Optimisation problems: turning a word problem into a derivative

Optimisation problems: turning a word problem into a derivative

Optimisation problems in IB Maths are a test of more than just calculus. They are often a test of problem comprehension, variable definition, and strategic thinking. My students, when they first encounter these problems, sometimes feel overwhelmed by the wall of text before they even get to the maths. They see the words and immediately want to jump to differentiating something. This is a common pitfall. The core of an optimisation problem is rarely the differentiation itself, but rather the translation from a real-world scenario into a solvable mathematical model. That translation is what we will focus on here.

Over my ten years teaching IB Maths, I have observed that students who master optimisation problems are not necessarily the ones who are fastest at differentiation. They are the ones who can systematically break down a complex problem into manageable parts. They understand that before any calculus can happen, a function representing the quantity to be optimised must be constructed. This article will guide you through that construction process, from the initial reading of the problem to setting up the derivative, with a focus on practical steps you can take in your own study.

Step 1: Understand the Goal and Identify Variables

The first step in any optimisation problem is to clearly understand what you are being asked to optimise. Is it area, volume, cost, time, profit, distance? This target quantity is what you will ultimately need to express as a function. Once you know what to optimise, identify all the quantities that can vary. These are your variables. In my classroom, I encourage students to write these down explicitly, perhaps drawing a diagram if appropriate. For instance, if you are optimising the area of a rectangle, your variables might be length $L$ and width $W$.

Consider a classic problem: "A farmer wants to fence off a rectangular plot of land adjacent to a river. No fence is needed along the river. He has $200 \text{ m}$ of fencing. What dimensions will maximise the area of the plot?"

• What to optimise? Area of the rectangular plot, $A$.
• What are the variables? Length of the sides of the rectangle. Let $x$ be the length of the sides perpendicular to the river, and $y$ be the length of the side parallel to the river.

Drawing a quick sketch helps here. Imagine the river as one boundary, and the three sides of the fence making up the other three boundaries. The fencing material represents a constraint, which leads us to the next step.

Step 2: Formulate the Primary and Secondary Equations

Once you have identified what to optimise and what your variables are, the next crucial step is to write down the mathematical relationships. You will typically have two types of equations:

1. Primary Equation (Objective Function): This is the equation for the quantity you want to optimise. It will usually involve two or more variables.
2. Secondary Equation (Constraint Equation): This equation relates the variables based on the given constraints in the problem. This is critical for reducing the number of variables in your primary equation.

Let's return to our farmer problem:

• Primary Equation: We want to maximise the area $A$. For a rectangle, $A = xy$. This is our objective function.
• Secondary Equation: The farmer has $200 \text{ m}$ of fencing. The fence covers two sides of length $x$ and one side of length $y$. So, $2x + y = 200$. This is our constraint.

My students often try to differentiate the primary equation $A=xy$ at this stage. But $A$ is a function of two variables, $x$ and $y$. In IB Maths, we typically only differentiate functions of a single variable. This is where the secondary equation comes in. You need to use the secondary equation to express one variable in terms of the other, and then substitute this into the primary equation.

Step 3: Reduce the Objective Function to a Single Variable

This step is often where students either succeed or get stuck. Using the constraint equation, solve for one variable in terms of the other. The goal is to substitute this expression into your primary equation, making it a function of a single variable. Which variable should you solve for? Choose the one that simplifies the substitution process. Sometimes, it makes no difference; other times, one choice leads to a much cleaner algebraic expression.

For the farmer problem:

From the secondary equation, $2x + y = 200$, we can easily solve for $y$: $y = 200 - 2x$.

Now, substitute this expression for $y$ into the primary equation $A = xy$:

$A(x) = x(200 - 2x)$

$A(x) = 200x - 2x^2$

Now we have our objective function, $A(x)$, expressed solely in terms of $x$. This is the function we are ready to differentiate. This process applies to problems across both SL Applications and Interpretation and HL Applications and Interpretation, and often involves algebraic manipulation that needs precision.

Step 4: Determine the Domain of the Function

Before differentiating, it's good practice to consider the realistic domain of your function. What are the possible values for $x$? This is often overlooked but can be crucial for interpreting your final answer, especially if the maximum or minimum occurs at an endpoint of the domain rather than a critical point.

For $A(x) = 200x - 2x^2$:

• Length $x$ must be positive, so $x > 0$.
• Length $y = 200 - 2x$ must also be positive, so $200 - 2x > 0 \Rightarrow 200 > 2x \Rightarrow x < 100$.

So, the domain for $x$ is $0 < x < 100$. This means that any optimal $x$ must lie within this range. If we find a critical point outside this range, it's not a valid solution for the physical problem.

Step 5: Differentiate and Find Critical Points

With $A(x) = 200x - 2x^2$, we can now apply calculus. To find the maximum area, we need to find the critical points by setting the first derivative equal to zero.

First, find the derivative with respect to $x$:

$\frac{dA}{dx} = \frac{d}{dx}(200x - 2x^2)$

$\frac{dA}{dx} = 200 - 4x$

Next, set the derivative to zero and solve for $x$:

$200 - 4x = 0$

$4x = 200$

$x = 50$

This value of $x=50$ is a critical point. It falls within our domain $0 < x < 100$, so it's a plausible solution.

Step 6: Verify the Nature of the Critical Point and Answer the Question

After finding critical points, you must determine if they correspond to a maximum or minimum. You can use either the first derivative test (checking the sign of $\frac{dA}{dx}$ around $x=50$) or the second derivative test.

Using the second derivative test:

$\frac{d^2A}{dx^2} = \frac{d}{dx}(200 - 4x) = -4$

Since $\frac{d^2A}{dx^2} = -4 < 0$, the critical point $x=50$ corresponds to a local maximum. This confirms we have found the dimensions that maximise the area.

Finally, reread the question to ensure you answer exactly what was asked. The question asked for "What dimensions will maximise the area?"

• We found $x=50 \text{ m}$.
• Substitute $x=50$ back into the expression for $y$: $y = 200 - 2(50) = 200 - 100 = 100 \text{ m}$.

So, the dimensions that maximise the area are $50 \text{ m}$ by $100 \text{ m}$.

If the question had asked for the maximum area itself, we would substitute $x=50$ and $y=100$ into $A=xy$: $A = 50 \times 100 = 5000 \text{ m}^2$.

This systematic approach, moving from understanding the problem to defining variables, setting up equations, reducing variables, determining domain, differentiating, and finally verifying, is what I teach my students. It takes the guesswork out of these multi-step problems. Practicing with a variety of IB Maths notes and example problems is key to solidifying these steps. Remember to review your algebra skills, as errors in simplification can derail an otherwise perfectly understood calculus problem.

Optimisation problems might seem daunting at first glance, but they are a fantastic way to link abstract mathematical concepts to real-world applications. By focusing on the structured process of translating the word problem into a single-variable function, you unlock the power of calculus. Don't rush to differentiate; take your time with the setup. This deliberate approach will not only earn you more marks but also build a deeper understanding of mathematical modelling. Keep practicing these steps, and you'll find these problems become far less intimidating and much more manageable.