- Topic
- Optimization
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Optimization practice question 6
A rectangular plot is fenced on three sides, with a straight river forming the fourth side. With 240 m of fencing available, what is the largest area that can be enclosed?
Give the area in square metres.
Name the two side lengths. The river side needs no fence: which sides does the 240 m of fencing have to cover?
Let be each side perpendicular to the river and the side parallel to it. Only three sides are fenced, so .
Use the fencing constraint to write in terms of , then express the area as a function of alone.
A whole number.
Theory UBC Math 100 §8CLP-1 §3.5Key results
Worked solution
Strategy
Write the quantity to be maximised, use the constraint to reduce it to one variable, differentiate, and confirm the critical point really is the maximum.
Working
The constraint. Only three sides need fencing because the river bounds the fourth.
Neither side can be negative, so x runs from 0 to 120.
Differentiate and solve for the critical point.
A is concave down on the whole domain, so its only critical point, x = 60, gives the absolute maximum.
Back-substitute for the other dimension and the area.
Answer
The answer is 7200 m².
Check
Sides of 50 m or 70 m give 50 × 140 = 7000 m² and 70 × 100 = 7000 m², both less than 7200 m².
Where this goes wrong
Using the constraint 2x + 2y = 240 out of habit, as though all four sides were fenced. That gives a 60 × 60 square of area 3600 and misses half the available area.
Leans on
Writing a function from a word problem, from Functions, graphs and inverses
Saved in this browser only.