Optimization practice question 11
An open-topped box with a square base must hold cm³. Let be the side of the base, in centimetres. Write the area of material the box needs, in cm², as a function of alone.
Let be the height. The volume gives one equation in and ; list the faces an open-topped box has, and the area of each.
Solve the volume equation for , and put it into .
A formula in x. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §8.3CLP-1 §3.5Key results
Worked solution
Strategy
Name every length, write the quantity to be made small in terms of them, then use the constraint to remove all but one. This is the first part of every optimization problem.
Working
The volume fixes the height for each base.
One square base and four sides, each x by h; there is no lid.
Put in h, so that A depends on x alone.
Answer
The answer is A equals x squared plus 128 000 over x.
Check
With x = 40 the height is 20, and the faces give 1600 + 4 · 800 = 4800, which is 40² + 128 000/40.
Where this goes wrong
Counting a lid, or only one side. List the faces of the box before adding: one base and four sides, each x by h.
The idea behind it
The optimization lesson, a figure to move, in a few steps
Leans on
Writing a function from a word problem, from Functions, graphs and inverses
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