Question : x<y<z and x,y,z are natural numbers. x+2y+3z=74 ---> what is x+y+z max value? answer:36
How can i solve that?
How can i solve that?
I thought L was a genious.
You stemmaxxing bro?
nigga who give a fuck
Is 36 the max value of x+y+z?
yeah i learnt now.x= 11 y=12 z= 13
low iq
Yes it is I just wanted to make sure
G
i actually have no idea i guess i suck at math
why waste your time
I'd make a permutation table in google sheets. You could also plot it as a 3d surface in some graph program/3d engine. It usually gets easier when you visualize your data.
But he lost
It reminded me how I hate math
G
Put the answer in a spoiler I was going to work it out 
@BoltzmanOscillate @blackcel are highest IQ mathcels on this forum
he was a fucking retard show is rarted
N
Fuck education. Only motivator to succeed in school for bluepillers it to betabux whore, for incels there is none. Just get a job as a cashier or something if you want one
Let's guess a value for Z and X
If x=1 and z=0, therefore;
1+2y=74
y=36.5 ( it's a rational number, so this can't be our answer)
if x=2 and z=0, then
2+2y=74
y=36 ( ok, it's a natural number. But let's keep going)
if x=3 and z=0
3+2y=74
y=35,5 ( again a rational number)
if x=4 and z=0
4+2y=74
y=35 ( it's a natural number, but 35<36)
Note that for every natural number greater than 1, the value of Y will decrease. Therefore, the maximum value for Y is 36. Plugging the value in x+2y+3z=74, we have 2+2x36+0=74.
If x=1 and z=0, therefore;
1+2y=74
y=36.5 ( it's a rational number, so this can't be our answer)
if x=2 and z=0, then
2+2y=74
y=36 ( ok, it's a natural number. But let's keep going)
if x=3 and z=0
3+2y=74
y=35,5 ( again a rational number)
if x=4 and z=0
4+2y=74
y=35 ( it's a natural number, but 35<36)
Note that for every natural number greater than 1, the value of Y will decrease. Therefore, the maximum value for Y is 36. Plugging the value in x+2y+3z=74, we have 2+2x36+0=74.
I only know every math an undergraduate electrical engineer would need to know. A mathmajorcel would be able to surpass me in abstract math, easily.
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