There are several applications of derivatives. One of them is finding extreme values (maxima or minima). And methods for finding maxima or minima values have practical applications in many areas of life. The problems of minimizing costs, lengths, times, etc. and maximizing profits, volumes, areas, etc. can be dealt with by these methods. I find students struggling with these optimization problems. The foremost struggle is to transform the word problems into a mathematical problem. Here’s a strategy I’ve developed over the years to tackle these types of problems effectively.
Observe, Plan, and Act
I’ve been teaching this approach for decades, and it’s consistently well-received by students. Let’s break the entire process into three simple steps:
1) Observe
In this step, we read the question attentively. Our observant eyes must search for G.O.D.
I recommend not writing anything at this stage. Just read, reflect, and observe.
- What is Given? Given is the quantity that does not change. Some times, it is mentioned there that this is given. Or, in some questions, they are just numerals.
- What to Optimize? Just identify the quantity that has to be maximized or minimized and the variables related to it.
- What to Determine? Identify what to be concluded at the end. This can be to find something, to deduce some relation, to prove something, etc.
2) Plan
In this step, we decide how to handle the quantities we observed in Step 1.
This is the time to explore possible paths, rough work is highly encouraged.
So, grab your pen and paper..
- The Given quantity is treated as a constant. Decide to assign such symbols to the given quantity (if not provided in question) that are used for constants.
- What to Optimize is the main function to be differentiated. Decide the convincing symbol for it and the related variables.
- What to Determine is the most important part. So, it is better to jot down the exact mathematical form of what is asked. If any relation is to established or something to prove, it is convenient to write these mathematically beforehand.
3) Act
So, now you are ready to present your solution. The following guidelines will help you to go through.
- Carefully. draw a diagram if needed. Depict the quantities on the diagram.
- Express the function to be optimized in terms of other variables. Let us call it the main function.
- If the main function has more than one (independent) variables, transform it into a function of one (independent) variable by eliminating other variables. For this, use the provided information to relate the independent variables.
- Do not forget to identify the domain of the main function once it is expressed as a function of a single variable.
- Differentiate the main function and find the critical numbers by setting the derivative equals zero.
- Accept only the appropriate critical numbers that are in the domain of the main function.
- Use the second derivative test or first derivative test, whatever be suitable, to decide whether the critical number corresponds to the maxima or to the minima.
- Once the decision is made, proceed for What to Determine.
Let’s Apply the Three-Step Strategy
To see this method in action, let’s solve a problem using our three-step approach.
The following question is taken from James Stewart’s Essential Calculus: Early Transcendentals.
Question: A cone-shaped drinking cup is made from a circular piece of paper of radius \(R\) by cutting out a sector and joining the edges \(CA\) and \(CB\). Find the maximum capacity of such a cup. Also, find the angle ACB when the volume is maximum.
Observe:
- We are given the radius \(R\).
- We have to maximize the capacity that is the volume of the cone. And it depends mainly on the radius and the height of the cone.
- We have to determine the maximum volume.
Plan:
- The given quantity is the radius which has already a symbol assigned to as \(R\).
- The volume is to be maximized, we decide to assign the symbol \(V\) to the volume function which has to be differentiated.
- Here, we have to find the maximum volume so nothing more to do. Just make sure, the final answer must contain \(R\).
Act: Now, let us write the solution.
Solution:
The volume of a cone, \(V = \frac{1}{3} \, \pi \, r^{2} h \)
Eliminating \(r^2\) by using the relation \(R^{2} = r^{2} + h^{2}\)we get the volume \(V\) expressed in a single variable \( h \) as
\[ V(h) = \frac{1}{3} \, \pi (R^{2} – h^{2}) h \]
which can be written as
\[ V(h) = \frac{1}{3} \, \pi (R^{2} h – h^{3}) \tag{1} \]
\[ V'(h) = \frac{1}{3} \, \pi (R^{2} – 3 h^{2}) \tag{2} \]
for \[ V'(h) = 0 \quad \text{implies} \quad \frac{1}{3} \, \pi (R^{2} – 3 h^{2}) =0 \]
\[ h = \frac{R}{\sqrt{3}} \tag{neglecting the negative value} \]
\( \text{from (2)} \quad V” (h) = -2\pi h \) \( \text{so} \quad V” \big( \frac{R}{\sqrt{3}} \big) = -2\pi R/\sqrt{3} < 0 \)
Thus at \(h = \frac{R}{\sqrt{3}}\), the volume is maximum.
And the maximum volume from (1) is
\[ V_{max} = \frac{2 \pi R^{3}}{9 \sqrt{3}} \]
To find the angle ACB, \(\theta\) we proceed as follow.
Since the length of the major arc (un-cut arc) is equal to the circumference of the base of the cone, we can write \( 2 \pi R – R \theta = 2 \pi \, r \)
\[ \text{or} \quad 2 \pi R – R \theta = 2 \pi \sqrt{R^{2} – h^{2}}\tag{using $ R^{2} = r^2 + h^2 $ } \]
Plugging \(h = \frac{R}{\sqrt{3}}\) and solving for \( \theta \) we get
\[ \theta = 2 \pi \big( 1 – \sqrt{\frac{2}{3}} \big) \]