Optimisation problems

What you will be able to do

• turn a word problem into a function of one variable;
• find its domain from the conditions of the problem;
• find the maximum or minimum and show that it is the best value;
• solve problems about boxes, fences, tins, costs and distances.

From a story to a function

A farmer has a fixed length of fence. A company wants a tin that uses the least metal. A cyclist wants the shortest route. In each case one quantity is to be made as large or as small as possible, and the variables are tied by a condition.

The method.

1. Draw a picture. Name the variables and write their units.
2. Name the quantity to be made largest or smallest. This is the objective.
3. Use the condition to write the objective in one variable.
4. Find the domain. The variable cannot make a length negative.
5. Differentiate and solve $f'(x) = 0$.
6. Show that it is a maximum or a minimum. Use $f''$, or compare with the end points of the domain.
7. Answer the question, with units. Check that the size is sensible.

The fence against a wall

A farmer has $120\ \text{m}$ of fence. She makes a rectangular pen against a long straight wall, so only three sides need fence. What size gives the greatest area?

Let each short side be $x$ metres. The long side is $120 - 2x$ metres, so

Then $A' = 120 - 4x = 0$ gives $x = 30$. Since $A'' = -4 < 0$, this is a maximum. The pen is $30\ \text{m}$ by $60\ \text{m}$, and its area is $1800\ \text{m}^2$. The long side is twice the short side.

The open box

A square sheet of side $24\ \text{cm}$ has a square of side $x$ cut from each corner. The sides are folded up to make a box with no lid. The base is $(24 - 2x)$ by $(24 - 2x)$ and the height is $x$, so

Then $V' = (24 - 2x)^2 - 4x(24 - 2x) = (24 - 2x)(24 - 6x)$. The zeros are $x = 12$, which is the end of the domain and gives $V = 0$, and $x = 4$. The second derivative is $V'' = 24x - 192$, which is $-96 < 0$ at $x = 4$. So the best cut is $x = 4\ \text{cm}$, and $V = 4 \times 16^2 = 1024\ \text{cm}^3$.

Checking that the answer is the best

The condition $f'(x) = 0$ only finds candidates. Check one of these:

• The second derivative. $f'' < 0$ gives a maximum and $f'' > 0$ gives a minimum.
• The first derivative test. The sign of $f'$ changes in the right way.
• The end points. On a closed domain, compare with the values at the ends. On an open domain, a single critical point that is a local maximum is the greatest value.

A worked example

A rectangle has a perimeter of $40\ \text{cm}$. Find the sides that give the greatest area.

1. The variables. Let the sides be $x$ and $y$ centimetres, so $2x + 2y = 40$ and $y = 20 - x$.
2. The objective. $A = xy = x(20 - x)$, for $0 < x < 20$.
3. Differentiate. $A' = 20 - 2x = 0$ gives $x = 10$, and $A'' = -2 < 0$.
4. The answer. $x = 10$ and $y = 10$. The best rectangle is a square with area $100\ \text{cm}^2$.

Practise with new numbers

Try it in Python

The program tries many cut sizes for the box and keeps the best. It then compares the answer with the result from calculus.

Common mistakes

• Leaving two variables in the function. Use the condition to remove one of them before differentiating.
• Forgetting the domain. A side of a box cannot be negative.
• Not showing that the answer is the best. A critical point may be a minimum when you want a maximum.
• Answering with $x$ only. Give the quantity the question asks for, with units.
• Rounding too early. Keep exact values until the last step.

Remember these

• Write the objective in one variable, using the condition. Find the domain.
• Solve $f'(x) = 0$ and check the end points and $f''$.
• A rectangle with a fixed perimeter has the greatest area when it is a square.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Two positive numbers add up to $10$. What is the greatest possible product?

2. A rectangle has a perimeter of $60\ \text{cm}$. What is the greatest area?

3. Why should you compare with the end points of the domain?

4. A square sheet of side $30\ \text{cm}$ is made into an open box by cutting equal squares from the corners. Find the best cut and the volume.

5. In an optimisation problem you find $f'(c) = 0$. What must you still check?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 6 (Application of Derivatives), the section on maxima and minima.
• OpenStax, Calculus Volume 1, Section 4.7 (Applied Optimization Problems).