Optimal decision for perfect substitutes utility function ? Given $u(x_1,x_2)=4x_1+14x_2$ and $m=\frac{1}{2}x_1+\frac{3}{2}x_2$, I shall choose the optimal decision among:
$a)(2m,\frac{2m}{3})$
$b)(2m,0)$
$c)(\frac{m}{2},0)$
$d)(0,\frac{2m}{3})$
The correct answer is $(d)$

But not sure how to find this, what I did:
$$m=\frac{1}{2}x_1+\frac{3}{2}x_2 \Leftrightarrow x_2=\frac{2m}{3}-\frac{1}{3}x_1 $$
And the marginal rate of substitution is $-\frac{2}{7}$,so we have :

Obviously $a)$ isn't the right one and $c)$ doesn't seem to be the best one... but really don't know 'how to think', some explaination would be great.
Thanks in advance
Utility maximization problem is:

$\max\limits_{(x_1, x_2)\in\mathbb{R}^2_+} 4x_1 + 14x_2 \\ \text{s.t.} \ \ m = \frac{1}{2}x_1 + \frac{3}{2}x_2$

This problem can be converted into an optimization problem involving just one variable $x_1$:

$\max\limits_{x_1} 4x_1 + 14\left[\frac{2}{3}\left(m - \frac{1}{2}x_1\right)\right] \\ \text{s.t.} \ \ 0 \leq x_1 \leq 2m$

Differentiating the objective $4x_1 + 14\left[\frac{2}{3}\left(m - \frac{1}{2}x_1\right)\right]$ with respect to $x_1$, we get $\left(4- \frac{14}{3}\right) = \frac{-2}{3} &lt; 0$. So we observe that the objective is decreasing in $x_1$. Therefore optimal choice must be the smallest $x_1$ we can have i.e. $x_1 = 0$ and hence $x_2 = \frac{2m}{3}$.