To interpret an optimum, check the stationary value against the allowed domain, then state the answer with units and a sentence in context. If the stationary value is outside the domain, the optimum is at an endpoint.
This lesson combines forming an objective function and checking endpoints. It completes stationary points and optimisation.
What does “domain” mean in a real problem?
In a pure maths question the domain may be stated: 0 ≤ x ≤ 5.
In a word problem you work it out from the situation.
Lengths are positive, time starts at zero, a wall has a fixed length, and items sold are whole numbers.
These limits turn a line of algebra into a real answer, and they are often what decides where the optimum lies.
How to finish an optimisation question, step by step
- Write the domain as an inequality from the context, including any limits given in the problem.
- Find the stationary x-value and check it is inside the domain.
- If it is inside, classify it. If it is outside, compare the endpoints.
- Calculate the optimum value with its units.
- Write a sentence naming the variable, the value and the units.
Worked example
A farmer uses 60 m of fencing on three sides of a pen, with the fourth side against a wall that is only 25 m long. Find the largest possible area.
Step 1, set-up: sides x m (two of them) and y m along the wall. The constraint is 2x + y = 60, so y = 60 − 2x, and A = x(60 − 2x) = 60x − 2x².
Step 2, domain: the wall is 25 m long, so y ≤ 25. That means 60 − 2x ≤ 25, so x ≥ 17.5. Also y > 0 means x < 30. The domain is 17.5 ≤ x ≤ 30.
Step 3, stationary point: dA/dx = 60 − 4x = 0 gives x = 15. But 15 is outside the domain, so it is not allowed.
Step 4, endpoints:
At x = 17.5: A = 60(17.5) − 2(17.5)² = 1050 − 612.5 = 437.5.
At x = 30: A = 1800 − 1800 = 0.
Step 5, conclusion: the largest area is 437.5 m², when x = 17.5 m and y = 25 m.
Check: 17.5 × 25 = 437.5. The most the farmer can do is use the whole wall.
The mistake to watch for
A common slip is to give the unrestricted optimum without testing it against the domain.
Mistaken answer: the maximum area is 450 m² when x = 15 m.
The student solved the calculus correctly but forgot the wall is only 25 m, and x = 15 would need y = 30 m.
The correction is to write the domain as a separate line before finishing. Substitute the stationary value back into every condition in the question, including the ones you did not use to form the function.
Check yourself
Try these without a calculator, then open each answer.
1. A ball’s height is h = 30t − 5t² metres after t seconds, for 0 ≤ t ≤ 5. Find the maximum height and when it occurs.
Show answer
dh/dt = 30 − 10t = 0 gives t = 3, inside the domain. h = 90 − 45 = 45. The second derivative is −10, so it is a maximum. Check the ends: h(0) = 0 and h(5) = 150 − 125 = 25.
The maximum height is 45 m, reached after 3 seconds.
2. A shop’s daily profit is P = 40x − 2x² − 50 ringgit when x items are sold, where 0 ≤ x ≤ 8. Find the greatest profit.
Show answer
dP/dx = 40 − 4x = 0 gives x = 10, which is outside 0 ≤ x ≤ 8. Compare endpoints: P(0) = −50 and P(8) = 320 − 128 − 50 = 142.
The greatest profit is RM142, when 8 items are sold.
3. The profit on x whole items is P = −x² + 7x − 3 ringgit. Find the whole-number x that gives the greatest profit and the profit.
Show answer
dP/dx = −2x + 7 = 0 gives x = 3.5, which is not whole. Compare the nearest whole numbers: P(3) = −9 + 21 − 3 = 9 and P(4) = −16 + 28 − 3 = 9.
Selling 3 items or 4 items both give the greatest profit of RM9.
Where this leads next
Put the whole module to work in the mixed practice set. The non-calculator working trainer and the quadratic structure explorer give you extra ways to test the arithmetic.
If a question that mixes set-up, calculus and domain still feels like too many steps at once, online one-to-one Additional Mathematics tuition lets a teacher split it into parts with you.