Application of Derivative Case Study Questions – Free Online Test Preparation
The Application of Derivative Case Study Questions are a vital component of Class 12 mathematics. These problems combine calculus theory with practical applications. They are often framed as math case study questions that require both calculation and interpretation skills.
In board exams, math case study questions for class 12 frequently focus on optimization, maxima and minima, rate of change, and related rates. Moreover, they often represent real-life scenarios such as designing containers, calculating profit, or predicting motion. Therefore, students should master the concepts before attempting complex problems.
Solving class 12 math case study questions builds analytical thinking and improves exam performance. However, understanding the question’s context is as important as applying formulas. For example, an optimization problem may require creating an equation from a practical description before differentiating.
Furthermore, regular practice through free online tests strengthens time management skills. These tests also expose students to various question patterns. Finally, combining solved examples with timed exercises ensures a balanced preparation strategy for tackling Application of Derivative Case Study Questions effectively.
Consistent practice with math case study questions, math case study questions for class 12, and class 12 math case study questions will not only boost marks but also prepare learners for competitive exams and higher studies in mathematics-related fields.
Case Study 3
A manufacturing company produces cylindrical containers for packaging food. The total surface area of each container is fixed at 100 cm2, and the company wants to determine the dimensions (radius and height) that would maximize the volume. The volume \( V \) of a cylinder is given by \( V = \pi r^2 h \), and the surface area is \( 2\pi r^2 + 2\pi rh = 100 \). Using these equations, the design team applies the concept of maxima and minima to derive the most efficient packaging design. This helps reduce cost and increase storage efficiency. The application of calculus through the first derivative test enables them to find the point at which the volume is maximized, providing a real-world use of optimization in business.
MCQ Questions:
1. What is the expression for the height h in terms of r using the surface area constraint?
Given \( 2\pi r^2 + 2\pi rh = 100 \), solving for \( h \): \( 2\pi rh = 100 – 2\pi r^2 \Rightarrow h = \frac{100 – 2\pi r^2}{2\pi r} \)
2. Substitute the expression of \( h \) into volume \( V \) to get \( V(r) \):
\( V = \pi r^2 h = \pi r^2 \cdot \frac{100 – 2\pi r^2}{2\pi r} = \frac{r(100 – 2\pi r^2)}{2} \)
3. What is the first derivative \( \frac{dV}{dr} \) of \( V = \frac{r(100 – 2\pi r^2)}{2} \)?
Using product rule: \( V = \frac{1}{2}(100r – 2\pi r^3) \)
\( \frac{dV}{dr} = \frac{1}{2}(100 – 6\pi r^2) = 50 – 3\pi r^2 \)
4. Find the critical point by solving \( \frac{dV}{dr} = 0 \).
\( 50 – 3\pi r^2 = 0 \Rightarrow r^2 = \frac{50}{3\pi} \Rightarrow r = \sqrt{\frac{50}{3\pi}} \) Note: This value is not in the given options — so answer is “None of the above.”
5. What does the second derivative tell us about this critical point?
\( \frac{d^2V}{dr^2} = -6\pi r \), which is negative for \( r > 0 \), indicating a maximum point.