Optimization practice question 8
A rectangular vegetable bed is to be edged all the way round and split into three equal strips by two inner rows of edging, each parallel to one pair of sides. The outer edging costs $4 per metre and the inner edging $2 per metre, and the budget is $120. Find the largest total area the bed can have.
Give the answer in square metres.
Name the side the inner rows run along and the side across them. Which lengths of edging does each price apply to?
Let the inner rows have length and the other sides length . The outer edging is metres at $4 and the inner metres at $2, so .
Then , and the area is a function of alone. Maximise it on .
A number.
Theory UBC Math 100 §8.3CLP-1 §3.5Key results
Worked solution
Strategy
The budget is the constraint: price each length of edging, use the cost equation to write one side in terms of the other, and maximise the area as a function of one variable, justifying the maximum.
Working
The two inner rows each have length y, parallel to one pair of sides.
Both sides must be positive, which bounds y.
The only critical point in the domain, and A is concave down everywhere, so it is the maximum.
A bed 7.5 m by 5 m, split into three strips 2.5 m wide.
Answer
The answer is 37.5 square metres.
Check
The 7.5 m by 5 m bed costs 4(15 + 10) + 2(10) = $120, and on that budget a 4.8 m or 5.2 m inner row gives only 37.44 m².
Where this goes wrong
Pricing every metre of edging the same, or leaving the inner rows out of the cost. The cost equation must charge each length exactly as described.
The idea behind it
The optimization lesson, a figure to move, in a few steps
Leans on
Area and perimeter of a rectangle, from Functions, graphs and inverses
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