Explanation:
(C) : Given,
\(z=5 x+2 y\), subject to the constraints \(\mathrm{x}+\mathrm{y} \leq 7, \mathrm{x}+2 \mathrm{y} \leq 10, \mathrm{x}, \mathrm{y} \geq 0\)
| constraints | Equation | $\mathrm{x}$ | v | $(x, y)$ | Region |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $x+y \leq 7$ | $x+y=7$ | 7 | 0 | $(7,0)$ | Origin side |
| | | 0 | 7 | $(0,7)$ | |
| $x+2$ | $x+2 y=10$ | 10 | 0 | $(10$, \lt br> $0)$ | side |
The feasible region is shaded. \(\mathrm{B}\) is the point of intersection of \(x+y=7\) and \(x+2 y=10\) solving,
B \(\equiv(3,4)\).

The corner points of feasible region are \(\mathrm{O}(0,0), \mathrm{A}(7,0), \mathrm{B}(4,3)\) and \(\mathrm{C}(0,5)\).
| Points | $\mathbf{Z}=\mathbf{5 x}+\mathbf{2 y}$ |
| :---: | :---: |
| $\mathrm{O}(0,0)$ | 0 |
| $\mathrm{A}(7,0)$ | 35 |
| $\mathrm{B}(3,4)$ | 23 |
| $\mathrm{C}(0,5)$ | 10 |
\(\therefore\) Maximum value of \(\mathrm{z}\) is 35 .