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: Find the price at 5 units.
Put \(x=5\) in the demand equation.
\[ p=150+60-25=185 \] Note that 185 is the price, not the marginal revenue. This is a common trap, and option 4 is exactly this number.

Step 2: Use the product rule on R = px.
Since \(R=px\) and \(p\) depends on \(x\),
\[ MR=\frac{d(px)}{dx}=p+x\frac{dp}{dx} \]

Step 3: Find dp/dx.
From \(p=150+12x-x^2\), \(\frac{dp}{dx}=12-2x\). At \(x=5\) this is \(12-10=2\).

Step 4: Combine.
\[ MR=185+5\times 2=195 \]

Final Answer:
So MR at 5 units equals 195. \[ \boxed{195} \]