An optimisation problem asks for the best value of something subject to a restriction. Calculus finds it exactly.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Two equations
The objective is what you are maximising or minimising. The constraint is the restriction. Almost every problem has both.
Reduce to one variable
Solve the constraint for one variable and substitute into the objective. Only then can you differentiate.
Then the usual work
Differentiate, set to zero, and solve. Confirm with the second derivative or a sign chart.
Do not forget the ends
On a closed interval the extremum may sit at an endpoint, where the derivative is not zero at all.
Check it makes sense
A negative length or a fractional number of people means an algebra slip or an ignored domain restriction.
The procedure
Write the quantity to be optimised, use a constraint to reduce it to one variable, differentiate, find the critical points, and verify which is the extremum. The modelling step is usually harder than the calculus.
The constraint is what makes it solvable
Maximising area with unlimited fencing has no answer. The fixed perimeter is what creates the trade-off and the optimum. Identifying the constraint is the first thing to do in any optimisation problem.
The domain is set by the situation
A length cannot be negative, and a side cannot exceed the available material. Endpoints of that physical domain must be checked alongside the critical points, since the optimum sometimes sits at one.
Verify it is the extremum you wanted
Use the first or second derivative test, or compare values at all candidates. Reporting a critical point without verifying it maximises rather than minimises is a common and complete failure of the answer.
Step 2: Try It Yourself
Tap and try it out.
- Point(2, 4)
- Slope of the tangent0
Step 3: Watch an Example
One step at a time.
Watch Marcus Fence the Largest Field
Marcus has 40 m of fence for a rectangular field and wants the greatest area.
- Step 1
The constraint is the perimeter: 2w + 2h = 40, so h = 20 − w.
Step 4: Your Turn
Practice makes it stick.
The Fence
Problem 1 of 2
With 40 m of fence, what is the largest rectangular area, in square metres?
m²
The Numbers
Problem 2 of 2
Two numbers add to 20. What is their greatest possible product?
Find the Best
1 of 8
With 24 m of fence, what is the largest rectangular area?
2 of 8
Two numbers add to 10. Greatest product?
3 of 8
A = 20w − w². What is dA/dw at w = 4?
4 of 8
A = 20w − w². At which w is dA/dw zero?
5 of 8
Two numbers add to 30. Greatest product?
6 of 8
Can the extremum sit at an endpoint of a closed interval? 1 for yes, 0 for no.
7 of 8
Put the optimisation method in order.
- 1Use the constraint to remove one variable.
- 2Differentiate and set the result to zero.
- 3Confirm it is the extremum you wanted, and check the endpoints.
- 4Identify the objective and the constraint.
8 of 8
With 100 m of fence, what is the largest rectangular area?
Step 5: Quick Check
Show what you know.
Question 1 of 2
Two numbers add to 16. What is their greatest product?
Question 2 of 2
Why must the constraint be used before differentiating?
What You Learned
- Optimisation pairs an objective with a constraint.
- Use the constraint to reduce the objective to one variable, then differentiate.
- Check endpoints, and check the answer makes physical sense.