Show that First order conditions are necessary and sufficient for utility maximization

Question

I have a budget set

$$B=\{x=(x_1,x_2)\in R^2_+ \mid 2\sqrt{x_1}+x_2\le y\}$$

where $y>0$ 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.

Answer 1

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 $y > 2$. So the consumer's problem is : \begin{eqnarray*} \max_{x_1, x_2} & x_1 + x_2 \\ \text{s.t.} & \ \ 2\sqrt{x_1} + x_2 \leq y \\ & 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} & = & 1 - \dfrac{\lambda}{\sqrt{x_1}} + \mu_1= 0 \\ \dfrac{\partial\mathcal{L}}{\partial x_2} & = & 1 - \lambda + \mu_2= 0 \\ \dfrac{\partial\mathcal{L}}{\partial \lambda} & = & 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} & = & x_1 \geq 0, \ \mu_1\geq 0, \ \mu_1x_1 = 0 \\ \dfrac{\partial\mathcal{L}}{\partial \mu_2} & = & 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.

Answer 2

Hi game I see that quite well! I don't know if you have already seen Khun Tucker Conditions as you use a Lagrangian. because lagrangian is with equalities and Khun Tucker with inequalities, and as you see that's inequality and the matrix conditions change, but let's assume you didn't saw Khun Tucker and Lagrangian is okay, I find okey your answer, but that's because the Utility is convex too! And maybe having in consideration that the set is compact and close! you can show it in a graph.