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$p$,$q$,$r$ are $3$ non-negative real numbers less than or equal to $1.5$ such that $p+q+r = 3$, what will be the maximum of $p^3 + q^3 + r^3 + 4pqr$ ?


I tried AM-GM on $p,q,r$ to get the maximum of $pqr$ as $1$, but on doing it for $p^3 + q^3 + r^3 $, I get the minimum, so I won't be able to combine them.

I tried to assume symmetry $p=q=r=1$, and the maximum value to be 7, but putting $p=1.1$, $q=1.1$ and $r=0.8$, I get the value to be $7.046 (> 7) $

How do I solve this with concepts like AM-GM?

PS: I do not know multivariable calculus concepts, however if it's not possible to solve it without using it, please show how to solve it with these concepts.

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    $\begingroup$ AM-GM's only equality case is when all terms are equal, so applying AM-GM on an inequality with an equality case where not all terms are equal will cause the inequality to fail. Maybe you might want to find what p,q,r should be for that to be maximal. $\endgroup$ – the4seasons Dec 17 '17 at 16:31
  • $\begingroup$ I suggest you look into partial differentiation, multivariate calculus and their applications to finding Maxima & minima of functions. $\endgroup$ – unseen_rider Dec 17 '17 at 17:01
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To maximize $$ p^3+q^3+r^3+4pqr\tag1 $$ while maintaining $$ p+q+r=3\tag2 $$ Means that we need to find $p,q,r$ so that for all variations so that $$ \delta p+\delta q+\delta r=0\tag3 $$ we also have $$ \left(3p^2+4qr\right)\delta p+\left(3q^2+4pr\right)\delta q+\left(3r^2+4pq\right)\delta r=0\tag4 $$ Orthogonality requires that $$ 3p^2+4qr=3q^2+4pr=3r^2+4pq\tag5 $$ If $p\ne q$, then $$ \begin{align} 3\left(p^2-q^2\right)&=4r(p-q)\\ \frac34(p+q)&=r\tag6 \end{align} $$ Since $p\ne q$, either $p\ne r$ or $q\ne r$. WLOG assume $p\ne r$. Then, as in $(6)$, we have $$ \frac34(p+r)=q\tag7 $$ Comparing $(6)$ and $(7)$ yields $$ 3p=q=r\tag8 $$


Noting symmetries, and the solution $p=q=r$, we get the solutions $$ \left\{(1,1,1),\left(\frac37,\frac97,\frac97\right),\left(\frac97,\frac37,\frac97\right),\left(\frac97,\frac97,\frac37\right)\right\}\tag9 $$ which give the values in $(1)$ to be $$ \left\{7,\frac{351}{49},\frac{351}{49},\frac{351}{49}\right\}\tag{10} $$ The point $(1,1,1)$ is a local minimum and the other three are saddle points.


There are also the edge cases of $p=0$ or $q=0$ or $r=0$. WLOG assume $r=0$. Equations $(5)$ become $p=q$. Symmetry gives the cases $$ \left\{\left(\frac32,\frac32,0\right),\left(\frac32,0,\frac32\right),\left(0,\frac32,\frac32\right)\right\}\tag{11} $$ which give the values in $(1)$ to be $$ \left\{\frac{27}4,\frac{27}4,\frac{27}4\right\}\tag{12} $$


Then there are the corner cases $$ \{(3,0,0),(0,3,0),(0,0,3)\}\tag{13} $$ which give the values in $(1)$ to be $$ \{27,27,27\}\tag{14} $$


Given the constraint that $p,q,r\le\frac32$, we have three more edge cases symmetric to $r=\frac32$ and $p+q=\frac32$. Equations $(5)$ become $3p^2+6q=3q^2+6p$, which, if $p\ne q$, implies $p+q=2$; thus, $p=q=\frac34$. Symmetry gives the cases $$ \left\{\left(\frac32,\frac34,\frac34\right),\left(\frac34,\frac32,\frac34\right),\left(\frac34,\frac34,\frac32\right)\right\}\tag{15} $$ which give the values in $(1)$ to be $$ \left\{\frac{243}{32},\frac{243}{32},\frac{243}{32}\right\}\tag{16} $$


The greatest value inside the $p,q,r\le\frac32$ constraint is attained by case $(15)$. This gives the maximum of $$ \bbox[5px,border:2px solid #C0A000]{\frac{243}{32}}\tag{17} $$


Here is a plot of $p^3+q^3+r^3+4pqr$ where $r=3-p-q$. The edges are where $p=\frac32$, $q=\frac32$, and $p+q=\frac32$. The points in $(15)$ are colored red. The interior points are those in $(9)$ and the corners are those in $(11)$.

enter image description here

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  • $\begingroup$ Why the downvote? I believe all cases were covered. $\endgroup$ – robjohn Dec 18 '17 at 6:28
  • $\begingroup$ You said 3p=q=r, but then gave solutions with p=3q=3r $\endgroup$ – Michael Dec 18 '17 at 12:11
  • $\begingroup$ @Michael: Fixed. It didn't change the overall answer, but those points are now inside the $\frac32$ limit. Thanks! $\endgroup$ – robjohn Dec 18 '17 at 13:02
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A hint before the answer:

$p^3+q^3+r^3+4pqr=p^3+(q+r)^3+4pqr-3(q+r)(qr)$

Try to then eliminate both $q$ and $r$ to make this in terms of $p$.


