Cogito
Calculus · Chapter 6 · Lesson 3
Optimisation
The largest, the cheapest, the fastest.
12 problems · about 23 minutes · FUN-4.C
What this lesson teaches
The student solves optimisation problems by building a function of one variable and locating its extremum.
- 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.
Warm Up
Straightforward practice. Get the method working first.
5 problemsTwo numbers add to 16. What is their greatest product?
Answer 64
Why 64.
Why must the constraint be used before differentiating?
Answer The objective must be a function of a single variable first.
Why You can only differentiate a one-variable function here.
With 24 m of fence, what is the largest rectangular area?
Answer 36
Why A 6 by 6 square.
Two numbers add to 10. Greatest product?
Answer 25
Why 5 × 5.
A = 20w − w². What is dA/dw at w = 4?
Answer 12
Why 20 − 8.
Build It Up
The same ideas with more to keep track of.
3 problemsA = 20w − w². At which w is dA/dw zero?
Answer 10
Why 20 − 2w = 0.
Two numbers add to 30. Greatest product?
Answer 225
Why 15 × 15.
Can the extremum sit at an endpoint of a closed interval? 1 for yes, 0 for no.
Answer 1
Why The derivative need not be zero there.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
4 problemsPut the optimisation method in order.
Answer 1. Identify the objective and the constraint. 2. Use the constraint to remove one variable. 3. Differentiate and set the result to zero. 4. Confirm it is the extremum you wanted, and check the endpoints.
Why Nothing can be differentiated until one variable remains.
With 100 m of fence, what is the largest rectangular area?
Answer 625
Why A 25 by 25 square.
The Fence: With 40 m of fence, what is the largest rectangular area, in square metres?
Answer 100 m²
Why 100 m².
The Numbers: Two numbers add to 20. What is their greatest possible product?
Answer 100
Why 100.