Introduction to Optimization Problems: Study Guide (Calculus AB 5.10)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation and words

“Optimise” means find the largest (maximum) or smallest (minimum) value. The maximum value is a value of the function, such as 2700 m². The place where it happens, such as x = 30 m, is the location of the maximum. Questions often ask for both, so read carefully.

Intervals: [a, b] includes the endpoints and (a, b) does not.

What an optimization problem is

You already know how to find the absolute maximum and minimum of a function on an interval. In Topic 5.5 you used the candidates test: on a closed interval [a, b], a continuous function takes its extreme values only at critical points or at endpoints, so you evaluate the function at those points and compare.

An optimization problem is that same task with one extra step at the start. The function is not handed to you in finished form. Sometimes it is given as a model (“the number of visitors is N(t) = …”). Often you must build it from a description (“a fence of fixed length encloses the largest possible area”). Once you have a function of one variable on an interval, everything is Topic 5.5 to 5.7 again.

Why derivatives help: if a differentiable function has its largest value at a point inside the interval, the graph must be flat there, so the derivative is 0. That is why the search begins with critical points.

The common structure

Problems that look very different share one skeleton. Learn to see it before you calculate.

Part of the skeleton	Fencing problem (Worked example 2)	Number problem	Shape under a curve
Quantity to optimise	Enclosed area A	Sum S of two positive numbers	Area A of a rectangle
Variables	Fence lengths x and y	The numbers x and y	Half-width x, height y
Constraint	Total fencing is 180 m	Product xy = 64	Top corners lie on y = 9 − x²
One-variable function	A(x) = x(180 − 3x)	S(x) = x + 64/x	A(x) = 2x(9 − x²)
Interval	0 < x < 60	x > 0	0 < x < 3

The three rows “constraint”, “one-variable function” and “interval” are where the thinking happens. The calculus that follows is the same in every column: differentiate, find critical points, test them.

Where the interval comes from. Lengths must be positive. Here, 180 − 3x > 0 gives x < 60, and the curve 9 − x² is above the x-axis only for x < 3. A question about a model usually states the interval directly, for example 0 ≤ t ≤ 10.

The method

1. Name the quantity to optimise and say whether you want a maximum or a minimum.
2. Draw and label a diagram if there is a shape. Give every changing length a letter.
3. Write the quantity as a formula. It may use two variables at first.
4. Use the constraint to replace one variable, so the quantity is a function of a single variable.
5. State the interval of allowed values.
6. Differentiate and find the critical points inside the interval.
7. Prove the extreme value. Closed interval: candidates test. Open interval: show there is only one critical point and use the first or second derivative test to show it is a maximum (or minimum). A continuous function with a single critical point that is a relative maximum on an interval has its absolute maximum there (Topic 5.7).
8. Answer the question asked: the value, the location, or both.

Step 7 is the step most often skipped. Finding f′(x) = 0 only finds a candidate. It could be a maximum, a minimum or neither.

Worked example 1: a model on a closed interval

Question. A gallery records how many visitors are inside during a 10-hour day. The number is modelled by

N(t) = t³ − 18t² + 81t + 40, for 0 ≤ t ≤ 10,

where t is hours after opening. (The data are fictional.) Find the maximum and minimum number of visitors predicted by the model, and when each occurs.

Solution.

1. The function and interval are given. N is a polynomial, so it is continuous on the closed interval [0, 10]. The Extreme Value Theorem guarantees a maximum and a minimum, and the candidates test applies.
2. Differentiate. N′(t) = 3t² − 36t + 81 = 3(t² − 12t + 27) = 3(t − 3)(t − 9).
3. Critical points. N′(t) = 0 at t = 3 and t = 9. Both lie inside [0, 10]. N′ exists everywhere, so there are no others.
4. Candidates table.

t	0 (endpoint)	3 (critical)	9 (critical)	10 (endpoint)
N(t)	40	148	40	50

Working: N(3) = 27 − 162 + 243 + 40 = 148. N(9) = 729 − 1458 + 729 + 40 = 40. N(10) = 1000 − 1800 + 810 + 40 = 50.

5. Compare. The largest value is 148 and the smallest is 40.

Answer. The model predicts a maximum of 148 visitors, at t = 3 (3 hours after opening), and a minimum of 40 visitors, which occurs twice: at opening (t = 0) and at t = 9.

Check. N″(t) = 6t − 36, so N″(3) = −18 < 0 (a relative maximum) and N″(9) = 18 > 0 (a relative minimum). That agrees with the table. Notice the endpoint t = 0 ties with the interior minimum, and the endpoint t = 10 (value 50) is higher than the local minimum at t = 9. Skipping endpoints would have missed the tie.

Worked example 2: building the function yourself

Question. A sports club sets up a rectangular warm-up area against a long straight wall. The wall forms one side, so no fencing is needed there. Portable fencing is used for the other three sides and for one extra fence, perpendicular to the wall, that splits the area into two parts. The club has 180 m of fencing. Find the largest possible total area, and the dimensions that give it.

Solution.

1. Quantity to optimise: the total area A, in m². We want a maximum.
2. Variables: let x be the length of each fence perpendicular to the wall and y the length of the fence parallel to the wall (Figure 2).
3. Formula: A = xy.
4. Constraint: 3x + y = 180, so y = 180 − 3x. Substitute: A(x) = x(180 − 3x) = 180x − 3x².
5. Interval: x > 0 and y = 180 − 3x > 0, so 0 < x < 60. This is an open interval, so the candidates test does not apply directly.
6. Critical points: A′(x) = 180 − 6x = 0 gives x = 30, which is inside (0, 60).
7. Prove it is the maximum. A″(x) = −6 < 0, so x = 30 is a relative maximum. It is the only critical point in (0, 60), and A is continuous there, so the relative maximum is the absolute maximum on the interval. (Or: A′ > 0 for x < 30 and A′ < 0 for x > 30, so A increases up to x = 30 and decreases after it.)
8. Answer: x = 30 m and y = 180 − 90 = 90 m. The largest area is 30 × 90 = 2700 m², with each perpendicular fence 30 m long and the fence parallel to the wall 90 m long.

Check. Try a nearby value: A(29) = 29 × 93 = 2697 and A(31) = 31 × 87 = 2697. Both are a little smaller than 2700, as expected. Also, 3 × 30 + 90 = 180, so all the fencing is used.

Notice the structure. Exactly the same steps would solve a problem about a box, a can or a pair of numbers. Only steps 2 to 5 change.

Common misconceptions

• Differentiating with two variables still in the formula. Use the constraint first. A = xy cannot be optimised until y is written in terms of x.
• Stopping at f′(x) = 0. A critical point is a candidate, not an answer. Show it is a maximum or minimum.
• Forgetting endpoints on a closed interval. In Worked example 1 the minimum also occurs at t = 0, an endpoint.
• Using the second derivative test alone for an absolute extremum. f″(c) < 0 gives a relative maximum. To call it absolute on an open interval, you also need “it is the only critical point” (or a candidates comparison).
• Giving the location when the value is asked, or the reverse. “The maximum is at x = 30” does not answer “what is the largest area?”.
• Ignoring the interval. A critical point outside the allowed values (a negative length, say) must be rejected.
• Putting the constraint value into the wrong place. In Worked example 2, the fencing total includes three lengths x, not two, because of the dividing fence. A labelled diagram prevents this.

Where this leads

You now have a method that works for any optimization problem. In Topic 5.11, Solving Optimization Problems, you apply it to fuller situations and explain what the maximum or minimum means in context, with units. The graph-reading skills from Topic 5.9 help you check that a critical point really is the kind of extremum you need. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.