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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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).

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!

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.

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!

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

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!

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

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