MCQ
The feasible solution for a LPP is shown in Figure Let $z=3 x-4 y$ be the objective function. (Maximum value of $z+$ Minimum value of $z$ ) is equal to $....$
- A$13$
- B$01$
- C$-13$
- ✓$-17$
| 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 ) |
we have (Maximum value of $z$ ) $+$ (Minimum value of $z$ ) $=15-32=-17$.
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