Show that First order conditions are necessary and sufficient for utility maximization I have a budget set
$$B=\{x=(x_1,x_2)\in R^2_+ \mid 2\sqrt{x_1}+x_2\le y\}$$
where $y&gt;0$ is income.
Assuming the preferences are strictly monotonic and convex, I want to show that first order conditions are necessary and sufficient for an interior solution to the utility maximization problem
**My solution **
First step : I maximize utility subject to the given budget set
Second step: I calculate first order conditions
Third step : I use hessian matrix.
If the determinant of this hessian matrix H is negative, then I can say that the FOCs are necessary is sufficient. (Is this true?)
My solution is like that 
where $$u_{11}={\partial^2 u(x_1, u_2)\over \partial x_1^2}$$
$$u_{22}={\partial^2 u(x_1, u_2)\over \partial x_2^2}$$
$$u_{12}={\partial^2 u(x_1, u_2)\over \partial x_1 \partial x_2}$$
$$u_{21}={\partial^2 u(x_1, u_2)\over \partial x_1 \partial x_2}$$
I don’t know this answer is enough and correct. Because this solution seems not to be sufficient. Please share me your ideas about my solution.
Thanks a lot.
Consider the following situation :
Preferences of the consumer are represented by the utility function $u(x_1, x_2) = x_1 + x_2$. Note that preferences are strictly monotonic and convex. Now let income be any $y &gt; 2$. So the consumer's problem is :
\begin{eqnarray*} \max_{x_1, x_2} &amp; x_1 + x_2 \\ \text{s.t.} &amp;  \ \ 2\sqrt{x_1} + x_2 \leq y \\ &amp; x_1\geq 0, \ x_2 \geq 0 \end{eqnarray*}
Now let us set up the Lagrangian :
\begin{eqnarray*} \mathcal{L}(x_1, x_2) = x_1+x_2 - \lambda(2\sqrt{x_1} + x_2 - y) +\mu_1x_1 + \mu_2x_2 \end{eqnarray*} 
Here are the first order necessary conditions:
\begin{eqnarray*} \dfrac{\partial\mathcal{L}}{\partial x_1} &amp; = &amp; 1 - \dfrac{\lambda}{\sqrt{x_1}} + \mu_1= 0 \\ \dfrac{\partial\mathcal{L}}{\partial x_2} &amp; = &amp; 1 - \lambda + \mu_2= 0 \\ \dfrac{\partial\mathcal{L}}{\partial \lambda} &amp; = &amp; y -2\sqrt{x_1} - x_2  \geq 0, \ \lambda\geq 0, \ \lambda(2\sqrt{x_1} + x_2 - y) = 0 \\ \dfrac{\partial\mathcal{L}}{\partial \mu_1} &amp; = &amp; x_1  \geq 0, \ \mu_1\geq 0, \ \mu_1x_1 = 0 \\ \dfrac{\partial\mathcal{L}}{\partial \mu_2} &amp; = &amp; x_2  \geq 0, \ \mu_2\geq 0, \ \mu_2x_2 = 0 \end{eqnarray*}

Check that one of the solutions to the above set of conditions is $x_1^* = 1, \ x_2^* = y -2, \ \lambda^*=1, \ \mu_1^* = 0, \ \mu_2^* = 0$, 

But it is not the solution to the utility maximization problem stated above. This is because utility from $(x_1^*, x_2^*) = (1, y-2)$, i.e. $u(x_1^*, x_2^*) = 1 + y - 2= y-1$, is less than the utility from another affordable consumption bundle $(x^{**}_1, x^{**}_2) = (0, y)$ as $u(x_1^{**}, x_2^{**}) = 0 + y = y$ and $2\sqrt{x_1^{**}} + x_2^{**} =y$. 

Therefore, first order conditions are not sufficient for utility maximization.