CBSEGrade 12MathematicsApplication of Derivatives

Maximum-Minimum Values?

The demand function for a product is given by D(x) = 1000 - 3x^2 + 2x. Find the critical points of D(x) and determine the maximum and minimum values of the demand function, if any, in the interval 0 ≤ x ≤ 5.

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📌 CONCEPT: The application of derivatives helps in finding the maximum and minimum values of a function by identifying its critical points, which are the points where the function's derivative is zero or undefined.

📐 RULE / FORMULA: To find the critical points, we need to find the derivative of the function and set it equal to zero, then solve for x. The second derivative test can be used to determine whether the critical point corresponds to a maximum or minimum value.

💡 WORKED EXAMPLE: Let's consider the function D(x) = 1000 - 3x^2 + 2x. To find the critical points, we first find the derivative D'(x) = -6x + 2. Setting D'(x) = 0 gives us -6x + 2 = 0. Solving for x, we get x = 1/3. To determine whether this point corresponds to a maximum or minimum, we use the second derivative test. The second derivative is D''(x) = -6, which is negative, indicating that x = 1/3 corresponds to a maximum value.

⚠️ COMMON MISTAKE: Students often forget to check the second derivative to determine whether the critical point corresponds to a maximum or minimum value. They may also overlook the fact that the critical point must lie within the given interval.

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