$a\cdot b\cdot c = 60$, What is the maximum value of $a + b + c$?

Question

$a$, $b$ and $c$ are positive integers with different numbers.

$a\cdot b\cdot c = 60$

What is the maximum value of $a+b+c$?

I'm trying to find the best method to solve this question. What are the methods we might use? Also, trial and error seems a bit useless for this question.

Answer 1

A way to solve this is to factorize $60$ giving $2\cdot 2\cdot 3\cdot 5$.

Now it should be fairly obvious that if we want to maximize the sum we want to group the biggest factors together. The best way to do this is just distributing the smallest factors (after $1$) into all but one of the variables, and clumping all the biggest ones into just one:

$$a = 1$$ $$b = 2$$ $$c = 2\cdot 3 \cdot 5 = 30$$

Giving maximum sum $33$.

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In general an algorithm for splitting up a number $n$ into $k$ different variables with the greatest sum is to repeatedly assign the smallest divisor $d$ of $n$ not assigned before to a variable, and continuing with $n/d$ and $k-1$ variables.

Answer 2

Let $a>b>c$.

Now, $a\leq30$ because if $a>30$ then $a=60$, which gives $b=c=1$, which is impossible.

Since $f(x)=e^x$ is a convex function and $$(\ln30,\ln2,\ln1)\succ(\ln{a},\ln{b},\ln{c}),$$ by Karamata we obtain: $$a+b+c=e^{\ln{a}}+b^{\ln{b}}+c^{\ln{c}}\leq e^{\ln30}+e^{\ln2}+e^{\ln1}=33.$$ The equality occurs for $(a,b,c)=(30,2,1)$, which says that we got a maximal value.

Done!

The minimal value of $a+b+c$ we can get by AM-GM.

Indeed, $$a+b+c\geq3\sqrt[3]{abc}=3\sqrt[3]{60}=11.7...,$$ which says that $$a+b+c\geq12.$$ The equality occurs for $(a,b,c)=(5,4,3)$, which says that $12$ is a minimal value.

Answer 3

Let us focus on the property that $a,b,c$ are all integers which are different from each other.

We realise that to make $c$ as large as possible, $a$ and $b$ have to be as small as possible to satisfy $a \cdot b \cdot c = 60$. To do this, we let $a=1$, and $b$ to be the next largest factor $2$. From this, $c=30$.

Therefore, $a=1$, $b=2$, and $c=30$, which has the sum $33$.

This strategy, along with @orlp's only works when the numbers are all integers. Otherwise, we could set $a$ and $b$ as close to $0$ as we want, and $c$ will just keep getting larger and larger.