Practise the method.
Build models, test stationary points and compare feasible endpoints.
Example 1
A rectangular growing area uses a wall for one side. The other three sides use 20m of fencing: two widths x and one length y, so 2x+y=20. Site rules require 1≤x≤4 metres.
Select the area model after eliminating y.
| Quantity | Given constraint |
|---|---|
| Fence (m) | 2x+y=20 |
| Width x(m) | [1,4] |
| Area (m2) | xy |
Show worked answer
Answer: A(x)=x(20−2x)
- The fence constraint gives y=20−2x.
- Area is xy, hence A(x)=x(20−2x).
Example 2
A machine setting x is a positive dimensionless quantity with 5≤x≤7. Its daily operating cost is C(x)=3x+x48 dollars.
a Select the correct marginal-cost function.
Find the permitted setting with the lowest daily cost.
Show worked answer
Answer: 5
- C′=0 gives x2=348=16; only x=4 is positive.
- C′′(x)=x396>0 for x>0, so cost decreases then increases.
- Compare the permitted stationary point with x=5 and x=7. The constrained minimum is x=5.
Example 3
An open-top box has square base side x metres and height h metres. Its available material area is 48m2, so x2+4xh=48. Only 1≤x≤3 is allowed. Ignore material thickness and waste.
a Find h when x=1. Give an exact fraction if needed.
b Find the permitted base side x giving the greatest volume.
Find the maximum volume. Give an exact fraction if needed.
Show worked answer
Answer: 4117
- The optimal height is 4×348−32=3.25m.
- Therefore V=448×3−33=4117m3.