RD Sharma Class 12 Solutions Chapter 18 - Maxima and Minima (Ex 18.4) Exercise 18.4 - Free PDF
FAQs on RD Sharma Class 12 Solutions Chapter 18 - Maxima and Minima (Ex 18.4) Exercise 18.4
1. What types of problems are primarily featured in RD Sharma Class 12 Solutions Chapter 18, Exercise 18.4?
Exercise 18.4 of Chapter 18 - Maxima and Minima focuses predominantly on application-based word problems. These questions require you to translate a real-world scenario into a mathematical function and then use calculus to find the maximum or minimum value of a specific quantity, such as maximising the volume of a box or minimising the surface area of a cylinder.
2. What is the fundamental step-by-step method for solving the maxima and minima word problems in Exercise 18.4?
To solve the problems in this exercise correctly, follow these key steps:
- Step 1: Carefully read the problem to identify the quantity that needs to be maximised or minimised and all given constraints.
- Step 2: Formulate a mathematical function for this quantity in terms of a single variable.
- Step 3: Calculate the first derivative of the function (f'(x)) and set it to zero to find the critical points.
- Step 4: Calculate the second derivative (f''(x)) and use the Second Derivative Test to determine if a critical point corresponds to a maximum (f''(x) < 0) or a minimum (f''(x) > 0).
- Step 5: State the final answer clearly in the context of the original problem.
3. Why is the Second Derivative Test generally preferred for solving problems in RD Sharma's Exercise 18.4?
The Second Derivative Test is crucial because finding the critical points using the first derivative (f'(x) = 0) only tells you where a potential maximum or minimum might occur. It doesn't distinguish between them. The second derivative checks the concavity of the function at that point. A negative second derivative indicates the graph is concave down, signifying a maximum, while a positive value indicates it is concave up, signifying a minimum. This provides a definitive way to classify the nature of the extreme point.
4. How can I formulate the correct function from a word problem in this exercise?
The key is to first identify the primary quantity you need to optimize (e.g., Area 'A' or Volume 'V'). Then, write an equation for it using the variables given in the problem. Usually, this equation will have more than one variable. Use the secondary information or constraints provided in the problem (e.g., fixed perimeter, given surface area) to create a second equation. Solve this second equation to express one variable in terms of the other, and substitute it back into your primary equation to get a function of a single variable.
5. Are the solutions for RD Sharma Class 12 Ex 18.4 on Vedantu aligned with the latest CBSE syllabus for 2025-26?
Yes, all solutions for RD Sharma Class 12 Maths Chapter 18, Exercise 18.4, are meticulously prepared by subject matter experts at Vedantu. They are fully aligned with the latest CBSE 2025-26 syllabus and marking scheme, ensuring that you learn the correct methods required to score full marks in your board examinations.
6. What is a common mistake to avoid when solving problems involving geometric shapes in Ex 18.4?
A common mistake is incorrectly formulating the initial function for the quantity to be optimised, such as the volume or surface area of a shape. For example, when dealing with an open-top box, students might accidentally use the formula for a closed box. Always double-check the problem statement for such details and ensure you are using the correct geometric formulas and applying the given constraints accurately.
7. Do the solutions for this exercise cover both local and absolute maxima/minima concepts?
While the techniques used (like the second derivative test) find local maxima or minima, the context of the word problems in Exercise 18.4 almost always implies finding the absolute maximum or minimum practical value. The solutions guide you to find the single critical point that makes practical sense for the problem, which in a well-defined physical scenario, corresponds to the absolute extremum required.
