Q.Refer to Exercise 27. (Maximum value of $Z$ + Minimum value of $Z$) is equal to
(A) $13$
(B) $1$
(C) $-13$
(D) $-17$

The constraints are $2x+3y\ge 6$, $x\ge 0$, $y\ge 0$. The boundary line $2x+3y=6$ meets the axes at $(3,0)$ and $(0,2)$; the feasible region is the unbounded region above this line in the first quadrant. …

Constraints: $3x+y\le 14$, $x+3y\le 10$, $x,y\ge 0$.

• Origin: $(0,0)$.
• $x$-axis intercept of $3x+y=14$: $(14/3,0)$.
• $y$-axis intercept of $x+3y=10$: $(0,10/3)$. …

The constraint lines are $L_1:x+2y=12$, $L_2:2x+y=12$, $L_3:4x+5y=20$, plus the axes.

Find the vertices of the feasible region (satisfying all constraints):

• $L_3\cap x$-axis: $(5,0)$ — check: $5\le12$, $10\le12$ ✓
• $L_2\cap x$-axis: $(6,0)$ — check: $24\ge20$, $6\le12$ ✓
• $L_1\cap L_2$: solving gives $(4,4)$ — check: $36\ge20$ ✓
• $L_1\cap y$-axis: $(0,6)$ — check: $6\le12$, $30\ge20$ ✓ …

Concept and Intuition

The feasible region is a polygon bounded by the constraint lines and axes; its vertices are the corner points.

Step-by-Step Solution

1. Origin $(0,0)$ is a corner.
2. On $y=0$: $2x\le60$ (binding) and $2x\le120$, so $x=30\Rightarrow(30,0)$.
3. On $x=0$: $3y\le120$ (binding) and $y\le60$, so $y=40\Rightarrow(0,40)$.
4. Intersect $2x+3y=120$ and $2x+y=60$: subtract to get $2y=60\Rightarrow y=30$, then $2x=30\Rightarrow x=15$, giving $(15,30)$. …

Constraints: $x\le 4,\ y\le 6,\ x+y\le 6,\ x\ge 0,\ y\ge 0$.

• $(0,0)$: all satisfied.
• $(4,0)$: $x=4\le4$, $x+y=4\le6$ — OK.
• $(4,2)$: $x=4$, $y=2$, $x+y=6\le6$ — OK. …