Given the product and sum of three variables , how can we find the maximum possible value that any variable can attain?

Question

We've three numbers $a, b, c$ such that $a+ b+c =13$ and $abc=36$. What can the largest value of a variable amongst the three be in this case?

I tried to use calculus but that didn't work neither does eliminating a variable help. Id be glad if someone pointed out or hinted out anything. Thanks

Answer 1

If you want to do it using Lagrange multiplier method, here is how you may -

We want to maximize $f(a) = a$

Given the constraint $ab (13 - a - b) = 36 \,$ ....(i)

Using Lagrange multiplier method,

$a = \lambda (13ab-a^2b-ab^2-36)$

Differentiating wrt to $a, b$,

$1 = \lambda (13b - 2ab - b^2)$

$0 = \lambda (13a - a^2 - 2ab)$

From second equation, we have $13a - a^2 - 2ab = 0$ ($\lambda \ne 0$)

i.e $a(13-a-2b) = 0$ leads to $a = 0, a + 2b = 13$.

As $a \ne 0$ given our constraint, $a + 2b = 13 \implies b = c, b = \frac{13-a}{2}$.

Substituting back in our constraint (i),

$a \times \frac{13-a}{2} \times \frac{13-a}{2} = 36$

or $\, a^3 -26a^2 + 169a - 144 = 0$

or $\, (a-1)(a-9)(a-16) = 0$.

Testing these values, we get max and min of $9, 1$ for $a, b, c \gt 0$. $a = 16, b = c = -\frac{3}{2}$ is a solution though in real.

Answer 2

Let's suppose $a,b,c \in \Bbb N$ and $1 \le a \le b \le c$.

By the first equation we have $c \le 11$. Let's do it manually: $11 \not \mid 36$ so $c \neq 11$, moving on to the next value, $10 \not \mid 36$ so $c \neq 10$, but $9 \mid 36$ so $c = 9$ could work: we have $ab=4$ and $a+b=4$ so $a=b=2$. Thus the largest value of a variable amongst the three is $9$.