Step-by-Step Solution
Topic: Maxima and Minima
Given Function
The revenue function is given by:
$$ y = 40,00,000 – (x – 2000)^2 $$
This is a downward-opening parabola (quadratic equation) because of the negative sign before the squared term.
(A) UNITS FOR MAXIMUM REVENUE
Method 1: Logical Analysis
The term (x – 2000)2 is a perfect square, so its value is always ≥ 0.
The revenue y is maximized when the quantity subtracted from 40,00,000 is minimized.
The minimum value of (x – 2000)2 is 0.
This occurs when x – 2000 = 0 ⇒ x = 2000.
Method 2: Derivatives (Calculus)
Differentiate y with respect to x:
$$ \frac{dy}{dx} = 0 – 2(x – 2000)(1) $$
$$ \frac{dy}{dx} = -2(x – 2000) $$
For maxima, set dy/dx = 0:
-2(x – 2000) = 0 ⇒ x = 2000.
Check 2nd Derivative: d2y/dx2 = -2 (Negative ⇒ Maxima).
(B) MAXIMUM REVENUE AMOUNT
Substitute x = 2000 into the revenue equation:
y = 40,00,000 – (2000 – 2000)2
y = 40,00,000 – 0
(C) REVENUE FOR 2500 UNITS
Substitute x = 2500 into the revenue equation:
y = 40,00,000 – (2500 – 2000)2
y = 40,00,000 – (500)2
Calculation:
| 500 × 500 | 2,50,000 |
y = 40,00,000 – 2,50,000
y = 37,50,000
⚠️ Silly Mistake Audit: Squaring Differences
Don’t forget the square!
A common error in Part (c) is calculating 40,00,000 – (2500 – 2000) and simply subtracting 500. Remember the formula has a squared term: (500)2 = 2,50,000. Always evaluate the exponent before subtracting.