Question

The demand function for a certain product is represented by the equation \(p=150+12x-x^2\), where \(x\) is the number of units demanded and \(p\) is the price per unit. The marginal revenue when 5 units are sold is

• A. Rs. 195
• B. Rs. 282
• C. Rs. 91
• D. Rs. 185

Hint:

Revenue \(R=px\). Marginal revenue is \(\frac{dR}{dx}\) at \(x=5\).

Answer 1

The Correct Option is A

Step 1: Understand the terms.
Revenue \(R\) is the money received from selling \(x\) units, so \(R=p\cdot x\). Marginal revenue (MR) is the rate of change of revenue with the number of units, so \(MR=\frac{dR}{dx}\).

Step 2: Write the revenue function.
Multiply the price by \(x\).
\[ R=x(150+12x-x^2)=150x+12x^2-x^3 \]

Step 3: Differentiate.
\[ MR=\frac{dR}{dx}=150+24x-3x^2 \]

Step 4: Put x = 5.
\[ MR=150+24(5)-3(25)=150+120-75=195 \]

Step 5: Check the other options.
Option 2 (282) and option 3 (91) do not come from \(150+24x-3x^2\) at \(x=5\). Option 4 (185) is the price per unit at \(x=5\), not the marginal revenue. Option 1 is correct.

Final Answer:
The marginal revenue at 5 units is Rs. 195, option 1. \[ \boxed{195} \]