Optimisation and rate models: VCE Mathematical Methods Units 3 & 4

Loav · VCE Mathematical Methods · Units 3 & 4

D05 · Differential calculus · Current curriculum

Turn a practical question into a one-variable function, respect its feasible domain, and distinguish the largest amount from the fastest change.

1. Decide what is being optimised

An optimisation problem asks for the largest or smallest possible value of a quantity under stated restrictions. The target might be area, volume, cost, profit, time or a rate. Write that target explicitly before differentiating. A formula for perimeter does not automatically answer a maximum-area question, and a formula for total amount does not directly answer when the amount is increasing fastest.

Choose one independent variable and define it with units. Translate every physical restriction into a feasible domain. A length is normally positive; a time may be limited to an observation interval; a production count may be a whole number. These restrictions are part of the mathematical model, rather than a final cosmetic step. A stationary solution outside the feasible domain cannot be the requested optimum.

The word “maximum” also needs a reference set. A local maximum compares nearby inputs, while the greatest feasible value compares the entire allowed domain. On a closed interval, endpoints can be optimal even when their derivative is non-zero. On an open domain, a best value may not be attained at all. Decide which situation applies before collecting candidates.

2. Use the constraint to obtain one variable

Many geometry problems begin with two dimensions. Use a perimeter, volume or material constraint to express one dimension in terms of another, then substitute into the objective. This produces a function of a single variable that can be differentiated using ordinary rules. State the allowable interval from the requirement that every dimension remains physically meaningful.

It helps to keep the constraint and objective on separate lines. The constraint says which designs are possible; the objective says how good each possible design is. Substituting the objective into itself, or differentiating while treating a constrained dimension as constant, changes the problem. Once the one-variable formula is established, check its units as a quick test of the algebra.

3. Verify optimality instead of stopping at a derivative zero

Solving F′(x) = 0 gives candidates. Use a derivative sign change, a complete-square argument, or a comparison of all valid candidates to show which is optimal. When the domain is closed and the function continuous, include endpoints. If the function has an interior corner or other non-differentiable input, evaluate it as another candidate. The absence of a derivative there does not prevent an optimum.

For an unbounded positive-length domain, inspect what happens as the variable approaches zero and as it becomes large. A single stationary point can be a global minimum when the objective decreases before it and increases after it throughout the entire domain. A graph can help you see the pattern, but the written solution should identify the evidence supporting the conclusion.

Completing the square is a useful independent check for quadratic objectives. The fence example can be written A(x) = 364.5 − 2(x − 13.5)². Since the square is non-negative, no feasible input can exceed 364.5. Calculus and algebra tell the same story, which strengthens confidence in the model and the arithmetic.

4. Revenue, cost and profit are different quantities

Revenue is money received from sales; cost is money spent; profit is revenue minus cost. If the sale price changes with quantity, revenue is quantity multiplied by that price, not simply the price function. Fixed costs contribute a constant to profit. They change the achieved profit but do not change the derivative in a smooth model with otherwise unchanged assumptions.

Interpret the model’s units carefully. If q counts individual items, a derivative of profit with respect to q is measured in dollars per additional item. If q represents hundreds of items, the numerical optimum and derivative units change accordingly. A continuous model can guide an integer decision, but nearby feasible whole-number choices should be compared before giving a final production quantity.

5. Maximum amount and maximum rate answer different questions

If Q(t) represents an amount, then Q′(t) is its instantaneous rate of change. To maximise the amount, analyse Q. To maximise its rate, analyse the new function r(t) = Q′(t). A time when Q′(t) = 0 is a stationary time for the amount, not generally the time of fastest increase. The quantity can still be increasing when its rate has already started to decline.

A locally greatest or least gradient corresponds to a point of inflection of a smooth quantity graph. If that gradient is non-zero, the point is non-stationary. You can locate the rate’s extremum from its formula or graph; a quadratic rate is especially clear after completing the square. Always report the rate in amount-per-time units and the amount in its original units.

6. Motion is another quantity-and-rate model

If s(t) is position along a line, its derivative v(t) is velocity. Positive velocity means motion in the chosen positive direction; negative velocity means motion in the opposite direction. Speed is |v(t)|, so it cannot be negative. For s(t) = t² − 6t + 4 metres on 0 ≤ t ≤ 5 seconds, v(t) = 2t − 6 m/s. At t = 1, position is −1 m, velocity is −4 m/s and speed is 4 m/s. These are three distinct descriptions.

The object changes direction at t = 3 because velocity changes from negative to positive there. A zero velocity alone would not establish a direction change; inspect its signs. On a position-time graph, the slope gives velocity, while the vertical coordinate gives position. An average velocity uses change in position divided by elapsed time. Distance travelled and displacement differ when direction changes.

7. Handle integer choices and numerical answers honestly

If an optimisation variable must be an integer, a non-integer stationary input is not a feasible final answer. Use the continuous result to identify nearby integers, evaluate the objective at those choices and compare other required candidates. For a simple single-peaked quadratic, the two adjacent integers are the natural comparison. More complicated functions require checking their actual behaviour rather than assuming ordinary rounding always works.

When the optimum is numerical, retain full precision in intermediate calculations. Rounding a dimension before calculating a volume can alter the reported objective value. Give exact forms where appropriate, then a decimal with units and stated accuracy if requested. Avoid claiming greater practical precision than the model’s data supports: a theoretical optimum is a consequence of assumptions, not a guarantee of real-world performance.

8. A modelling answer has three stages

First formulate: define variables, write the constraint, form the objective and state the feasible domain. Then solve: differentiate, identify candidates and establish the optimum. Finally interpret: give the requested dimensions, time, quantity or rate with units and explain any relevant boundary or integer issue. Leaving out the final interpretation can make a correct symbolic solution incomplete.

Check the chosen design against the original constraint, not just the reduced formula. Check that every quantity is meaningful, that no excluded input was used, and that the answer corresponds to the target named in the question. A quick substitution into the original context often catches an algebraic error that a derivative calculation would not reveal.

Practice: seven original questions

1. A rectangle has perimeter 80 m. Find its greatest possible area and dimensions.
2. A rectangular enclosure beside a straight wall uses 70 m of fencing for its other three sides. Find the dimensions giving maximum area.
3. A business has R(q) = q(50 − 0.1q) and C(q) = 10q + 300, in dollars, for 0 ≤ q ≤ 400. Find the quantity and maximum profit.
4. Q(t) = 100 + 12t − t² litres on 0 ≤ t ≤ 8 minutes. Find the greatest amount and the rate at t = 2.
5. A rate is r(t) = 18t − 3t² L/min on 0 ≤ t ≤ 6. Find the maximum rate and when it occurs. Does this alone specify the amount in the tank at that time?
6. A closed cylinder has volume 54π cm³. Its surface area is S = 2πr² + 2πrh. Find the positive radius and height giving minimum surface area.
7. An integer production choice n ∈ {0, 1, …, 12} has profit P(n) = −2n² + 29n − 20 dollars. Use the continuous model to identify and verify the best integer choice.