Minimum volume for cuboid with specific properties in relation to other cuboids

Question

I have lately started to practice some math for fun. I have come across the following exercise:
"There are three cuboids with the volumes $V_1=10x_1y_1$, $V_2=10x_2y_2$ and $V_3=10x_3y_3$. These must meet the following criterias $V_1\geq V_2$, $V_1\geq V_3$ and $V_1+V_2+V_3=1000$ with $x_n$ and $y_n$ being natural numbers $0\leq x_n$, $y_n\leq 10$. What is the smallest value for $V_1$?"
Firstly I tried an inelegant exhaustive search through a small script I wrote. With that I came up with the solution of $V_1=350$, which seems plausible but i would like something more elegant than that. I have also thought of representing the volumes like this: $V_1=1000-(V_2+V_3)$, $V_2=1000-(V_1+V_3)$, $V_3=1000-(V_1+V_2)$. I hoped to make progress with that but it seems like a dead end to me because all the expressions are dependent on each other so you can't really graph how they affect each other until you already know two volumes.
So I am asking what might be a more elegant approach to the exercise above?

Answer 1

You want $V_1$ to be as small as possible, but you also want it to be greater than $V_2$ and $V_3$. Therefore, it must be at least a third of the combined volume ($\ge 333.\bar{3}$). Since it must be a multiple of $10$, you can check $V_1 = 340, 350, ...$ whether it would work. For $V_1 = 10z$, it must be true that $z$ can equal $xy$ with $x, y \in [1, 10]$. Therefore, only $V_1 = 350, 360, 400, 420, 450, ...$ would work.

For $V_1 = 350$, you have that $V_2 + V_3 = 650$, but since $V_1 \ge V_2, V_3$, it must be true that $(V_2, V_3)$ is in $\{(300, 350), (310, 340), (320, 330)\}$. $V_2, V_3 = 300, 350$ would work because $300 = 10\cdot 5 \cdot 6$ and $350 = 10\cdot 5\cdot 7$. Therefore the minimum $V_1$ is $350$.

Answer 2

$V_1 = 10 x_1 y_1, V_2 = 10 x_1 y_1, V_3 = 10 x_1 y_1$

where $0 \leq x_n, y_n \leq 10\in \mathbb{Z+}$

$V_1 + V_2 + V_3 = 1000$

So reduce the problem to finding $min(x_1y_1)$

and we write for ease of working, $a = x_1 y_1, b = x_2 y_2, c = x_3 y_3$.

So we have $a + b + c = 100$ with $a \geq b, a \geq c$

So $a \geq 34$

Now you should look for two factors of $a$, both $\leq 10$.

So yes the first one is $35 = 5 \times 7$ and we are left with $b + c = 65$ and $b, c \geq 30$ (If any of them is $\lt 30$, the other one becomes $\gt a \, ( = 35)$ which is not allowed. So the possible values with both factors $\leq 10$ are $30 = 3 \times 10$ and $35 = 5 \times 7$ and they meet the criteria of $a \geq b, a \geq c$.

Hence $a = 35$ works with $min (V_1) = 350$.