## Spring 2018, problem 58

Let $a$ and $b$ be two real numbers and let $M(a,b)=\max\left \{ 3a^{2}+2b; 3b^{2}+2a\right \}$. Find the values of $a$ and $b$ for which $M(a,b)$ is minimal.

3 months ago

Call this $M=\text{max}\{u; v\}$,

now with $(s,p)=(a+b,ab)$, AM-GM's $\Rightarrow~4p\le s^2\Rightarrow~~$

$2M(a,b)\ge u+v=3(s^2-2p)+2s\ge \frac 32s^2+2s$

$=\frac 32(s+\frac 23)^2-\frac 23\ge -\frac 23=2M(-\frac 13; -\frac 13),$ the answer.

Any chance of expanding this answer with some explanatory notes ?

I'm probably not the only one who's not familiar with the terminology.

For example what does "AM-GM's" mean?

Thanks, Phil.

philboyd2 3 months ago
3 months ago

The minimality of $M(a,b)$ forces $3a^2+2b=3b^2+2a$, say $3a^2-2a=3b^2-2b$.

E.g. solving in $a$ we get $a=\frac{1\pm(3b-1)}3$ say $a=b$ or $a=\frac23-b$.

The first case bring to minimize $3b^2+2b$ and choosing $b=-\frac13$ we get the value $-\frac13$.

The second case asks for the minimum of $3b^2-2b+\frac43$ and choosing $b=\frac13$ we get the value 1.

Thus $\min M(a,b)=1$, reached with $a=b=\frac13$.

Check that $M(-\frac 13; -\frac 13)=-\frac 13$, and how do you show that minimality yields equality of the terms ?

Hubert 3 months ago

modified on february 11

@Hubert: thank you for signalling my mistakes.

In particular I forget to say that functions must agree unless the minimum is reached in a local minimum of one of them.

Claudio 3 months ago
3 months ago

The function $M(a,b)$ is minimized when both of its arguments equal each other. That is:

$3a^2 + 2b = 3b^2 + 2a \Rightarrow 3a^2 - 3b^2 + 2b - 2a = 0 \Rightarrow 3(a+b)(a-b) - 2(a-b) = 0 \Rightarrow (a-b)[3(a+b) - 2] = 0$

or when: $a = b$ (i), $a+b = \frac{2}{3}$ (ii). Substituting (i) into either argument results in the quadratic equation:

$f(a) = 3a^2 + 2a$

whose first & second derivavtives yield:

$f'(a) = 6a + 2 = 0 \Rightarrow a = -\frac{1}{3}$;

$f''(-\frac{1}{3}) = 6 > 0$ (hence, a global minimum)

which produces the critical pair $P_{1}(a,b) = (-\frac{1}{3}, -\frac{1}{3})$. Likewise, substituting (ii) produces:

$g(a) = 3a^2 + 2(\frac{2}{3} - a) = 3a^2 - 2a + \frac{4}{3}$;

$g'(a) = 6a - 2 = 0 \Rightarrow a = \frac{1}{3}$;

$g''(\frac{1}{3}) = 6 > 0$ (also a global minimum)

and the critical pair $P_{2}(a,b) = (\frac{1}{3}, \frac{1}{3})$. Checking both critical pairs against $M(a,b)$ now results in:

$M(-\frac{1}{3}, -\frac{1}{3}) = 3(-\frac{1}{3})^2 + 2(-\frac{1}{3}) = \boxed{-\frac{1}{3}}$.

$M(\frac{1}{3}, \frac{1}{3}) = 3(\frac{1}{3})^2 + 2(\frac{1}{3}) = \boxed{1}$.

Since $M_{P_{1}} \le M_{P_{2}}$, $\boxed{a = b = -\frac{1}{3}}$ are the critical values.

Now this is a solution I can understand !

However, the rather bold statement at the beginning surely needs some explaining.

i.e. "The function M(a,b) is minimized when both of its arguments equal each other"

Phil.

philboyd2 3 months ago

Apart from the obvious mistake (a wrong choice between the two extremal points) this solution is identical to mine; perhaps more detailed, but still without a justification for the equal values of the functions at the minimum point. As explained in my reply to Hubert, if a function is bigger than the other one near a minimum point for $M(a, b)$, such a function should have a local minimum in such a point; however none of our functions can have local minima.

Claudio 3 months ago