Consumer behavior with constrained demand Suppose there is a linear demand function $D(p)$ describing the demand of an individual consumer. This is derived from a utility function or is just observed, but let us assume it is correct. Now suppose that the product is available at price $p$, but the store insists that each consumer buys a minimum amount that is in fact larger than $D(p)$, but only by very little.
 Say it is $D(p) + \epsilon$, where $\epsilon &lt; D(p)/100$.

Can I say without additional assumptions that given the options of buying a quantity $q$ where $q \in \left\{0\right\} \bigcup \left[D(p) + \epsilon,\infty\right)$, the consumer will choose to buy $D(p) + \epsilon$?

My argument is that this will maximize her consumer surplus given the constraint on $q$. As I specified that this is a linear demand function and $\epsilon &lt; D(p)/100$ you can do the math, but it is easy to see that the surplus will be positive for any continuous demand function given a small enough $\epsilon$.

Is this an accepted line of argumentation?

In most models it is accepted that consumers seek to maximize their utility. I am not sure however if one can argue absent a utility function that surplus maximization is the goal.
 A workaround would be specifying a quasi-linear utility function supporting $D(p)$ but I would like to avoid this if possible.
In a two-good space, initially the consumer maximizes $U(x,z)\;\; s.t. \;\;p_xx+p_zz =I$ and we assume it obtains the solution $(x^*, z^*)$ as a function of prices and income.  

In the constrained case, the consumer will either choose $(0, \tilde z)$ or $(x^*+\epsilon, z'$), for some $\epsilon &gt;0 $ always exhausting its budget, so in particular, $\tilde z = I/p_z$.
In order for the consumer to still choose to buy a strictly positive quantity of $x$, it must be the case that 

$$U(x^*+\epsilon, z') &gt; U(0, \tilde z)$$

Apply a first order approximation around $(x^*, z^*)$ without ignoring the remainders, we want 

\begin{align}
 U(x^*, z^*) + U_x(x^*)\cdot \epsilon + U_z(z^*)(z'-z^*) + R_{\epsilon} &amp;\\&gt; U(x^*, z^*) + U_x(x^*)(-x^*) + U_z(z^*)(\tilde z-z^*) + R_0
\end{align}

Simplify and re-arrange, we want

$$U_x(x^*)(x^*+\epsilon) + R_{\epsilon} &gt; U_z(z^*)(\tilde z-z') + R_0 $$

We know that from the unconstrained optimization, $U_x(x^*)/U_z(z^*) = p_x/p_z$  so 

$$\frac {p_x}{p_z}(x^*+\epsilon) + \frac {R_{\epsilon}}{U_z(z^*)} &gt; \left(\frac{I}{p_z}-z'\right) + \frac {R_0}{U_z(z^*)} $$

Multiply throughout by $p_z$,

$$p_x(x^*+\epsilon) + p_z\frac {R_{\epsilon}}{U_z(z^*)} &gt; I - p_zz'  + p_z\frac {R_0}{U_z(z^*)} $$

but $p_x(x^*+\epsilon) + p_zz' = I \implies p_x(x^*+\epsilon) = I-p_zz'$ so we are left with the requirement that (ignoring positive terms)

$$R_{\epsilon} &gt;  R_0 $$

in order for the consumer to choose $x^*+ \epsilon$ and not $0$ for $x$.  

Note that the above take into account also the signs of the remainders, it is not just about their absolute magnitudes. 

Now let's go back to our first-order expansions. We know that both candidate bundles yield utilities lower than $U(x^*, z^*)$, because they were feasible in the unconstrained case, and they weren't chosen. 

Looking at the expansion of $U(0, \tilde z)$ we then conclude that we have 

$$U_x(x^*)(-x^*) + U_z(z^*)(\tilde z-z^*) + R_0 &lt; 0 $$

$$\implies U_z(z^*)\cdot \Big[(U_x(x^*)/U_z(z^*))\cdot(-x^*) + \tilde z-z^*\Big] + R_0 &lt; 0$$

$$\implies \frac {U_z(z^*)}{p_z}\cdot \Big[-p_xx^* + p_z\tilde z-p_zz^*\Big] + R_0 &lt; 0$$

But $-p_xx^* -p_zz^* = -I$ and $p_z\tilde z =I$ so the term in brackets is zero. Therefore we conclude that 

$$R_0 &lt;0$$

Looking now at the expansion of $U(x^*+\epsilon, z')$, we know we have

$$U_x(x^*)\cdot \epsilon + U_z(z^*)(z'-z^*) + R_{\epsilon} &lt; 0$$

Performing the same manipuations as before we obtain here too that

$$R_{\epsilon} &lt; 0$$

So the condition to buy $x^*+\epsilon$ can be re-written as

$$|R_{\epsilon}| &lt; |R_0|$$

This formalizes somewhat the notion that if $\epsilon$ is "sufficiently small", $R_{\epsilon}$ will be smaller in absolute terms than  $R_0$, since the approximation to the same function will be "better", and so we will observe $x^*+\epsilon$ and not $0$. But it also tells us what the graphs in the other answer told us too, that there is not a single general answer to the matter.