Motivation

Firstly one might try $p=q=r=1$ and get $p^3+q^3+r^3+4pqr=7$

But the trying of more cases will reveal that this is not the maximum. One might then try to fix a term and see what is the maximum with the fixed term.

e.g. Set $p=3/2$. Then $p^3+q^3+r^3+4pqr=27/8+q^3+r^3+6qr$, $q+r=3/2$

The $q^3+r^3$ motivates writing $q^3+r^3+6qr$ in the form $(q+r)^3-something$ as $(q+r)^3=27/8$, dealing with the $q^3$ and $r^3$ terms.

i.e. $q^3+r^3+6qr=q^3+r^3+3(q+r)(qr)+(6-3q-3r)qr=(3/2)^3+(3/2)qr$

Then we are left to maximise $qr$, which is simple by AM-GM. (i.e. $q=r=3/4$) Trying to do this without fixing p at the start leads to the solution.


Solution $$p^3+q^3+r^3+4pqr$$ $$=p^3+q^3+r^3+3(q+r)qr+(4p-3q-3r)qr$$ $$=p^3+(3-p)^3+(4p-(9-3p))qr$$ $$=p^3+(3-p)^3+(7p-9)qr$$ This is
less than or equal to $p^3+(3-p)^3+(7p-9)(\frac{3-p}{2})^2$, equality when $q=r$ if $p\geq 9/7$ and
less than or equal to $p^3+(3-p)^3+(7p-9)(1.5)(1.5-p)$, equality when one of $q, r$ equals $1.5$ if $p\leq 9/7$.

Note that among $p, q, r$, the smallest number has to be less than or equal to $1$. Without loss of generality $p\leq q \leq r$. Then $p\leq 9/7$. For $p^3+q^3+r^3+4pqr$ to be maximal, one of $q, r$ is 1.5. As $r$ is the largest, $r=1.5$.

Finally, note that $3/2>9/7$, so we can apply the other case, replacing $r$ with $p$ (which is fine since $p^3+q^3+r^3+4pqr$ is symmetric over the 3 terms), concluding that if $r=3/2$, $p^3+q^3+r^3+4pqr$ is maximal when $p=q=3/4$.

Therefore, the maximum achievable is when one of $p, q, r$ is $3/2$ and the other 2 are $3/4$, to get a value of $\frac{243}{32}$.

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  • $\begingroup$ Sure. q+r = 5-p, but what about qr? $\endgroup$ – Rick Dec 17 '17 at 17:39
  • $\begingroup$ You should take some time to think about it. AM-GM might be useful. $\endgroup$ – the4seasons Dec 17 '17 at 17:43
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    $\begingroup$ @the4seasons I wanted to down-vote the Macavity's solution and erroneously down-voted your. I tried to change it, but it's impossible. I am sorry! $\endgroup$ – Michael Rozenberg Dec 18 '17 at 6:58
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Let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.

Hence, the condition does not depend on $w^3$ and we need to find a maximal value of $f,$ where $$f(w^3)=27u^3-27uv^2+7w^3.$$ We see that $f$ increases,

which says that it's enough to solve our problem for the maximal value of $w^3$.

Now, $p$, $q$ and $r$ are three non-negative roots of the following equation. $$(x-p)(x-q)(x-r)=0$$ or $$x^3-3ux^2+3v^2x-w^3=0$$ or $$x^3-3ux^2+3v^2x=w^3.$$ Let $u$ and $v^2$ will be constants and $w^3$ is changing.

Since the line $y=w^3$ and the graph of $y=x^3-3ux^2+3v^2x$ have three common points,

we see that $w^3$ will get a maximal value,

when a line $y=w^3$ is a tangent line to the graph of $y=x^3-3ux^2+3v^2x$,

which happens for equality case of two variables.

Also, we need to check the case, when one of our variables is equal to $\frac{3}{2}.$

  1. Two variables are equal.

Since our inequality is symmetric, we can assume $b=a$.

Thus, $c=3-2a\leq\frac{3}{2},$ which gives $\frac{3}{4}\leq a\leq\frac{3}{2}$ and we need to find a maximum of $g$, where $$g(a)=2a^3+(3-2a)^3+4a^2(3-2a)$$ and easy to see that $$\max_{\frac{3}{4}\leq a\leq \frac{3}{2}}g=g\left(\frac{3}{4}\right)=\frac{243}{32}.$$ 2. One of variables is equal to $\frac{3}{2}.$

Let $c=\frac{3}{2}.$

Hence, $b=\frac{3}{2}-a$ and we need to find a maximal value of $h$, where $$h(a)=a^3+\left(\frac{3}{2}-a\right)^3+\frac{27}{8}+3a(3-2a)$$ and easy to see that $$\max_{0\leq a\leq\frac{3}{2}}h=h\left(\frac{3}{4}\right)=\frac{243}{32},$$ which gives the answer: $\frac{243}{32}.$

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