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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

juantheron
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  • Can't we use Lagrange's multipliers? –  Nov 15 '16 at 08:20
  • @Muralidharan - won't necessarily give integer solutions, and then there is no guarantee nearest integers to the solution will optimize in general. If you seek real solutions, then necessarily the maximum is at one of $A, B, C$ as the objective function is quadratic and the domain closed & convex. – Macavity Nov 15 '16 at 10:47
  • Note that if $0 \in \mathbb N$ the answer is different than if $0\not\in\mathbb N$, so the exact definition of $\mathbb N$ is important. See http://math.stackexchange.com/questions/283/is-0-a-natural-number. – David K Nov 15 '16 at 14:32

1 Answers1

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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 $a=1, b=2k=30, c=2$.

Macavity
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