Mathematical tools: a hammer vs a microscope

hammer-vs-microscope We congratulate everyone who celebrates the Knowledge Day on 1st September! This post is based on real student stories and illustrates a powerful principle about using appropriate mathematical tools.

A problem in multivariable calculus

The following problem was the first introductory question in a recent exam on multivariable calculus at a top UK university. This question was designed for students who had found the course hard. Indeed you could get a full mark within few seconds using only basic maths from school.

Q1. Find the minimum value of the function \(f(x,y)=x^2+2y^2-4x-4y\) over all real \(x,y\).

Derivatives may not be needed at all

Rather surprisingly, only about 2% of the 200+ class remembered the lecturer’s advice to use appropriate tools. The remaining 98% preferred the hard way finding extreme points through partial derivatives. Similarly to 1-variable functions, as we discussed in the post on global extrema, we may often find extreme values of a function without using any derivatives.

Moreover, about 50% of all students claimed that their critical point is a local minimum without justifications, hence lost almost a half of the full mark. The other 48% spent a lot of time justifying that the only critical point (where both 1st order partial derivatives vanish) is indeed a local minimum by using the Hessian of four 2nd order partial derivatives. About 10% of the class actually wrote more than 2 pages on this first question, though only a couple of lines was enough.

Here is the full solution:

\(f(x,y)=x^2+2y^2-4x-4y=(x-2)^2+2(y-1)^2-6\geq -6\) for all real \(x,y\).
Hence the minimum value is \(f(2,1)=-6\).

This simple technique of completing a square illustrates the powerful principle that mastering the basics is more important than trying an advanced method without proper understanding.

Completing a square vs differentiation

Here is the simple mnemonic rule: What can you do with a quadratic polynomial? – Complete a square, of course! Most students usually try to directly write the roots of a quadratic polynomial by using the so-called quadratic formula. Actually, this formula is proved by completing a square, which is a good exercise especially if you have never done it.

From the computational point of view, the quadratic formula is equivalent to completing a square, so we essentially make the same computations in both cases. However, after completing a square, a quadratic polynomial becomes structured (similar to the simplest form \(x^2\)).

Moreover, the same idea of completing a binom \((x+a)^n\) works for any higher degree polynomial, but the quadratic formula doesn’t. Actually many real problems in mathematics are about putting various objects (say, functions or matrices or groups) into a normal or structured form.

So completing a square is simple and efficient like a hammer, while differentiation is powerful and delicate like a microscope. A popular student question: “Can I still use a microscope, because I like it?” Our answer: “Yes, you can, though it may look a bit unprofessional.”

  • Riddle 19: find the maximum value of \(f(x)=6x^3-x^6\) over all real \(x\).
  • How to submit: to write your full answer, submit a comment.
  • Hint: find a point \(a\) such that \(f(x)\leq f(a)\) for any real \(x\).
  • Warning: using derivatives is possible, but is hard to justify.
  • Prize: free 1-year access to one of our interactive web tutorials.
  • Restriction: only the first correct public answer will be rewarded.
  • Update: Carlo has solved the problem, see attempts 1 and 2.

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8 thoughts on “Mathematical tools: a hammer vs a microscope

  1. Jakub

    After factorizing, f(x)=x^3(6-x^3)
    Looking at that, I would maximize the value for (6-x^3) keeping it positive.
    Numbers from x=0.000…1 to x=1 satisfy that.
    But x=1 also gives the biggest value for x^3 among them.

    Then the maximum value of the function over all reals is f(1).

    Reply
    1. Master Maths blogger Post author

      Dear Jakub, thank you for your attempt. If you wish to “maximize the value for (6-x^3) keeping it positive”, this function has no maximum value, because 6-x^3 goes to +infinity when x tends to -infinity. The question was to find the maximum over all real x, so you should consider x beyond the interval [0,1], e.g. f(1.5)>f(1)=5. We give you one more chance to submit a correct solution. However hurry up before someone else solves the riddle faster. Good luck!

      Reply
  2. Jasturan

    f(x)=6x^3-x^6
    =>f'(x)=18x^2-6x^5=6x^2(3-x^3)
    Stationary point when f'(x)=0
    => Stationary points at x=0 and x=3^(1/3)
    Plugging these in we find 3^(1/3) has the largest value of 9.
    We can confirm this to be a maximum point by looking at f”(3^(1/3))
    This is equal to -54(3^(1/3)) which implies it is a maximum point as f”(x)<0.

    Reply
    1. Master Maths blogger Post author

      Dear Jasturan, thank you for your attempt at this question. Your implicit claim that a maximum is achieved at one of stationary points fails for f(x)=x^2, which has the stationary point at x=0, but not a maximum at x=0. The more complicated example f(x)=x^3-x shows that f(x) may not have a global extremum (maximum or minimum over all real x) at any stationary point. Your argument about the second derivative indeed confirms that that x=3^{1/3} is a local maximum. However, the question was to find a global maximum over all real x, while derivatives can justify only local extrema. As a one-off exception, you may have another attempt. The hint is to re-read the post about completing squares. Good luck!

      Reply
  3. Carlo Russian

    Well, I see no one has answered the riddle correctly in 4 years, so I’ll have a go.

    Make the substitution y=x^3 , then 6x^3 – x^6 = 6y – y^2 = – (y – 3)^2 + 9 <= 9 since – (y-3)^2 <= 0

    Reply
    1. Master Maths blogger Post author

      Dear Carlo, yes. Your solution is almost complete. You have probably meant that 9 is the maximum value. You have indeed proved that 6x^4-x^6<=9. However, the correct inequality -x^2<=1 doesn't imply that (-x^2) has the maximum value of 1. It remains to give x where the function attains the expected maximum value of 9. We hope that you will complete your solution soon. Good luck!

      Reply
      1. Carlo

        Right, should have known better. Let x^3=3, then 6x^3 – x^6 = 6*3 – 3^2 = 9, which completes the proof.

        Reply
        1. Master Maths blogger Post author

          Dear Carlo, yes. Now you have completed riddle 19 and can claim your prize. Congratulations!

          Reply

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