Find the maximum value of $A$

Question

Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$

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WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$

So i need to prove $A\le 36$. Or I will prove

$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$

Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2-12abc-b^2c-bc^2)\ge 0$

Then Im stuck here, help me solve it.

Answer 1

For $(a,b,c)\rightarrow(0,6,0)$ we have $A\rightarrow72$ and since $$72-a^2bc-a^2-2b^2-2c^2\geq0$$ it's $$2(a+b+c)^4-(a+b+c)^2(a^2+2b^2+2c^2)-36a^2bc\geq0$$ or $$a^4+6a^3b+6a^3c+9a^2b^2+9a^2c^2-14a^2bc+4ab^3+4ac^3+20b^2ac+20c^2ab+4ab^3+4ac^3+8b^2c^2\geq0,$$ which is obvious, we see that the maximum does not exist, but the supremum is $72$.

Answer 2

Given that $a+b+c=6$ we can substitute $a=6-b-c$ into $A$ to get $$A=(6-b-c)^2(bc+1)+2(b^2+c^2),$$ where $b,c>0$ and $b+c<6$. Differentiating with respect to $b$ and $c$ yields $$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2c+4b,$$ $$\frac{\partial A}{\partial b}=-2(6-b-c)(bc+1)+(6-b-c)^2b+4c.$$ If there is a maximum then both derivatives must equal 0 in such a point, and hence $$0=\frac{\partial A}{\partial b}-\frac{\partial A}{\partial c}=(c-b)((6-b-c)^2-4),$$ so either $b=c$ or $b+c=4$. In the latter case we get $c=4-b$ and so $$A=2(bc+1)+2(b^2+c^2)=2(b(4-b)+1)+2(b^2+(4-b)^2)=2b^2-8b+34,$$ which does not assume a maximum on $(0,4)$. In the former case where $b=c$ we get $$A=(6-2b)(b^2+1)+4b^2=-2b^3+10b^2-2b+6,$$ which does not assume a maximum on $(0,3)$. So the maximum does not exist.

Answer 3

With the help of the Lagrange multipliers and $f(a,b,c) = a^2 b c + a^2 + 2 b^2 + 2 c^2$

$$ L(a,b,c,\lambda) =f(a,b,c)+\lambda(a+b+c-6) $$

the stationary points are the solutions for

$$ \nabla L = \left\{ \begin{array}{c} 2 b c a+2 a+\lambda =0 \\ c a^2+4 b+\lambda =0 \\ b a^2+4 c+\lambda =0 \\ a+b+c-6=0 \\ \end{array} \right. $$

giving

$$ \left[ \begin{array}{ccccc} a & b & c & \lambda & f(a,b,c)\\ -2 & -1 & 9 & -32 & \color{red}{132} \\ -2 & 9 & -1 & -32 & \color{red}{132} \\ 2 & 1 & 3 & -16 & \color{green}{36} \\ 2 & 3 & 1 & -16 & \color{green}{36} \\ 3 & \frac{3}{2} & \frac{3}{2} & -\frac{39}{2} & \color{green}{\frac{153}{4}} \\ 3-\sqrt{5} & \frac{1}{2} \left(3+\sqrt{5}\right) & \frac{1}{2} \left(3+\sqrt{5}\right) & -12 & \color{green}{32} \\ 3+\sqrt{5} & \frac{1}{2} \left(3-\sqrt{5}\right) & \frac{1}{2} \left(3-\sqrt{5}\right) & -12 & \color{green}{32} \\ \end{array} \right] $$

In red non feasible solutions and in green feasible solutions. The maximum corresponds to $\frac{153}{4}= 38.25$

NOTE

The same solution can be obtained under the tangency condition between $f(a,b,c)-\mu=0$ and $a+b+c-6=0$ by calculating the values for $\mu$

Answer 4

Next to the usage of Lagrange multipliers, one can also solve it directly function-wise by assuming $c(a,b)=6-a-b$, which leads to the following derivatives to find the extrema:

$$\left\{\begin{align} \frac{\partial A}{\partial a} &= (ab-2)(12-3a-2b)&=0 \\ \frac{\partial A}{\partial b} &= (a^2-4)(6-a-2b)&=0 \\ \end{align}\right.$$

This gives us various solutions:

1. ${\partial A}/{\partial b}=0$ gives us $a=2$ leading to:

   • $(a,b,c)=(2,1,3)$ with A = 36
   • $(a,b,c)=(2,3,1)$ with A = 36

2. ${\partial A}/{\partial b}=0$ gives us $2b=6-a$ resembling $b=c$, leading to:

   • $6-2a=0$ or $(a,b,c)=(3,3/2,3/2)$ with A = 38.25
   • $ab=2$ which creates a quadratic eqution having : $(a,b,c)=(3\pm\sqrt5,(3\mp\sqrt 5)/2,(3\mp\sqrt 5)/2)$ with A = 32

From all these five points, the Hessian matrix shows that they are saddle points, with two exceptions, $(a,b,c)=(3-\sqrt5,(3+\sqrt 5)/2,(3+\sqrt 5)/2)$ which is a local minimum and $(a,b,c)=(3,3/2,3/2)$ which is a local maximum.

As the domain of investigation is $a>0,b>0,c>0$ we should also investigate these various planes for extrema:

1. $a=0$ gives that $A=72-24b +4b^2$ which has a local minimum in $b=3$ with A = 36
2. $b=0$ gives that $A=72-24a+3a^2$ which has a local minimum in $a=4$ with A = 24

Note: these points are not true extrema of the function $A(a,b,c(a,b))$

The real maximal value, however, is reached when $(a,b,c)$ tends to $(0,0,6)$ with A=72.

Note: this point is the highest value reached, which lays just outside of the domain and is a supremum.