A Non-Symmetric Expression?

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

Let $A=ab(72-3a-4b)$ .

If $72-3a-4b\le0$ , $A\le0$

If $72-3a-4b>0$ , applying the AM-GM inequality, we have:

$A=\dfrac{1}{12}\cdot3a\cdot4b\cdot(72-3a-4b)\le\dfrac{1}{12}\cdot\left(\dfrac{3a+4b+(72-3a-4b)}{3}\right)^3=1152$

The equality holds iff $3a=4b=72-3a-4b\Leftrightarrow \left\{\begin{array}{l}a=8\\b=6\end{array}\right.$ .

So $\max A=\boxed{1152}$ , when $a=8; b=6$

I will be using a slight different method of multivariable calculus,

Let $y=72ab-3ab^2-4ab^2$ ,

Partial differentiation of $y$ with respect to $a$ and $b$ yields,

$\frac{\partial y}{\partial a} = 72b-6ab-4b^2$ and $\frac{\partial y}{\partial b} = 72a-3a^2-8ab$ and for minimum or maximum value $\frac{\partial y}{\partial a} \text{ and } \frac{\partial y}{\partial b} = 0$ .

Now we have got two variables and two equations we can solve for $a$ and $b$ , after solving the equations note that for maximum value of $y$ we must take $a=8 \text{ and } b=6$ instead of $a=0 \text{ and } b=0$ (as it would minimum value of y).

Hence,

$y|_{max}=72(8)(6)-3(8)^2(6)-4(8)(6)^2=\boxed{1152}$ .

Using AM GM for 3a+4b we get -(3a +4b)<=2(√12ab). Now put ab=m and we get ab= 12/√3. And finally answer

It is non-symmetric. Let us make it symmetric by the transformation $a' \Rightarrow \frac{a}{4}$ and $b'\Rightarrow \frac{b}{3}$ . The expression becomes $144a'*b'(6-a'-b')$ which is symmetric.Now if we interchange the new variables the value of the expression remains unchanged.It is also a polynomial in two variables and hence smooth and differentiable infinitely.Therefore, it follows that the extrema will occur along the line $a'=b'$ .One quickly finds the value by converting it into a one dimensional polynomial and finds the extrema.