MCQ
The feasible solution for a $LPP$ is shown in Figure Let $z=3 x-4 y$ be the objective function. Minimum of $Z$ occurs at $....$
- A$(0,0)$
- ✓$(0,8)$
- C$(5,0)$
- 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 ) |
Minimum value of the objective function is $-32$
$\therefore$ Minimum value exists at point $(0,8).$
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If it intersects $y$-axis at $y=-1$, and the intersection point of the curve with $x$-axis is $(\alpha, 0)$ the $\mathrm{e}^{\alpha}$ is equal to $.....$