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Longhand

MATH 100 optimization practice

An optimization question hands you a situation and asks for the best possible version of it: the largest area, the cheapest can, the closest point. The calculus involved is usually just a derivative set to zero. The marks are in everything around that.

Every one of these problems has the same skeleton. Name the quantity being optimised, write it as a function, use the constraint to reduce it to a single variable, differentiate, and then justify that the critical point you found really is the maximum or minimum on the domain the situation allows.

That last step is the one most often skipped and most often worth marks. A critical point is a candidate, not an answer, and a closed interval has endpoints that are candidates too.

Begin practice sessionBrowse all 5 questions

What you should be able to do

  • Separate the objective from the constraint

    One quantity is being optimised and another is fixed. Identifying which is which before writing anything down prevents the most expensive error in the topic.

  • Reduce to a single variable

    Solve the constraint for whichever variable is easiest to remove, substitute, and simplify. You cannot differentiate usefully until the objective depends on one thing.

  • Establish the domain from the situation

    A length cannot be negative and a cut-out corner cannot exceed half the sheet. The physical setup, not the algebra, decides which critical points are admissible.

  • Justify the extreme value

    Use the second derivative test, a first derivative sign chart, or the closed-interval method. Stating which test you used and what it showed is part of a complete answer.

  • Answer the question that was asked

    Some questions want the dimensions, some want the optimal value, some want both. Read the final line again before writing your answer down.

Where marks are usually lost

  • Optimising the constraint by accident

    In a fixed-volume can problem the volume is given, so it cannot be what you maximise. Surface area is the objective. Write down which is which before differentiating.

  • Assuming all four sides need fencing

    When a river, wall or existing fence bounds one side, the constraint has three terms, not four. Sketching the region first makes the correct constraint obvious.

  • Keeping a critical point outside the domain

    Solving V′(x) = 0 for a box made from a 30 cm sheet gives x = 5 and x = 15. At x = 15 the box has no base at all, so it is rejected on physical grounds rather than algebraic ones.

  • Ignoring the endpoints on a closed interval

    On [a, b] the absolute extremes can occur at either end. Evaluating only the critical points is the standard way to lose a closed-interval question.

  • Stopping at the critical point

    Without a second derivative, a sign chart or an endpoint comparison, nothing so far distinguishes a maximum from a minimum or a point of inflection.

Practice questions

Every question below is original material written for Longhand. Open a solution to read the full working, or answer it yourself to have it counted towards your progress.

  • Standard3 marks · 5 min

    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?

    Answer this question yourself
  • Standard3 marks · 5 min

    An open-topped box is made from a 30 cm by 30 cm square sheet of card by cutting a square of side x from each corner and folding up the sides. Which value of x gives the largest volume?

    V(x)=x(30−2x)2,0<x<15V(x) = x(30 - 2x)^{2}, \qquad 0 < x < 15
    Answer this question yourself
  • Challenging4 marks · 7 min

    Find the xx-coordinate of the point on the curve y=xy = \sqrt{x} closest to the point (4,0)(4, 0).

    Answer this question yourself
  • Challenging5 marks · 8 min

    A closed cylindrical can must hold 1000 cm³. Find the radius and height that use the least material, and state the relationship between them.

    V=πr2h,A=2πr2+2πrhV = \pi r^{2} h, \qquad A = 2\pi r^{2} + 2\pi r h
    Answer this question yourself
  • Standard2 marks · 4 min

    Find the absolute maximum value of f(x)=x3−6x2+9x+2f(x) = x^{3} - 6x^{2} + 9x + 2 on the closed interval [0,4][0, 4].

    Answer this question yourself

Common questions

Which justification should I use, the first or second derivative test?
Either is acceptable when it applies. The second derivative test is quickest when the second derivative is easy to compute and non-zero at the critical point. A first derivative sign chart always works and is the safer choice when the second derivative is awkward or vanishes.
Do I have to check endpoints every time?
Whenever the domain is a closed interval, yes. When the domain is open — a length that must be strictly positive, for example — you instead check the behaviour as the variable approaches each end, which is often enough to rule those ends out.
Is it acceptable to minimise the squared distance instead of the distance?
Yes, and it is usually the better choice. Squaring is an increasing function on non-negative values, so the two are minimised at the same point, and the algebra is considerably lighter.

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