The maximum value of $z=5 x+2 y$ subject to constraints $x+y \leq 7, x+2 y \leq 10, x, y \geq 0$

Change the inequalities into equations and draw the graph of lines, thus we get the required feasible region as shown below.


The region bounded by the vertices $A(0,5), B(4,3), C(7,0)$
The objective function is maximum at $\mathrm{C}(7,0)$ and $\operatorname{Max} z=5 \times 7+2 \times 0=35$.