CBSE Board Examination 2023 · Section C · 3 marks

Q29: Maximise $z = 6x + 3y$ graphically

Board 2023Ch 12 · Linear Programming

The question

Solve graphically the following linear programming problem :

Maximise $z = 6x + 3y$ ⁠,

subject to the constraints

$4x + y \ge 80$ ⁠,

$3x + 2y \le 150$ ⁠,

$x + 5y \ge 115$ ⁠,

$x \ge 0$ ⁠, $y \ge 0$ ⁠.

Exam answer

Draw the lines (graph below):

$L_1$ ⁠: $4x + y = 80$ through $(20, 0)$ and $(0, 80)$ ⁠;

$L_2$ ⁠: $3x + 2y = 150$ through $(50, 0)$ and $(0, 75)$ ⁠;

$L_3$ ⁠: $x + 5y = 115$ through $(115, 0)$ and $(0, 23)$ ⁠.

Why?

Each inequality is first drawn as its boundary line, with the sign replaced by "=". A line is fixed by two points, and the easiest two are where it cuts the axes.

For $L_1$ ⁠: $y = 0$ gives $4x = 80$ ⁠, so $x = \frac{80}{4} = 20$ ⁠. And $x = 0$ gives $y = 80$ ⁠.

For $L_2$ ⁠: $y = 0$ gives $3x = 150$ ⁠, so $x = \frac{150}{3} = 50$ ⁠. And $x = 0$ gives $2y = 150$ ⁠, so $y = \frac{150}{2} = 75$ ⁠.

For $L_3$ ⁠: $y = 0$ gives $x = 115$ ⁠, and $x = 0$ gives $5y = 115$ ⁠, so $y = 23$ ⁠.

At $(0, 0)$ ⁠: $0 \ge 80$ is false, $0 \le 150$ is true and $0 \ge 115$ is false. So shade away from the origin for $L_1$ and $L_3$ ⁠, and towards it for $L_2$ ⁠. With $x, y \ge 0$ ⁠, the feasible region is the shaded triangle. It is bounded.

Why?

To find which side of a line to shade, test a point that is not on it. The origin is the easiest.

If the origin satisfies the inequality, shade the side that contains it. If not, shade the other side.

The feasible region is where all five shadings overlap. It is bounded: it is a closed triangle and does not run off to infinity.

Corner points: $L_1$ and $L_2$ meet at $(2, 72)$ ⁠, $L_1$ and $L_3$ at $(15, 20)$ ⁠, $L_2$ and $L_3$ at $(40, 15)$ ⁠.

Why?

Each corner is where two boundary lines cross, so solve their equations together.

$L_1$ and $L_2$ ⁠: put $y = 80 - 4x$ in $3x + 2y = 150$ ⁠:

\begin{aligned} 3x + 2(80 - 4x) &= 150 \\ 3x + 160 - 8x &= 150 \\ -5x + 160 &= 150 \\ -5x &= 150 - 160 = -10 \\ x &= \frac{-10}{-5} = 2 \\ y &= 80 - 4(2) \\ &= 80 - 8 = 72 \end{aligned}

$L_1$ and $L_3$ ⁠: put $y = 80 - 4x$ in $x + 5y = 115$ ⁠:

\begin{aligned} x + 5(80 - 4x) &= 115 \\ x + 400 - 20x &= 115 \\ -19x + 400 &= 115 \\ -19x &= 115 - 400 = -285 \\ x &= \frac{-285}{-19} = 15 \\ y &= 80 - 4(15) \\ &= 80 - 60 = 20 \end{aligned}

$L_2$ and $L_3$ ⁠: put $x = 115 - 5y$ in $3x + 2y = 150$ ⁠:

\begin{aligned} 3(115 - 5y) + 2y &= 150 \\ 345 - 15y + 2y &= 150 \\ 345 - 13y &= 150 \\ -13y &= 150 - 345 = -195 \\ y &= \frac{-195}{-13} = 15 \\ x &= 115 - 5(15) \\ &= 115 - 75 = 40 \end{aligned}

The feasible region (shaded), bounded by

L_1

⁠:

4x + y = 80

⁠,

L_2

⁠:

3x + 2y = 150

and

L_3

⁠:

x + 5y = 115

⁠.

\begin{array}{c|c} \text{Corner point} & z = 6x + 3y \\ \hline (2, 72) & 12 + 216 = 228 \\ (15, 20) & 90 + 60 = 150 \\ (40, 15) & 240 + 45 = 285 \end{array}

Why?

Work out $z = 6x + 3y$ at each corner. For example, at $(2, 72)$ ⁠:

\begin{aligned} &\text{At } (2, 72)\text{: } 6(2) + 3(72) \\ &= 12 + 216 = 228 \\[4pt] &\text{At } (15, 20)\text{: } 6(15) + 3(20) \\ &= 90 + 60 = 150 \\[4pt] &\text{At } (40, 15)\text{: } 6(40) + 3(15) \\ &= 240 + 45 = 285 \end{aligned}

The region is bounded, so the largest value in the table is the maximum of $z$ ⁠.

Why?

Corner point theorem: if the feasible region is bounded, $z$ has a maximum and a minimum, and each one occurs at a corner point.

So only the three corners need checking, and the largest value, 285, is at $(40, 15)$ ⁠.

Watch out: if the region were unbounded, the largest corner value would need an extra check before it could be called the maximum.

$\therefore$ Maximum $z = 285$ at $x = 40$ ⁠, $y = 15$ ⁠.

Where this comes from

Chapter 12, Linear Programming: LPP: maximise graphically, bounded region.

Question type: A " $\ge$ " rule

How to spot it: click to see

Type 2. A " $\ge$ " rule: shade away from the origin

Spot it: a rule like $x + 2y \ge 10$ ⁠, often with "Minimise $Z$ ". The region no longer touches the origin.

Here: $4x + y \ge 80$ and $x + 5y \ge 115$ fail at $(0, 0)$ ⁠, so shade away from the origin for both; only $3x + 2y \le 150$ keeps the origin's side. That leaves a bounded triangle, and the largest table value, 285, is the maximum.

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Q29: Maximise z=6x+3yz = 6x + 3yz=6x+3y graphically

Q29: Maximise $z = 6x + 3y$ graphically