Question

A product costs the manufacturer ₹20 per unit. The demand function is given by p(x) = 1000-20x, then the quantity for maximum profit is:

• A. 25 units
• B. 50 units
• C. 49 units
• D. \(\frac{49}{2}\)

Answer 1

The Correct Option is D

To determine the quantity that maximizes profit, we first need to establish the profit function. Given the cost per unit to the manufacturer is ₹20, and the demand function is \(p(x) = 1000 - 20x\), the selling price per unit as a function of quantity \(x\) is \(p(x) = 1000 - 20x\).

The revenue \(R(x)\) is given by the product of the price per unit and the quantity sold:

\[ R(x) = x \cdot p(x) = x(1000 - 20x) = 1000x - 20x^2 \]

The cost \(C(x)\) is the cost per unit multiplied by the quantity:

\[ C(x) = 20x \]

Thus, the profit \(P(x)\) is the difference between revenue and cost:

\[ P(x) = R(x) - C(x) = (1000x - 20x^2) - 20x = 1000x - 20x^2 - 20x = 980x - 20x^2 \]

The profit function is a quadratic function of the form \(ax^2 + bx + c\), where \(a = -20\), \(b = 980\), and \(c = 0\). The maximum profit occurs at the vertex of this parabola.

The vertex \(x\) of a parabola of the form \(ax^2 + bx + c\) is given by:

\[ x = -\frac{b}{2a} \]

Substituting the values:

\[ x = -\frac{980}{2 \times -20} = \frac{980}{40} = \frac{49}{2} \]

Therefore, the quantity for maximum profit is \(\frac{49}{2}\) units.