Sample 2022-23 Class 12 Maths Q11 Solution: Where Z = 4x + 6y is minimum on an unbounded region

The question

The corner points of the shaded unbounded feasible region of an LPP are (0, 4), (0.6, 1.6) and (3, 0) as shown in the figure. The minimum value of the objective function $Z = 4x + 6y$ occurs at

The given figure, redrawn.

(a) $(0.6, 1.6)$ only

(b) $(3, 0)$ only

(c) $(0.6, 1.6)$ and $(3, 0)$ only

(d) at every point of the line-segment joining the points $(0.6, 1.6)$ and $(3, 0)$

Exam answer

Values of $Z = 4x + 6y$ at the corner points:

\begin{aligned} \text{At } (0, 4)&: Z = 0 + 24 = 24 \\ \text{At } (0.6, 1.6)&: Z = 2.4 + 9.6 = 12 \\ \text{At } (3, 0)&: Z = 12 + 0 = 12 \end{aligned}

Why?

The corner point method: if Z has a minimum on the feasible region, it occurs at a corner point. So we first evaluate Z at every corner.

At $(0.6, 1.6)$ ⁠: $4(0.6) = 2.4$ and $6(1.6) = 9.6$ ⁠.

Put each corner into $Z = 4x + 6y$ ⁠:

\begin{aligned} (0, 4)&: 4(0) + 6(4) \\ &= 0 + 24 = 24 \\[4pt] (0.6, 1.6)&: 4(0.6) + 6(1.6) \\ &= 2.4 + 9.6 = 12 \\[4pt] (3, 0)&: 4(3) + 6(0) \\ &= 12 + 0 = 12 \end{aligned}

The smallest value is 12, at the two adjacent corners $(0.6, 1.6)$ and $(3, 0)$ ⁠. The region is unbounded, so check the half-plane $4x + 6y < 12$ ⁠, that is $2x + 3y < 6$ ⁠.

Why?

On an unbounded region the smallest corner value need not be the minimum. It is the minimum only if the open half-plane $Z < 12$ has no point in common with the region.

Dividing $4x + 6y < 12$ by 2 gives the simpler $2x + 3y < 6$ ⁠.

The boundary through $(0.6, 1.6)$ and $(3, 0)$ is the line $2x + 3y = 6$ ⁠, and the region lies on the side $2x + 3y \ge 6$ ⁠. So $4x + 6y < 12$ has no point in common with the region.

Why?

Both corners lie on $2x + 3y = 6$ ⁠:

\begin{aligned} 2(0.6) + 3(1.6) &= 1.2 + 4.8 \\ &= 6 \\ 2(3) + 3(0) &= 6 \end{aligned}

The origin gives $2(0) + 3(0) = 0 < 6$ ⁠, and the origin is outside the shaded region. So the region is on the other side, where $2x + 3y \ge 6$ ⁠.

So the minimum value 12 occurs at every point of the segment joining $(0.6, 1.6)$ and $(3, 0)$ ⁠.

Why?

The line $Z = 12$ ⁠, that is $4x + 6y = 12$ ⁠, is the same line as the edge $2x + 3y = 6$ ⁠. So Z equals 12 all along that edge, not just at its ends. For example, at the midpoint $(1.8, 0.8)$ ⁠:

\begin{aligned} x &= \frac{0.6 + 3}{2} = \frac{3.6}{2} = 1.8 \\ y &= \frac{1.6 + 0}{2} = \frac{1.6}{2} = 0.8 \end{aligned}

\begin{aligned} Z &= 4(1.8) + 6(0.8) \\ &= 7.2 + 4.8 = 12 \end{aligned}

That is why options (a), (b) and (c), which say "only", are wrong.

Answer: (d) at every point of the line-segment joining the points $(0.6, 1.6)$ and $(3, 0)$

Where this comes from

Chapter 12, Linear Programming: LPP: minimum on an unbounded feasible region.

Question type: An unbounded region + Both maximum and minimum, and a tie along an edge (combines 2 types)

How to spot it: click to see

Type 4. An unbounded region: check the minimum with a half-plane

Spot it: all the rules are " $\ge$ ", so the shaded region goes on for ever. The question asks for a minimum.

Here: The region is unbounded and the table gives $m = 12$ ⁠. The open half-plane $4x + 6y \lt 12$ lies below the boundary $2x + 3y = 6$ ⁠, so it has no point in the region, and 12 is the minimum.

Type 3. Both maximum and minimum, and a tie along an edge

Spot it: "Minimise and maximise $Z$ ". Often one rule, like $x - 2y \ge 0$ ⁠, is a line through the origin.

Here: The value 12 comes at two corners, $(0.6, 1.6)$ and $(3, 0)$ ⁠, so every point of the segment joining them gives it too: option (d).

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Q11: Where Z=4x+6yZ = 4x + 6yZ=4x+6y is minimum on an unbounded region

Q11: Where $Z = 4x + 6y$ is minimum on an unbounded region