Mathematics: Class 12 (Book 1) — Chapter 8: Mathematical Modelling

This is the transition from a real-world problem to a mathematical abstraction.

• Identify the Problem: Clearly define what needs to be solved.
• State Assumptions: Simplify the problem by keeping primary variables and discarding negligible ones.
• Define Variables & Parameters: Assign mathematical symbols (e.g., $x, y, t, k$) to represent real-world quantities.
• Construct the Model: Build mathematical equations, inequalities, or graphs connecting these variables (e.g., $y = mx + c$ or $\frac{dy}{dx} = ky$).

Once the mathematical equations are formed, we use algebraic, geometric, or calculus-based methods to solve them.

• This phase operates entirely within the realm of mathematics, independent of the real-world context.
• Techniques may involve solving matrices, integrating differential equations, or finding optimal points using derivatives.

The mathematical solutions obtained in Phase 2 must be translated back into the context of the original real-world problem.

• For example, if a quadratic equation yields $t = -5$ and $t = 5$ for a time-based problem, we discard $t = -5$ because time cannot be negative in classical contexts.

The interpreted results are compared with actual real-world data or observations.

• If the model is accurate: The cycle concludes, and the model is accepted.
• If the model is inaccurate: We return to the Formulation phase to refine our assumptions or include previously discarded variables.

Mathematical Problem Statement: Formulate a model to calculate the amount $A$ after $n$ years if a principal $P$ is invested at an annual interest rate $r$ (in percentage) compounded annually.

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Step-by-Step Algorithmic Solution Trace:

1. Formulation: Let the initial principal be $P$. The interest after the first year is $I_1 = P \times \frac{r}{100}$.
2. Amount after 1st year: $A_1 = P + P\left(\frac{r}{100}\right) = P\left(1 + \frac{r}{100}\right)$.
3. Amount after 2nd year: The new principal is $A_1$.$$A_2 = A_1 \left(1 + \frac{r}{100}\right) = P\left(1 + \frac{r}{100}\right)\left(1 + \frac{r}{100}\right) = P\left(1 + \frac{r}{100}\right)^2$$
4. Generalizing for $n$ years via inductive reasoning gives the model:$$A = P\left(1 + \frac{r}{100}\right)^n$$

Final Evaluation Output: The mathematical model for compound interest is $A = P\left(1 + \frac{r}{100}\right)^n$.

Mathematical Problem Statement: A manufacturer wants to construct a cylindrical can to hold $1000\text{ cm}^3$ of oil. Find the dimensions that minimize the cost of the metal to construct the can.

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Step-by-Step Algorithmic Solution Trace:

1. Formulation: Let $r$ be the radius and $h$ be the height. Volume $V = \pi r^2 h = 1000$. Surface Area $S = 2\pi r^2 + 2\pi r h$.
2. Express $h$ in terms of $r$:$$h = \frac{1000}{\pi r^2}$$
3. Substitute $h$ into the Surface Area function to create a single-variable model:$$S(r) = 2\pi r^2 + 2\pi r \left(\frac{1000}{\pi r^2}\right) = 2\pi r^2 + \frac{2000}{r}$$
4. Solution (Calculus): Differentiate $S$ with respect to $r$ to find the critical point:$$S'(r) = 4\pi r - \frac{2000}{r^2} = 0 \implies 4\pi r^3 = 2000 \implies r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42\text{ cm}$$
5. Calculate $h$:$$h = \frac{1000}{\pi (5.42)^2} \approx 10.84\text{ cm}$$
6. Interpretation & Validation: The dimensions that minimize the surface area (and thus material cost) occur when the height is equal to the diameter ($h = 2r$).

Final Evaluation Output: Optimal dimensions: $r \approx 5.42\text{ cm}$ and $h \approx 10.84\text{ cm}$.