Mathematical Modelling (Advanced) (Appendix)

This appendix builds in four movements, each anchored to one worked NCERT example.

Why modelling matters and what it demands. We open with the motivating question — why not just solve word problems directly? — and the eight numbered principles NCERT lists behind every model, from identifying the need and the parameters to the consistency checks and the room for improvement.

The five-step cycle itself, the compressed engine that all eight principles feed into, together with a flowchart figure showing how Step 5 can loop back into Step 2.

Four fully worked models, one per major tool already in your syllabus: trigonometric modelling for an inaccessible tower's height, matrix modelling for a manufacturer's raw-material planning, linear-programming modelling for a profit-maximising production schedule, and calculus modelling for salt concentration in a mixing tank. Two are developed as full step-by-step derivations, because the method — not just the final number — is the object being taught.

Honest limitations, closing the chapter: mathematical modelling is powerful but not omnipotent, and knowing where it strains is as much a part of mastering it as knowing how to build a model in the first place.

Why mathematical modelling is needed

You already know how to solve word problems in arithmetic, algebra, trigonometry and linear programming — someone poses a scenario, you translate it into an equation, you solve. Mathematical modelling is what happens before that translation is handed to you. Consider: finding the width of a river you cannot cross, the optimal launch angle in a shot-put throw, the height of a tower you cannot climb, the temperature at the Sun's surface, the mass of the Earth, or India's population in a future year without waiting for it to arrive. None of these comes with a pre-written equation. Each demands physical insight — introducing the right physical laws and symbols — before any solving can begin. That translation step, done with discipline rather than guesswork, is mathematical modelling.

The eight principles behind every model

NCERT frames mathematical modelling as a principled activity, not an ad-hoc guess, and lists the instructions that should guide the construction of any model: Identify the need for the model — what exactly are we looking for. List the parameters and variables required for the model. Identify the available relevant data — what is actually given. Identify the circumstances that can be applied, i.e., the assumptions. Identify the governing physical principles. Identify the equations that will be used, the calculations that will be made, and the solution that will follow. Identify tests that can check the consistency of the model and its utility. And identify the parameter values that can improve the model. These eight principles are almost philosophical — they are a checklist for thinking, not a formula. They compress, however, into a concrete five-step procedure that you can actually execute on paper.

The five-step cycle of mathematical modelling

Step 1: Identify the physical situation. State plainly what real-world problem is being solved, in its own terms, before touching any mathematics.

Step 2: Convert the physical situation into a mathematical model by introducing parameters and variables and invoking known physical laws and symbols. This is the formulation step — the situation becomes an equation, a matrix relation, an inequality system, or a differential equation.

Step 3: Find the solution of the mathematical problem. This is ordinary mathematics — trigonometric manipulation, matrix reduction, corner-point evaluation, or solving a differential equation — using whatever techniques the formulation demands.

Step 4: Interpret the result in terms of the original problem, and compare that result with observations or experiments. A mathematical answer is not useful until it is translated back into the language of the original situation and checked against what actually happens.

Step 5: If the result agrees well with observation, accept the model. If it does not, modify the hypotheses or assumptions to better reflect the physical situation, and return to Step 2. Modelling is therefore not a straight line but a loop: disagreement at Step 4 is information, not defeat, and it feeds directly back into a revised Step 2.

Model one: measuring an inaccessible tower with trigonometry

The simplest illustration is Example 1 of the NCERT text: find the height of a tower that cannot be climbed. Let $AB$ be the tower and $PQ$ an observer of eye-height $PQ = h$, standing a baseline distance $l = PC = QB$ from the tower, who measures an angle of elevation $\alpha$ to the top of the tower using an instrument such as a sextant. Right-triangle trigonometry on $\triangle PCA$ gives $\tan\alpha = \dfrac{AC}{PC} = \dfrac{H-h}{l}$, which rearranges to $H = h + l\tan\alpha$. Since $h$, $l$ and $\alpha$ are all measurable by the observer, this single relation is the solution. If even the foot of the tower is inaccessible — so $l$ itself is unknown — a second reading, the angle of depression $\beta$ from $P$ down to the foot $B$, supplies $\tan\beta = \dfrac{PQ}{QB} = \dfrac{h}{l}$, so $l = h\cot\beta$, which can be substituted back into $H = h + l\tan\alpha$. No refinement loop (Step 5) is needed here because $h$, $l$, $\alpha$ and $\beta$ are all known exactly to the precision of the instrument.

