Sample 2022-23 Class 12 Maths Q30 Solution: Maximise Z = 400x + 300y graphically

The question

Solve the following Linear Programming Problem graphically:

Maximize Z = 400x + 300y subject to $x + y \le 200$ ⁠, $x \le 40$ ⁠, $x \ge 20$ ⁠, $y \ge 0$

Exam answer

Draw the lines (graph below):

$x + y = 200$ through $(200, 0)$ and $(0, 200)$ ⁠;

Why?

To draw a line, two points are enough. Put $x = 0$ to get $y = 200$ ⁠, and put $y = 0$ to get $x = 200$ ⁠.

$x = 40$ and $x = 20$ ⁠, parallel to the y-axis; $y = 0$ is the x-axis.

$x + y \le 200$ is the side of $x + y = 200$ containing O; $x \le 40$ is on or to the left of $x = 40$ ⁠; $x \ge 20$ is on or to the right of $x = 20$ ⁠; $y \ge 0$ is on or above the x-axis. The feasible region is CDBA (shaded). It is bounded.

Why?

To find the side of a line that an inequality describes, test a point not on the line. O(0, 0) gives $0 + 0 \le 200$ ⁠, which is true, so $x + y \le 200$ is the side containing O.

The feasible region is the part of the plane that satisfies all four constraints at once. It is bounded because it fits inside a finite rectangle.

The feasible region CDBA (shaded), with corners C(20, 0), D(40, 0), B(40, 160) and A(20, 180).

Solving the boundary lines in pairs,

\begin{aligned} x = 20,\ x + y = 200 &\Rightarrow y = 180 \\ x = 40,\ x + y = 200 &\Rightarrow y = 160 \end{aligned}

Why?

Each corner is where two boundary lines meet. C and D lie on the x-axis, where $y = 0$ ⁠. A and B lie on $x + y = 200$ ⁠: put $x = 20$ and $x = 40$ in it.

\begin{aligned} 20 + y &= 200 \\ y &= 200 - 20 = 180 \\[4pt] 40 + y &= 200 \\ y &= 200 - 40 = 160 \end{aligned}

The corner points of the feasible region are C(20, 0), D(40, 0), B(40, 160) and A(20, 180).

\begin{array}{c|c} \text{Corner} & Z = 400x + 300y \\ \hline C(20, 0) & 8000 \\ D(40, 0) & 16000 \\ B(40, 160) & 64000 \leftarrow \text{max} \\ A(20, 180) & 62000 \end{array}

Why?

Corner point method: when the feasible region is bounded, $Z$ has a maximum, and it occurs at a corner point. So we only compare the values of $Z$ at the corners. For example, at B,

\begin{aligned} Z &= 400(40) + 300(160) \\ &= 16000 + 48000 \\ &= 64000 \end{aligned}

The other three corners the same way:

\begin{aligned} C&: 400(20) + 300(0) \\ &= 8000 + 0 = 8000 \\[4pt] D&: 400(40) + 300(0) \\ &= 16000 + 0 = 16000 \\[4pt] A&: 400(20) + 300(180) \\ &= 8000 + 54000 = 62000 \end{aligned}

The largest of 8000, 16000, 64000 and 62000 is 64000, at B.

$\therefore$ Maximum $Z = 64000$ at $x = 40$ ⁠, $y = 160$ ⁠.

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: $x \ge 20$ keeps only the side of $x = 20$ away from the origin, so the region is the closed strip CDBA between $x = 20$ and $x = 40$ ⁠, under $x + y = 200$ ⁠. The largest table value, 64000 at $(40, 160)$ ⁠, is the maximum.

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Q30: Maximise Z=400x+300yZ = 400x + 300yZ=400x+300y graphically

Q30: Maximise $Z = 400x + 300y$ graphically