Q17: Another minimum point of an LPP, given one at $(5, 0)$

In a bounded LPP, an optimum value is attained at a corner point. If it is attained at two corners, it is attained at every point of the segment joining them.

From the graph, the feasible region is the triangle with corners $(0, 15)$⁠, $(5, 0)$ and $(10, 0)$⁠. Two of its edges pass through $(5, 0)$⁠: one to $(0, 15)$ and one to $(10, 0)$⁠.

Intercept form $\frac{x}{a} + \frac{y}{b} = 1$ with $a = 5$ and $b = 15$⁠. Multiplying by 15 gives $3x + y = 15$⁠.

The other edge is the part of $y = 0$ from $x = 5$ to $x = 10$⁠. $(6, 6)$ and $(4, 7)$ are not on $y = 0$⁠, and $(0, 0)$ is not in the feasible region at all.

Also, $3x + y$ is $24$ at $(6, 6)$ and $19$ at $(4, 7)$⁠, not 15:

For example, $Z = 3x + y$ has the value 15 at $(5, 0)$⁠, $(0, 15)$ and $(2, 9)$⁠, and 30 at $(10, 0)$⁠.

The line $3x + 5y = 10$ splits the plane into two half planes. One test point not on the line decides which side the inequality describes, and the origin is the easiest.

So (B), which leaves out the origin, fails. (A) fails too: $(10, 10)$ gives $3(10) + 5(10) = 30 + 50 = 80$⁠, which is not less than 10, so the region is not the entire plane.

With $<$⁠, a point on the line gives exactly 10, which is not less than 10. So the line is left out, and the half plane is called open. With $\le$ the line would be included, as in (D).