Maximum value of algebraic expression 
If $2a+3b+4c=100$ and $a,b,c\in \mathbb{N}\;,$ Then $\max(2a^2+5b^2+8c^2)$

$\bf{My\; Try::}$ We can write it as $\displaystyle \frac{a}{50}+\frac{b}{\frac{100}{3}}+\frac{c}{25} = 1$
So it represent a plane which cut $x,y,z$ axis at $\displaystyle A(50,0,0)$ and $\displaystyle B\left(0,\frac{100}{3},0\right)$ and $C(0,0,25)$
Now how can i solve it after that, Help required, Thanks

First note that $2a+3b + 4c=100 \implies b \equiv 0 \pmod 2$ and $2a+c \equiv 1 \pmod 3$.
As $b$ is even, we may simplify a bit by letting $b = 2k$, then $a+3k+2c = 50$ and we maximize $V = a^2+10k^2+4c^2$.  
As $50^2 = a^2+9k^2+4c^2 + 2(3ak+6kc+2ca)$, we have 
$$V= k^2+50^2 - 2(3ak+6kc+2ca) = (k-(3a+6c))^2+50^2 - 4ca - (3a+6c)^2$$
So for any choice of $a, c$, setting $k=3a+6c$ is clearly the best choice.
We are left to minimise $4ca + (3a+6c)^2$, which is clearly increasing in $c, a$ with more sensitivity on $c$.  The first choice for minimum is thus $(a, c)=(1, 1)$ but this fails $\pmod 3$, the next best choice is $(2, 1)$ which also doesn't work, but $(a, c) = (1, 2)$ works.  So does $(3, 1)$ but this has a slightly higher value, so no further investigation is needed. 
Thus the maximum you seek is when $a=1, b=2k=30, c=2$.
