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

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

$\mathrm{A}(0,5),\mathrm{B}(4,3),\mathrm{C}(7,0)$.

The objective function is maximum at $\mathrm{C}(7,0)$ and 

Max $z=5\times 7+2\times 0=35$.