MCQ
The feasible solution for a $LPP$ is shown in Figure Let $z=3 x-4 y$ be the objective function. Maximum of $Z$ occurs at $......$


- ✓$(5,0)$
- B$(6,5)$
- C$(6,8)$
- D$(4,10)$

| Corner point |
Objective function $z=3 x-4 y$ |
| $(0,0)$ | $z=3(0)-4(0)=0$ |
| $(5,0)$ | $z=3(5)-4(0)=15$ (Maximum value) |
| $(6,5)$ | $z=3(6)-4(5)=18-20=-2$ |
| $(6,8)$ | $z=3(6)-4(8)=-14$ |
| $(4,10)$ | $z=3(4)-4(10)=-28$ |
| $(0,8)$ | $z=3(0)-4(8)=-32($ minimum value ) |
maximum value of objective function $z=3 x-4 y$ is $15$ which exists at point $(5,0).$
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