IB Applications and Interpretation SL · Differential and integral calculus · Section 5 of 7

Optimisation: maxima and minima in context problems

3 worked examples · By Álvaro López

Key idea

An optimisation problem asks for the value that makes a quantity a maximum or a minimum: the largest volume, the lowest cost or the highest profit. With derivatives it is always solved the same way: express the quantity in terms of one variable, differentiate, find where $f^{'} (x) = 0$ and check that this point is the maximum or minimum you need.

Setting up the function

The hardest part of these problems is usually writing the function, not differentiating it. The quantity you want to optimise is the objective function. The question also almost always gives a constraint, a condition that links the variables: a total length, a fixed volume or a number of units. Use it to eliminate one of the variables. Make one of them the subject of the constraint and substitute it into the objective function. This leaves you with a function of a single variable.

Checking whether it is a maximum or minimum

The solution of $f^{'} (x) = 0$ shows where the extremum may be, but it does not say whether it is a maximum or a minimum. To find out, look at the sign of $f^{'} (x)$ on each side of the point. If the derivative goes from positive to negative, the function stops increasing and starts decreasing: there is a maximum. If it goes from negative to positive, there is a minimum. You can also sketch the function on the GDC and read the coordinates of the highest or lowest point.

Graph of the profit function B(x) = −x³ + 12x² − 36x + 40 between x = 3 and x = 8, with the maximum at (6; 40) and the final value (8; 8), and the dashed vertical line at x = 6

Look at the graph of the bicycle factory in the third example: the maximum is at an interior point of the interval and the minimum at an endpoint, where the derivative is not zero.

Worked examples

Example 1 Basic

Pen next to a wall

A farmer wants to build a rectangular pen against a straight wall. The wall forms one of the sides, so he only needs fencing for the other three. He has $40$ m of fencing. Let $x$ be the length of each of the two sides perpendicular to the wall, and $y$ the length of the side parallel to the wall.

a) Write the area of the pen in terms of $x$ .

b) Find the dimensions that give the maximum area, and that area.

Solution

a) The fencing totals $40$ m: two sides of length $x$ and one side of length $y$ . So $2 x + y = 40$ , and making $y$ the subject gives $y = 40 - 2 x$ . The area is $A = x \cdot y = x (40 - 2 x) = 40 x - 2 x^{2}$

b) We differentiate and solve $A^{'} (x) = 0$ : $A^{'} (x) = 40 - 4 x = 0 \Rightarrow x = 10$ Then $y = 40 - 2 \cdot 10 = 20$ . We check the type of extremum: $A^{'} (5) = 20 > 0$ and $A^{'} (15) = - 20 < 0$ , so the area increases up to $x = 10$ and decreases after that. It is a maximum. The maximum area is $A (10) = 40 \cdot 10 - 2 \cdot 1 0^{2} = 200$

Answer: the pen has sides of $10$ m perpendicular to the wall and $20$ m parallel to it, with a maximum area of $200 m^{2}$ .

Example 2 Medium

Open box of fixed volume

A company makes open-top boxes with a square base. Each box must have a volume of $108$ cm³. Let $x$ be the side of the base, in cm, and $h$ the height, in cm.

a) Write the total area of card of the box in terms of $x$ . The area includes the base and the four lateral faces.

b) Find the side of the base and the height that minimise the amount of card, and that minimum amount.

Solution

a) The volume is $x^{2} h = 108$ , so $h = \frac{108}{x ^{2}}$ . The area is the base, $x^{2}$ , plus the four lateral faces, $4 x h$ : $S = x^{2} + 4 x h = x^{2} + 4 x \cdot \frac{108}{x ^{2}} = x^{2} + \frac{432}{x}$

b) We differentiate and solve $S^{'} (x) = 0$ : $S^{'} (x) = 2 x - \frac{432}{x ^{2}} = 0 \Rightarrow 2 x^{3} = 432 \Rightarrow x^{3} = 216 \Rightarrow x = 6$ The height is $h = \frac{108}{6 ^{2}} = 3$ . To check the minimum: $S^{'} (5) = 10 - \frac{432}{25} \approx - 7.3 < 0$ and $S^{'} (7) = 14 - \frac{432}{49} \approx 5.2 > 0$ . The derivative goes from negative to positive, so it is a minimum. The minimum area is $S (6) = 36 + \frac{432}{6} = 108$

Answer: the base has a side of $6$ cm and the height is $3$ cm; the minimum amount of card is $108$ cm².

Example 3 Hard

Bicycle factory with limited capacity

A factory making electric bicycles estimates its monthly profit, in thousands of euros, according to the number of bicycles it makes. If it makes $x$ hundreds of bicycles, the profit is $B (x) = - x^{3} + 12 x^{2} - 36 x + 40, 3 \leq x \leq 8$ The interval comes from the factory's capacity: it never makes fewer than $300$ or more than $800$ bicycles a month.

a) Find the values of $x$ for which $B^{'} (x) = 0$ and state which of them lie within the interval.

b) Find the number of bicycles that maximises the profit, and the maximum profit.

c) Find the minimum profit the factory can make within that interval, and the number of bicycles at which it occurs.

Solution

a) Differentiate and factorise: $B^{'} (x) = - 3 x^{2} + 24 x - 36 = - 3 (x^{2} - 8 x + 12) = - 3 (x - 2) (x - 6)$ The values that make the derivative zero are $x = 2$ and $x = 6$ . Only $x = 6$ lies within $[3, 8]$ .

b) Study the sign of $B^{'} (x)$ on the interval. For example, $B^{'} (4) = - 3 \cdot 2 \cdot (- 2) = 12 > 0$ and $B^{'} (7) = - 3 \cdot 5 \cdot 1 = - 15 < 0$ . The function increases up to $x = 6$ and decreases after that, so $x = 6$ is a maximum. Using this value: $B (6) = - 216 + 432 - 216 + 40 = 40$ The maximum profit is $40$ thousand euros, with $600$ bicycles.

c) Since $x = 2$ lies outside the interval, the minimum cannot be at a point with zero derivative inside $[3, 8]$ . We compare the endpoints and the maximum: $B (3) = - 27 + 108 - 108 + 40 = 13 B (6) = 40 B (8) = - 512 + 768 - 288 + 40 = 8$ The lowest value is $8$ , at $x = 8$ , that is, with $800$ bicycles. The minimum profit is $8$ thousand euros.

In the exam: the point $x = 2$ is a local minimum, but it lies outside the factory's range. If the interval has endpoints, always evaluate the endpoints as well as the points where $f^{'} (x) = 0$ .

Common mistakes

Summary

An optimisation problem is solved in four steps: objective function, constraint, derivative set equal to zero and checking.
With a constraint, reduce the function to a single variable before differentiating.
To find out whether $f^{'} (x) = 0$ gives a maximum or a minimum, look at the sign change of $f^{'} (x)$ , or use the graph on the GDC.
If the variable lies in a closed interval, compare the points where $f^{'} (x) = 0$ with the endpoints of the interval.
Answer in context, with units, and to the question being asked.

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