Let $m$ be any give postive ream numbers,and $a+b+c+d=m$,where $a,b,c,d\ge 0$ Find the maximum of the value $$a+ab+abc+abcd$$


when $m=1$,we have $a+ab+abc+abcd\le a(b+1)(c+1)(d+1)\le\left(\dfrac{a+b+c+d+3}{4}\right)^4=1$ when $a=1,b=c=d=0$

when $m=3$, I got the $$a+ab+abc+abcd=a+ab(1+c+cd)\le a+ab(1+c+d+cd)=a+ab(1+c)(1+d)\le a+a\left(\dfrac{b+1+c+1+d}{3}\right)^3=a+a\cdot\left(\dfrac{5-a}{3}\right)^3=4-\dfrac{1}{27}(a-2)^2(a^2-11a+27)\le 4$$ when $a=2,b=1,c=d=0$

for other any postive $m$,How to find it? partial derivative?

  • $\begingroup$ Yes, the Lagrange multipliers method helps, but the answer depend on $m$ . $\endgroup$ – Michael Rozenberg Aug 22 '17 at 13:01
  • $\begingroup$ Form a Lagrangian with multiplier $\lambda$. The first order conditions give $$ abc=ab(1+d)=a(1+c+cd)=1+b(1+c+cd)=\lambda.$$ From here $c=1+d$ and $b=(1+c^2)/c$ while $a = m-b-c-d$, so we may express $a,b,d$ as explicit functions of $c$. This leaves just a univariate optimization over $d$. $\endgroup$ – encore Aug 22 '17 at 13:01
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    $\begingroup$ I suspect that (a) with $0\le m \le 1$ you will achieve the maximum with $b=c=d=0$; (b) with $1\le m \le 3$ you will achieve the maximum with $c=d=0$ and $a=b+1$; (c) with $3\le m \le \frac{11}2$ you will achieve the maximum with $d=0$ and $b=c+1$ and $a=b+\frac{1}{b}$; and (d) $\frac{11}2 \le m$ you will achieve the maximum with $c=d+1$ and $b=c+\frac{1}{c}$ and $a=b+\frac{1}{bc}$ $\endgroup$ – Henry Aug 22 '17 at 13:02
  • $\begingroup$ Yes, the answer will naturally depend on $m$. $\endgroup$ – encore Aug 22 '17 at 13:03

This seems to me a classical problem of restricted optimisation. You want to maximise $ f(a,b,c,d)=a+ab+abc+abcd $ on the set $A=\{(a,b,c,d)|a,b,c,d\in\mathbb{R}_{\ge0}; a+b+c+d=m\}$ You first determine a (finite) set of candidates among which you will find the maximum. To get your candidates, you first guess that the maximum might be in the interior of $A$ and search look for candidates there, and then you check the boundary $\partial A$ of $A$.

Interior: You have the auxiliary condition $a+b+c+d-m=0$. We define the Lagrange function $L(a,b,c,d,\lambda):= f(a,b,c,d) + λ(a+b+c+d)$. An extreme point of $L$ is an extreme point of $f$ when subject to the auxiliary condition. We get the extreme points of $L$ by setting its gradient, i.e. all of its partial derivatives to zero: $$ \frac{\partial L}{\partial a} = 1+bc+bcd+\lambda =0$$ $$ \frac{\partial L}{\partial b} = a+ac+acd+\lambda=0$$ $$\frac{\partial L}{\partial c} = ab +abd+\lambda=0$$ $$\frac{\partial L}{\partial d}= abc+\lambda=0$$ $$\frac{\partial L}{\partial\lambda}=a+b+c+d-m=0$$ Solving this non-linear system of equations, I leave to you... (you can discard all solutions not in $\mathbb{R}_{\ge0}^4$)

Boundary: The boundary can be either of: $a=0,b=0,c=0$ or $d=0$. So to look for the remaining candidates on the boundary, we have to consider each coordinate on its own. Suppose we are considering the coordinate $x$ ($x$ being one of $a,b,c,d$) right now. We have two auxiliary conditions, namely $a+b+c+d-m=0$ and $x=0$, so we need two Lagrange multipliers, $\lambda$ and $\mu$. Our new Lagrange function looks as follows: $$ L(a,b,c,d,\lambda,\mu):= f(a,b,c,d) + \lambda(a+b+c+d-m) + \mu x.$$You proceed like above, except that you now have six equations and unknowns.

Finally, you evaluate $f$ at each candidate point and determine the maximum.


Denote the objective function by $f$. The domain under consideration is a $3$-dimensional simplex $\Sigma$ in ${\mathbb R}^4$. Since $f$ is nonhomogeneous the parameter $m>0$ occurring in the description of $\Sigma$ enters the discussion in an essential way. It is easy to see that at ${\rm argmax}$ of $f\restriction\Sigma$ we have $a\geq b\geq c\geq d\geq0$. This allows to restrict the analysis to the following strata of $\Sigma$: $$\eqalign{\rm{(i)} \quad &a>0,\qquad b=c=d=0;\cr \rm{(ii)} \quad &a>0,\quad b>0,\qquad c=d=0;\cr \rm{(iii)} \quad &a>0,\quad b>0,\quad c>0,\qquad d=0;\cr \rm{(iv)} \quad &a>0,\quad b>0,\quad c>0,\quad d>0\ .\cr}$$ Case (i) is trivial: One necessarily has $a=m$, hence $f=m$.

In case (ii) we have to consider the Lagrangian $$\Phi:=a+a b-\lambda(a+b)$$ and obtain the conditionally stationary point $$a_*={m+1\over2},\quad b_*={m-1\over2}\ .$$ The restriction $b>0$ implies that this point is only relevant when $m>1$. One obtains $$f(a_*,b_*,0,0)={1\over4}(m+1)^2\ ,$$ which is $>m$ when $m>1$.

In case (iii) we have to consider the Lagrangian $$\Phi:=a+a b+abc -\lambda(a+b+c)$$ and obtain after some calculation two conditionally stationary points $(a,b,c,0)$, whereby $$a={b^2+1\over b},\quad b={1\over6}\bigl(m+1\pm\sqrt{(m+1)^2-12}\bigr),\quad c=b-1\ .$$ Here the minus sign can be rejected, and the condition $c>0$ implies that the remaining point $(a_*,b_*,c_*,0)$ is only relevant when $m>3$. One then obtains $$f(a_*,b_*,c_*,0)={13 + \sqrt{(m+1)^2-12} + m \bigl(m+2 + \sqrt{(m+1)^2-12}\bigr)^2\over 54 (m+1 + \sqrt{(m+1)^2-12})}\ ,\tag{1}$$ which is $>{1\over4}(m+1)^2$ when $m>3$.

Case (iv) cannot be dealt with in a similar way since we are running into a fourth degree equation containing the parameter $m$. Therefore we have to stop here.

I'd say that when $m\leq1$ $\max f=m$, and that when $1<m\leq 3$ one has $\max f={1\over4}(m+1)^2$. When $3<m<?$ the maximum of $f$ is given by $(1)$, and starting at an unknown value of $m$ the ${\rm argmax} f$ will have all four coordinates $>0$.


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