Model two: matrices for raw-material planning

A firm making products $P_1, P_2, P_3$ from raw materials $R_1, R_2, R_3$ has purchase orders from two clients $F_1$ and $F_2$. Let matrix $A$ (order-by-product) record how many units of each product each client wants, and matrix $B$ (product-by-material) record how much raw material one unit of each product needs. The product $AB$ (order-by-material) then gives exactly the raw material each order requires, compared against stock to see if it suffices. NCERT works this concretely: $AB$ demands $335$ units of $R_1$, $467$ of $R_2$ and $147$ of $R_3$ against a stock of only $330$, $455$ and $140$ — the model fails Step 4's reality check. Step 5 then bites: either procure more raw material, or revise the orders downward (a smaller matrix $A_1$), which NCERT shows brings the requirement to $311$, $436$ and $138$ units — comfortably within stock. A companion question — given fixed stock and $B$, what quantities $x, y, z$ exactly exhaust the raw material — is answered in the first derivation below.

Model three: linear programming for production scheduling

A manufacturer of medicines $M_1$ and $M_2$ has raw material enough for $20000$ bottles of $M_1$ and $40000$ bottles of $M_2$, only $45000$ bottles total, $3$ hours to fill $1000$ bottles of $M_1$, $1$ hour to fill $1000$ bottles of $M_2$, and $66$ hours of operating time. With profits of Rupees $8$ per bottle of $M_1$ and Rupees $7$ per bottle of $M_2$, the objective function to maximise is $Z = 8x + 7y$, subject to $x \le 20000$, $y \le 40000$, $x + y \le 45000$, $3x + y \le 66000$, and $x, y \ge 0$ — a linear-programming model built entirely from the syllabus you met in the chapter on Linear Programming. Evaluating $Z$ at every corner point of the feasible region — $(0,0)$, $(20000,0)$, $(20000,6000)$, $(10500,34500)$, $(5000,40000)$, $(0,40000)$ — shows the maximum, Rupees $325500$, occurs at $(10500, 34500)$: produce $10500$ bottles of $M_1$ and $34500$ bottles of $M_2$.

Model four: a differential equation for a mixing tank

A tank holds $1000$ litres of brine with $250$ g of salt per litre — an initial salt content of $250$ kg. Fresh brine of concentration $200$ g per litre flows in at $25$ litres per minute, and the well-stirred mixture flows out at the same rate. Letting $y(t)$ be the salt (kg) at time $t$ (minutes), the inflow adds salt at $25 \times 200\text{ g} = 5$ kg per minute, while the outflow removes it at $25 \times \dfrac{y}{1000} = \dfrac{y}{40}$ kg per minute. Balancing these rates gives the governing equation $\dfrac{dy}{dt} = 5 - \dfrac{y}{40}$, a first-order linear equation solved fully in the second derivation below.

Board level - the width of a river using two elevation readings

Problem. A surveyor stands at point $P$ on one bank of a river and sights a tree $T$ directly across on the far bank, measuring an angle of elevation $\alpha = 30$ degrees to the top of the tree, whose height is known to be $h = 20$ m (the surveyor's own eye height is negligible for this estimate). Model the river's width $l$.

Step 1. Identify the situation: an inaccessible distance found from an angle of elevation and a known tree height, directly analogous to the tower model.

Step 2. Formulate. Let $l$ be the river's width, treating the tree's base as level with the surveyor (a stated, simplifying assumption). Right-triangle trigonometry gives $\tan\alpha = \dfrac{h}{l}$.

Step 3. Solve. $l = \dfrac{h}{\tan\alpha} = \dfrac{20}{\tan 30^\circ} = \dfrac{20}{1/\sqrt3} = 20\sqrt3 \approx 34.64$ m.

Step 4. Interpret. The river is approximately 34.6 m wide. This estimate is only as good as the assumption that the tree's base is level with the observer's eye; a real survey would use two readings (elevation and depression, as in the tower model) to remove that assumption, which is exactly the kind of refinement Step 5 of the cycle calls for.

