CBSEGrade 12MathematicsApplication of Derivatives

Maximizing a Profit Function?

A company manufactures and sells two products, A and B. The profit function, P, in terms of the number of units produced of each product, x and y, is given by P(x,y) = 100x - 2y + 2xy - 1000. A factory has a fixed constraint that the number of units produced of product B cannot exceed double the number of units produced of product A, i.e., y ≤ 2x. What should be the production levels of product A and product B to maximize the profit function under this constraint?

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📌 CONCEPT: The problem requires finding the maximum value of the profit function P(x,y) = 100x - 2y + 2xy - 1000, subject to the constraint y ≤ 2x, to determine the optimal production levels of product A and product B.

📐 RULE / FORMULA: To solve this constrained optimization problem, we will use the method of Lagrange multipliers. The Lagrangian function is given by L(x, y, λ) = P(x,y) - λ(g(x,y) - c), where g(x,y) is the constraint function and c is the constraint value. In this case, g(x,y) = y - 2x and c = 0.

💡 WORKED EXAMPLE: Given the profit function P(x,y) = 100x - 2y + 2xy - 1000 and the constraint y ≤ 2x, we set up the Lagrangian function as L(x, y, λ) = 100x - 2y + 2xy - 1000 - λ(y - 2x). To find the maximum value, we need to find the critical points of L(x, y, λ) by solving the system of partial derivatives.

⚠️ COMMON MISTAKE: Students often forget to check the second-order conditions to confirm that the critical point corresponds to a maximum.

20 Aug 26