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MTH6131 Chap.11 Applied Maximum and Minimum Problems

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Chapter 11 of 16 · MTH6131

Applied Maximum and Minimum Problems

The calculus is short; the setting up is not

The applied examples in the lectures are all stories: two ships on crossing courses, a corral fenced against a barn with interior fences, a box-shaped bag under a size limit, a ladder that must clear a fence. The calculus in each takes a few lines.

What takes the time, and what earns the marks, is turning the story into one function of one variable on a stated interval.

Six guidelines

The notes list six steps, which in short are: draw the situation and name what changes; write the quantity to be optimised as a formula; write every restriction as an equation; substitute until one variable is left; decide which values that variable can take; and only then differentiate and compare candidates.

Closed and open intervals

On a closed interval the answer comes from comparing the critical values with the endpoints.

On an open interval there are no endpoints to compare, so the argument has to come from the behaviour of the function near the ends or from a single critical point with the right sign change, as in the lecture example of the ships, where one critical value with a sign change from negative to positive is shown to be the absolute minimum.

Recurring devices

Several devices recur across the exercise set.

A distance is minimised by minimising its square, which has the same minimiser and no square root. Physically impossible critical values, such as a negative length, are rejected explicitly. Similar triangles supply the constraint in problems about shapes inscribed in cones, and a fixed volume supplies it in problems about least material.

Reading the answer back into the story, for instance noticing that the best closed can has height equal to its diameter, is the check that the model was right.

In this chapter

What this chapter covers

  • 01

    From a story to an objective and a constraint

  • 02

    The six guidelines in order

  • 03

    Eliminating all but one variable

  • 04

    The interval of interest

  • 05

    Comparing endpoints on a closed interval

  • 06

    Arguing on an open interval

  • 07

    Minimising a squared distance

  • 08

    Rejecting impossible critical values

Worked example · free

A product under a sum constraint

Q [6 marks]. AskSia-authored practice. Two positive numbers add up to 20. Find them if the product of the first and the square of the second is as large as possible. The marks shown are an AskSia study allocation, not the University's marking scheme.
Plan and objective side by side.Plan and objective side by side. The area function is computed from the constraint; its single peak at ⟦MK:eyJ0IjogInggPSAxMDAiLCAiZCI6IGZhbHNlfQ==⟧ beats both ends of the interval, where the area is zero.The pen and its area functionriver: no fence neededxxxy = 600 − 3x1002001500030000xApeak 30,000 m²
Plan and objective side by side. The area function is computed from the constraint; its single peak at ⟦MK:eyJ0IjogInggPSAxMDAiLCAiZCI6IGZhbHNlfQ==⟧ beats both ends of the interval, where the area is zero.
  • 2Name the variables, the objective and the constraint.
  • 2Reduce to one variable on its interval and differentiate.
  • 2Justify the maximum and state the numbers.
Let the numbers be x and y with x + y = 20, and maximise P = xy². Eliminating x gives P(y) = (20 − y)y² on the interval from 0 to 20. The derivative is P′(y) = 40y − 3y², which is zero at y = 0 and y = 40/3. At both ends of the interval P is zero, while at y = 40/3 it is (20/3)(1600/9) = 32000/27, about 1185.2. So the numbers are 20/3 and 40/3, with the larger one squared.
Sia tip — Write the interval of the single remaining variable before differentiating; the endpoints are candidates you must compare.
Glossary

Key terms

Objective Function
The quantity to be maximised or minimised, written in the variables of the problem.
Constraint
An equation linking the variables of an optimisation problem, used to reduce the objective to a single variable.
Interval of Interest
The set of values the single remaining variable can take in the physical problem, which supplies the endpoints to compare.
Squared Distance
The square of a distance, minimised instead of the distance because it has the same minimiser and no square root.
FAQ

Applied Maximum and Minimum Problems FAQ

How do I know whether an optimisation answer is a maximum or a minimum?

On a closed interval, compare the value at each critical point with the values at the endpoints. On an open interval, use the sign change of the derivative at the critical point, or the behaviour near the ends of the interval. If there is only one critical point and the derivative changes sign there, it is the absolute extremum.

Why minimise the square of a distance instead of the distance?

Because the square root is an increasing function, the distance and its square are smallest at the same point, and the square is far easier to differentiate. This only works for a single distance; a sum of distances has to be handled with the square roots kept in place.

What should I do with a critical value that makes no physical sense?

Reject it explicitly and say why, for instance because it gives a negative length or lies outside the interval of interest. The exercise solutions mark such values as rejected, and leaving one in an answer without comment suggests the physical setting was not checked.

Study strategy

Exam move

For each practice problem, write the six guideline steps as headings and fill them in before doing any calculus. Most applied answers go wrong at step four or five, in the elimination or the interval, and the headings make those steps impossible to skip.

Working through Applied Maximum and Minimum Problems in MTH6131? Sia is AskSia’s AI Mathematics tutor — ask any MTH6131 Applied Maximum and Minimum Problems question and get a clear, step-by-step explanation grounded in how MTH6131 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.

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