Practice Extremum problems and optimisation
An optimisation problem searches for the greatest or smallest possible value under given constraints.
Build a model, examine candidates and check the domain and boundary values.
Exercises with solutions
In an extremal problem you look for the largest or smallest value of a quantity. First write down the target function. Use the constraint to express it in a single variable, then differentiate and set the derivative to zero. Finally check the boundary values of the domain.
Exercise 1
A rectangle has perimeter 20 cm. How long are the sides for the largest area?
Show solution and working
5 cm and 5 cm, so a square
From a + b = 10 it follows that b = 10 − a. The area is A(a) = a · (10 − a) = 10a − a². Setting 10 − 2a = 0 gives a = 5, hence b = 5.
Exercise 2
Why does an extremal problem always need a constraint?
Show solution and working
to reduce the target function to one variable
Without a constraint the target function usually has infinitely many solutions because one quantity stays free.
Exercise 3
How do you show that it is a maximum and not a minimum?
Show solution and working
with the second derivative
If the second derivative at that point is negative it is a maximum; if it is positive, a minimum.
Exercise 4
A rectangle has perimeter 20 cm. What is the largest area?
Show solution and working
25 cm²
The square with side length 5 cm has area 5 · 5 = 25 cm².
Common mistake
Forgetting the constraint and leaving two variables in the target function. And setting the derivative to zero without using the second derivative to check whether it is a maximum or a minimum.
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