Board level - checking a revised purchase order against raw material stock

Problem. Using the raw-material requirement matrix $B = \begin{pmatrix} 3 & 4 & 0 \\ 7 & 9 & 3 \\ 5 & 12 & 7 \end{pmatrix}$ (rows $P_1, P_2, P_3$, columns $R_1, R_2, R_3$) and a revised order matrix $A_1 = \begin{pmatrix} 9 & 12 & 6 \end{pmatrix}$ (units of $P_1, P_2, P_3$ ordered by client $F_1$), verify whether this order can be met from stock of $330$ units of $R_1$, $455$ of $R_2$, $140$ of $R_3$.

Step 1. Identify the situation: check whether one client's order is within raw-material stock, i.e., compute $A_1 B$ and compare each entry to the corresponding stock figure.

Step 2. Formulate. Compute the row vector $A_1 B = \begin{pmatrix} 9 & 12 & 6 \end{pmatrix}\begin{pmatrix} 3 & 4 & 0 \\ 7 & 9 & 3 \\ 5 & 12 & 7 \end{pmatrix}$.

Step 3. Solve. The $R_1$ entry is $9(3) + 12(7) + 6(5) = 27 + 84 + 30 = 141$. The $R_2$ entry is $9(4) + 12(9) + 6(12) = 36 + 108 + 72 = 216$. The $R_3$ entry is $9(0) + 12(3) + 6(7) = 0 + 36 + 42 = 78$. So $A_1B = \begin{pmatrix} 141 & 216 & 78 \end{pmatrix}$.

Step 4. Interpret. The order requires 141 units of R1, 216 units of R2 and 78 units of R3, all comfortably below the stock of $330$, $455$ and $140$ respectively — the revised order is feasible, confirming Step 5's "accept the model" branch without any further refinement needed.

Higher-order thinking (HOTS) - two tanks in series, tracking salt into a second tank

Problem. Suppose the outflow of the brine tank from the chapter's core model feeds directly into a second, initially empty tank of the same $1000$-litre capacity, whose own outflow (also $25$ litres per minute, well-stirred) drains away and is discarded. Set up the differential equation governing $z(t)$, the salt content of the second tank, and confirm its behaviour is consistent at $t=0$. Timing note: allow about 12 to 15 minutes — chaining two rate-balance models together is more about correct setup than heavy computation.

Step 1. Identify the situation: this extends Model four by treating the first tank's outflow as the second tank's inflow — a two-stage mixing process, common in chemical engineering, built from the same rate-balance idea used once already.

Step 2. Formulate. Tank one's solution is already known: $y(t) = 200 + 50e^{-t/40}$, so it feeds tank two at rate $25 \times \dfrac{y(t)}{1000} = \dfrac{y(t)}{40}$ kg per minute. Tank two, starting empty ($z(0) = 0$), loses salt at rate $\dfrac{z(t)}{40}$ kg per minute (well-stirred, same in and out flow of $25$ litres per minute, so its volume stays fixed at $1000$ litres). The governing equation is $\dfrac{dz}{dt} = \dfrac{y(t)}{40} - \dfrac{z}{40} = \dfrac{1}{40}\left(200 + 50e^{-t/40}\right) - \dfrac{z}{40}$, i.e., $\dfrac{dz}{dt} + \dfrac{1}{40}z = 5 + \dfrac{5}{4}e^{-t/40}$, a first-order linear equation with a time-varying forcing term supplied by tank one's own solution.

Step 3. Confirm consistency at $t = 0$. Here $z(0) = 0$, so $\dfrac{dz}{dt}\Big|_{t=0} = \dfrac{y(0)}{40} - \dfrac{0}{40} = \dfrac{250}{40} = 6.25$ kg per minute — matching the expectation that with tank one at maximum concentration and tank two empty, salt should flow in at its fastest rate of the whole process.

Step 4. State the answer. The governing equation for tank two is $\dfrac{dz}{dt} + \dfrac{1}{40}z = 5 + \dfrac{5}{4}e^{-t/40}$ with $z(0)=0$, a linear equation solvable by the same integrating-factor method as the second derivation above, now with a $t$-dependent forcing term $Q(t)$ — the natural next step for a student who wants to push the model further.

