How can we minimize $z=2x_1+3x_2-x_1^2-2x_2^2$ subject to $x_1+3x_2\le 6$, $5x_1+2x_2\le 10$, and $x_1,x_2\ge 0$?
I need to know the steps to solve or at least the guidelines as I am really new to nonlinear equations and I am trying this on my own.
How can we minimize $z=2x_1+3x_2-x_1^2-2x_2^2$ subject to $x_1+3x_2\le 6$, $5x_1+2x_2\le 10$, and $x_1,x_2\ge 0$?
I need to know the steps to solve or at least the guidelines as I am really new to nonlinear equations and I am trying this on my own.
The constraints describe a closed region bounded by a quadrilateral with vertices $$(0,0),\ (0,2),\ (2,0),\ (18/13,20/13).$$ The objective function, using $x,y$ for your $x_1,x_2$, is $$z=2x+3y-x^2-2y^2.$$ The standard technique for such problems is to find the critical point(s) of $z$ which lie in the interior of the region, and then the min/max must occur either at such a critical point, or else at a boundary point of the constraint region.
NOTE: critical points are where both partials of $z$ w.r.t $x,y$ are zero, in case that idea is new to you.
This $z$ has one critical point at $(x,y)=(1,3/4)$ where $z=17/8$, but that isn't the min since at $(0,0)$ we have $z=0$. So the min is on the boundary.
The next step is to make up functions describing each segment on the boundary, plug each one into the $z$ formula and minimize. I won't do this part, but note that at $(x,y)=(0,2)$ we have $z=-2$, and at least this is the min among the vertices of the quadrilateral. There may be a smaller value along one of the sides of the quadrilateral.
This are just some thoughts. You can refer to coffemath's answer for an analytical approach. This is from an optimization point of view.
Your problem can be re-written as \begin{align} \min_{\mathbf{x}~\in~R^2}~&\mathbf{x}^T\mathbf{Q}\mathbf{x}+\mathbf{r}^T\mathbf{x} \\~subject~to~&\mathbf{Ax\leq b} \\ &\mathbf{x\geq 0} \end{align}
where \begin{align} \mathbf{ x = \begin{bmatrix} \textit{x}_1 \\ \textit{x}_2 \end{bmatrix}~~Q = \begin{bmatrix}-1 & 0 \\ 0 & -2\end{bmatrix}~~r = \begin{bmatrix}2 \\ 3 \end{bmatrix}~~A = \begin{bmatrix}1 & 3 \\ 5 & 2\end{bmatrix}~~b = \begin{bmatrix}6 \\ 10 \end{bmatrix} } \end{align}
This is the well-known Quadratic Programming with linear inequality constraints which you can find in many textbooks on optimization.