The demand function for commodity X, is \( Q_D = 300 - 20P \); where P is the price in rupees per unit and \( Q_D \) is the quantity demanded in units per period. Which of the following is the price level at which total revenue of a firm facing this demand function is maximised?
7.5
Option 2 — 7.5 is correct.
Total revenue is price multiplied by quantity sold. Substituting the demand curve into TR gives \( TR = P \times Q_D = P(300 - 20P) = 300P - 20P^2 \). This is a downward-opening parabola in P, so revenue rises, peaks, then falls as price increases.
To locate the peak, differentiate TR with respect to P and set the slope (marginal revenue in price terms) to zero: \( \frac{d(TR)}{dP} = 300 - 40P = 0 \). Solving, \( 40P = 300 \), so \( P = 7.5 \).
You can verify: at \( P = 7.5 \), \( Q_D = 300 - 20(7.5) = 150 \) units and \( TR = 7.5 \times 150 = 1125 \). A rupee more or less than 7.5 lowers TR, confirming the maximum. At this point demand is unit-elastic, the classic revenue-maximising condition on a linear demand curve.
Why the other options are wrong: 8.0, 10.5 and 12.5 all lie to the right of the turning point, where marginal revenue is negative and total revenue is already declining.
Takeaway: On a linear demand curve, revenue is maximised at the mid-price where MR = 0 and elasticity equals one.
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