given below:

The demand function (in Rs.) for a product is given by P=20−0.25xP = 20 - 0.25xP=20−0.25x, where P is the price per unit and x is the number of units sold, then the price of one unit, when the revenue is maximized, is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The demand function (in Rs.) for a product is given by $P = 20 - 0.25x$ , where P is the price per unit and x is the number of units sold, then the price of one unit, when the revenue is maximized, is: