CBSEGrade 12MathematicsApplication of Derivatives

Maximizing Profit in Industry?

A smartphone manufacturing company produces x units of its latest model in a day. The cost price of each unit is ₹150 and the selling price is ₹250. Using the concept of optimization, determine the production level that maximizes profit, assuming the cost and selling price remain constant.

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📌 CONCEPT: Maximizing profit in the smartphone manufacturing company involves finding the production level that results in the maximum possible profit, given the constant cost and selling prices.

📐 RULE / FORMULA: To maximize profit, we use the concept of optimization by finding the maximum value of the profit function P(x) = (Selling price - Cost price) × Number of units sold, which can be expressed as P(x) = (250 - 150)x = 100x.

💡 WORKED EXAMPLE: Suppose the company produces x units of smartphones. The cost price of each unit is ₹150 and the selling price is ₹250. We need to find the production level that maximizes profit. The profit function is P(x) = (250 - 150)x = 100x. To maximize profit, we find the critical point by taking the derivative of P(x) with respect to x, which is P'(x) = 100. Since P'(x) is a constant, the profit is maximized for all values of x. Therefore, the company should produce as many units as possible to maximize profit.

⚠️ COMMON MISTAKE: Students often get confused between the concept of optimization and the formula for profit, which is P(x) = (Selling price - Cost price) × Number of units sold. They may also fail to recognize that the profit function is a linear function and hence has no critical points.

23 Aug 26