A product has demand equation p = 100 - 0.5q, where p is price and q is quantity. What quantity maximises total revenue?
Correct answer: B. 100 units
- A. 50 units
- B. 100 units
- C. 150 units
- D. 200 units
Explanation
Revenue is R(q) = pq = 100q - 0.5q^2. Since R'(q) = 100 - q, the stationary point occurs at q = 100, and R''(q) = -1 confirms a maximum.
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About Differentiation and Optimisation
Differentiation covers gradients, standard derivative rules, the product and quotient rules, the chain rule, stationary points and rates of change. Optimisation uses first and second derivatives to classify maxima and minima, while applications include tangent and normal lines, marginal quantities and related rates.
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