Higher-order thinking (HOTS) - re-optimising the medicine production model under a new constraint

Problem. In the profit-maximising medicine model (Model three), suppose a new regulation caps the combined weekly output of $M_1$ and $M_2$ bottles at $40000$ instead of $45000$ (all other constraints unchanged: $x \le 20000$, $y \le 40000$, $3x + y \le 66000$, $x,y \ge 0$). Find the new profit-maximising production plan and profit, and identify which original corner point survives. Timing note: allow about 15 to 18 minutes — re-deriving vertices under one changed constraint and re-evaluating the objective function is the standard linear-programming workflow, but with fresh arithmetic.

Step 1. Identify the situation: this is Step 5 of the modelling cycle in action — a changed real-world circumstance requires returning to Step 2 and re-solving with a revised constraint, exactly as NCERT's own remark about revising purchase orders illustrates for the matrix model.

Step 2. Formulate. Objective function unchanged: $Z = 8x + 7y$. Constraints: $x \le 20000$, $y \le 40000$, $x + y \le 40000$, $3x + y \le 66000$, $x, y \ge 0$. Find the new corner points of the feasible region.

Step 3. Solve for the vertices. $x + y = 40000$ meets $3x + y = 66000$ where subtracting gives $2x = 26000$, so $x = 13000$, $y = 27000$; both satisfy $x \le 20000$ and $y \le 40000$, so this is a genuine vertex. The point $(20000, 20000)$, where $x+y=40000$ meets $x=20000$, fails $3(20000)+20000 = 80000 > 66000$ and is cut off — the old vertex $(20000, 6000)$ survives instead, since $x+y=26000 \le 40000$ there. So the candidate corner points are $(0,0)$, $(20000,0)$, $(20000,6000)$, $(13000,27000)$, $(0,40000)$.

Step 4. Evaluate $Z$ at each surviving vertex. $Z(0,0)=0$. $Z(20000,0)=160000$. $Z(20000,6000)=160000+42000=202000$. $Z(13000,27000)=8(13000)+7(27000)=104000+189000=293000$. $Z(0,40000)=280000$. The maximum is at $(13000, 27000)$.

Step 5. State the answer. The new profit-maximising plan is 13000 bottles of M1 and 27000 bottles of M2, giving a maximum profit of Rupees 293000 — lower than the unrestricted maximum of Rupees $325500$, exactly the expected direction: tightening a constraint can only reduce, never increase, the achievable maximum, a useful consistency check on the arithmetic itself.

A mathematical model is not the same thing as the physical reality it represents — treating its output as automatically exact, without returning to Step 4 to check it against observation, is the single most common misuse. Students also skip stating assumptions explicitly at Step 2, which makes it impossible to know later which assumption to revise if the prediction disagrees with reality. In the matrix model, a frequent slip is multiplying $A$ and $B$ in the wrong order — only $AB$, not $BA$, is the correctly dimensioned "orders times per-unit-requirement" computation here. In the linear-programming model, students forget to check that a computed intersection point lies within all the other constraints before calling it a corner point, exactly the check performed in the second Higher-order thinking example above. In the differential-equation model, a common slip is dropping the negative sign in the outflow term, writing $\dfrac{dy}{dt} = 5 + \dfrac{y}{40}$ instead of $5 - \dfrac{y}{40}$ — fix the direction of each rate (in is positive, out is negative) before writing the equation down.

• State the five-step mathematical modelling cycle from memory and explain what happens at each step.
• Explain why Step 5 sometimes loops back to Step 2, and give a concrete example of a revised assumption from this chapter.
• Set up and solve the tower height model $H = h + l\tan\alpha$, including the depression-angle variant $l = h\cot\beta$ when the base is inaccessible.
• Build a raw-material requirement matrix model, correctly compute $AB$, and interpret the result against available stock.
• Solve a $3\times3$ linear system arising from a raw-material allocation model using elementary row operations.
• Formulate a linear-programming model — objective function plus constraints — from a word description of a production-scheduling problem, and find the optimum by evaluating corner points.
• Formulate a first-order linear differential equation from a rate-balance description (inflow rate minus outflow rate) and solve it using an integrating factor.
• Interpret a solved differential-equation model physically, including identifying long-run limiting values.
• Name at least two genuine limitations of mathematical modelling and explain why they matter.

Ask the tutor: Want to work through re-deriving the medicine production model's corner points with a different profit ratio, or set up your own two-stage mixing-tank model from scratch?