Calculus Optimisation Problems

Optimisation is the mathematical engine behind countless real-world decisions, from designing fuel-efficient cars to maximizing corporate profits. In IB Mathematics, you are not just learning abstract calculus; you are mastering a powerful toolkit for solving practical problems by finding the best possible outcome—be it a maximum or a minimum. This skill hinges on your ability to translate a described scenario into a precise function, analyze it using derivatives, and interpret the results meaningfully within the original context.

From Words to Functions: The Art of Modelling

The most critical—and often most challenging—step in any optimisation problem is constructing the correct mathematical model. A mathematical model is a function, often single-variable, that represents the quantity you want to optimise (e.g., area, cost, profit). This rarely comes ready-made; you must derive it from the problem's constraints.

The process follows a clear strategy. First, identify the objective function. Ask yourself: "What quantity am I trying to maximize or minimize?" Label this variable, say $P$ for profit or $A$ for area. Second, identify all constraints and relationships. These are the fixed conditions of the problem, such as a fixed amount of fencing or a relationship between dimensions. Use these constraints to express your objective function in terms of a single independent variable. This often involves substitution.

For example, consider a classic problem: "You have 100 meters of fencing to enclose a rectangular plot adjacent to a river, so only three sides need fencing. Find the dimensions that maximize the area." Here, the objective function is area: $A=L\times W$. The constraint is the total fencing: $L+2W=100$. Solving the constraint for $L$ gives $L=100-2W$. Substituting this into the area function creates a model in one variable: $A(W)=(100-2W)\times W=100W-2W^2$. The problem is now purely mathematical: find the value of $W$ that maximizes $A(W)$.

Locating Candidates: Critical Points and Derivatives

Once you have your model $f(x)$, calculus takes center stage. To find where maximum or minimum values can occur, you must locate the function's critical points. A critical point occurs at a value $x=c$ if either $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined. For the smooth, continuous functions typical in IB optimisation, we primarily look where the first derivative equals zero.

The rationale is geometric: the derivative $f^{\prime }(x)$ represents the slope of the tangent line. At a peak (maximum) or a trough (minimum) of a smooth function, the tangent line is horizontal, meaning its slope is zero. Therefore, by finding where $f^{\prime }(x)=0$, you find all potential locations for local maxima and minima.

Using our fencing example, we have $A(W)=100W-2W^2$. The first derivative is $A^{\prime }(W)=100-4W$. Setting this equal to zero to find critical points: $$100-4W=0$$ $$4W=100$$ $$W=25$$ Thus, $W=25$ meters is a critical point and the prime candidate for maximizing the area. But is it truly a maximum? And is it the absolute best value? The derivative test helps answer the first question, while considering the domain answers the second.

Classifying Critical Points: The First and Second Derivative Tests

Finding a critical point tells you where something might be optimal; a derivative test confirms what it is. You have two reliable methods.

The First Derivative Test examines the sign of $f^{\prime }(x)$ around the critical point. You pick test values just to the left and right of the critical point $c$ and evaluate $f^{\prime }(x)$.

• If $f^{\prime }(x)$ changes from positive to negative at $c$, the function changes from increasing to decreasing, indicating a local maximum.
• If $f^{\prime }(x)$ changes from negative to positive at $c$, the function changes from decreasing to increasing, indicating a local minimum.
• If the sign does not change, $c$ is neither a max nor a min (it could be an inflection point).

The Second Derivative Test is often quicker for simple functions. You evaluate the second derivative at the critical point itself.

• If $f^{\prime \prime }(c)<0$, the graph is concave down at $c$, indicating a local maximum.
• If $f^{\prime \prime }(c)>0$, the graph is concave up at $c$, indicating a local minimum.
• If $f^{\prime \prime }(c)=0$, the test is inconclusive, and you must use the first derivative test.

Returning to $A(W)=100W-2W^2$ with critical point $W=25$, the second derivative is $A^{\prime \prime }(W)=-4$, which is always less than zero. Therefore, by the second derivative test, $W=25$ yields a local maximum. To find the corresponding length: $L=100-2(25)=50$ meters. The maximum area is $A(25)=100(25)-2(25{)}^2=1250$ m².

Contextual Interpretation and Applied Problem Types

The final, non-negotiable step is to translate your mathematical answer back into the language of the original problem. "The maximum area of 1250 m² is achieved with a width of 25 meters and a length of 50 meters." Always state your answer with units and context.

IB problems frequently fall into recognizable categories:

• Geometric Optimisation: Maximizing area or volume given a perimeter/surface area constraint (e.g., box volume, gutter cross-section).
• Economic Optimisation: Maximizing profit $P(x)=R(x)-C(x)$ or minimizing average cost $\frac{C(x)}{x}$. You often find the critical point by setting marginal revenue equal to marginal cost ($R^{\prime }(x)=C^{\prime }(x)$).
• Distance/Time Minimisation: Finding the point on a curve closest to a given point, or optimizing travel time across different media (e.g., rowing then walking).

For all types, you must consider the domain. Critical points are only valid if they lie within the problem's practical domain. For instance, a dimension cannot be negative. Always evaluate the objective function at the critical points and at the endpoints of the domain to find the absolute maximum or minimum.

Common Pitfalls

1. Forgetting to Check Endpoints: A function defined on a closed interval $[a,b]$ achieves its absolute maximum and minimum either at a critical point or at an endpoint $a$ or $b$. Neglecting to evaluate $f(a)$ and $f(b)$ is a common source of lost marks. Always state the domain of your variable based on the physical constraints.

1. Misinterpreting the Second Derivative Test: Remember, $f^{\prime \prime }(c)=0$ does not mean the critical point is not an extremum; it means the test fails. For example, $f(x)=x^4$ has a minimum at $x=0$, but $f^{\prime \prime }(0)=0$. You must revert to the first derivative test in these cases.

1. Poor Modelling from Word Problems: The single biggest error is incorrect translation from words to algebra. Carefully define your variables, write down every relationship, and ensure your final objective function is in one variable before differentiating. A misplaced square or incorrect substitution will derail the entire solution.

1. Neglecting Units and Context in the Final Answer: Presenting a naked number like "25" is incomplete. The final answer must be a full sentence that references the original question: "The manufacturer should produce 25 units per day to maximize profit." Omitting units (meters, dollars, etc.) or failing to specify what the variable represents shows a lack of contextual understanding.

Summary

• Modelling is Key: Success starts with accurately translating the word problem into a single-variable objective function using given constraints. This step defines the entire problem.
• Derivatives Locate Candidates: Find critical points by taking the first derivative and setting $f^{\prime }(x)=0$. These are the only interior points where local maxima or minima can occur.
• Tests Classify the Point: Use either the First Derivative Test (checking sign changes) or the Second Derivative Test (evaluating concavity) to determine if a critical point is a local maximum or minimum.
• Domain is Crucial: Always consider the practical domain of your variable. The absolute maximum or minimum on an interval is found by comparing the function's value at all critical points and at the endpoints.
• Interpret Your Answer: Conclude by stating your final numbers with proper units, clearly connected back to the context of the original problem. This demonstrates full-circle understanding.