$$\begin{array}{ll} \text{maximize} & f(x, y) := 2xy + 3x\\ \text{subject to} & x + 2y = 83\end{array}$$
I can apply, for example, the Lagrange theorem since its assumptions are satisfied and computing the Lagrange system I get the critical point is $x=43$, $y=20$. However, the Lagrange theorem gives only the necessary condition. How can I show that the function has a min/max at $x=43, y=20$?It looks like I cannot use Weirstrass theorem since the set $x + 2y = 83$ is closed but not bounded..
As suggested by @dmtri's comment, solving the constraint for $y$ yields $y= \frac12(83-x)$. Substituting this into the objective function yields $$f(x) = 2x\cdot\frac12(83-x) +3x = x(83-x) + 3x = x(86-x) = -x^2 + 86x.$$ Now, the graph $\{(x,f(x)): x\in\mathbb R\}$ is a descending parabola. We can write the quadratic $-x^2+86x$ in vertex form: $-x^2+86x = a(x-h)^2+k$ where the coordinates of the vertex are $(h,k)$. Expanding the vertex form, we have $ax^2 - 2ahx + ah^2+k$. Since $f(0)=0$, we immediately see that $ah^2+k=0$. Moreover, $a$ must match the coefficient of $x^2$, so $a=-1$. This yields $-x^2 +2hx = -x^2+86x$, and hence $h=43$. It follows then that $k= -ah^2 = 43^2 = 1849$. Since $f$ is concave, the maximum value that $f$ obtains is thus $1849$.
To solve this using calculus, we first note that $f$ is continuously differentiable and that $\lim_{x\to-\infty}f(x)=\lim_{x\to\infty} f(x)=-\infty$, so the maximum must occur at a point where $f'(x)=0$. Differentiating, we have $f'(x) = -2x +86$. Solving $f'(x)=0$ yields $x=43$. Now, $f'(x)>0$ for $x<43$ so $f$ is increasing on $(-\infty,43)$ and $f'(x)<0$ for $x>43$ so $f$ is decreasing on $(43,\infty)$, so $x=43$ is a local (and because $f$ is concave, global) maximum for $f$. We evaluate $f(43) = 43(86-43) = 43^2 = 1849$.
Although the above method is simpler, we can still use Lagrange multipliers to solve this problem. To solve \begin{align} \max&\quad f(x,y)\\ \mathrm{s.t.}&\quad g(x,y)=0 \end{align} we form the system of equations \begin{align} \nabla f(x,y) &= \lambda\nabla g(x,y)\\ g(x,y) &= 0, \end{align} where $\nabla$ denotes the gradient operator. We compute $$ \nabla f(x,y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2y+3, 2x) $$ and $$ \lambda\nabla g(x,y) = \lambda \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = (\lambda, 2\lambda), $$ which yields the system of equations: \begin{align} 2y + 3 &= \lambda\\ 2x &= 2\lambda\\ x + y - 83 &= 0. \end{align} By inspection, $x=\lambda$, whence $y = \frac12(x-3)$. Substituting into the third equation, we find that $$x = x + \frac12(x-3) - 83 = 0 \implies x=43.$$ Therefore $y = \frac12(43-3) = 20$, and the maximum of $f$ is obtained at $(43,20)$, with $f(43,20) = 1849$. To prove that $(43,20)$ is indeed a maximum, we consider the Lagrangian function $L(x,y) = f(x,y)-\lambda g(x,y) = 2xy+3x - \lambda(x+2y-83)$ and the bordered Hessian matrix $$ \mathcal HL(x,y) = \begin{pmatrix} \frac{\partial^2L}{\partial x^2}&\frac{\partial^2L}{\partial x\partial y}&\frac{\partial^2L}{\partial x\partial \lambda}\\ \frac{\partial^2L}{\partial y\partial x}&\frac{\partial^2L}{\partial y^2}&\frac{\partial^2L}{\partial y\partial\lambda}\\\frac{\partial^2L}{\partial \lambda\partial x}&\frac{\partial^2L}{\partial\lambda \partial y}&\frac{\partial^2L}{\partial\lambda^2} \end{pmatrix} = \begin{pmatrix}0&2&-1\\2&0&-2\\-1&-2&0\end{pmatrix}. $$ Since $\det(\mathcal HL(43, 20))=-8<0$, we conclude that $(43,20)$ is indeed a maximum.