Solve the following Linear Programming Problem graphically by using Iso-cost method: Minimize $Z = 4x - 2y$ Subject to the constraints: $x + y \leq 14$ $2x + y \leq 24$ $3x + 2y \geq 14$ and $x \geq 0, y \geq 0$

> [!TLDR] > We convert the minimization problem into a maximization of $-Z$ and use the iso-cost line method. The feasible region is bounded, and the minimum value of $Z$ occurs at the corner point $(0, 14)$, giving $Z_{\min} = -28$. --- ## 1. Understanding the problem and the iso-cost method We are asked to **minimize** $Z = 4x - 2y$ subject to linear constraints. The iso-cost method works by drawing the feasible region, then sliding a line of constant $Z$ (the objective function) in the direction that reduces $Z$ until it just touches the feasible region. For minimization, we move the line opposite to the direction of the gradient of $Z$. The gradient of $Z$ is $(4, -2)$, which points in the direction of increasing $Z$. So to minimize, we slide the line in the direction of $(-4, 2)$ — that is, decreasing $x$ and increasing $y$. --- ## 2. Plotting the constraints We first rewrite each constraint as an equation to draw the boundary lines. 1. $x + y = 14$ Intercepts: $(14, 0)$ and $(0, 14)$. The inequality is $\leq$, so the region is below this line. 2. $2x + y = 24$ Intercepts: $(12, 0)$ and $(0, 24)$. The inequality is $\leq$, so the region is below this line. 3. $3x + 2y = 14$ Intercepts: $(14/3, 0) \approx (4.67, 0)$ and $(0, 7)$. The inequality is $\geq$, so the region is **above** this line. 4. $x \geq 0$, $y \geq 0$ — the first quadrant. > [!WARNING] > A common mistake is to treat the third constraint ($\geq$) as a "below" region. Always check the inequality sign: $3x + 2y \geq 14$ means the half-plane containing the origin is **not** feasible (since $0 \geq 14$ is false). So the feasible region lies away from the origin relative to this line. --- ## 3. Finding the corner points of the feasible region The feasible region is the intersection of all half-planes. Let's find where the boundary lines intersect. **Intersection of $x + y = 14$ and $2x + y = 24$:** Subtract first from second: $(2x + y) - (x + y) = 24 - 14 \Rightarrow x

Solve the following Linear Programming Problem gra – Class 12 Applied Mathematics

Check Your Progress · Q7

Z = 4x - 2y

Sikkim CbseNCERTSubjective· 5mImportance★★★★★

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Concept understanding — Linear Programming Formulation

Linear Programming Formulation: From Intuition to Precision

Imagine you run a small factory that makes two products: chairs and tables. Each chair gives you ₹200 profit, each table gives you ₹300 profit. You have limited wood (240 units) and limited labour hours (100 hours). A chair needs 2 units of wood and 1 hour of labour; a table needs 4 units of wood and 3 hours of labour. You can't make negative chairs or tables. How many of each should you produce to maximise your profit?

This is the kind of problem Linear Programming (LP) is built to solve. The word "programming" here doesn't mean computer programming — it's an old term for "planning". So linear programming is about planning with linear relationships.

The Core Intuition

Every LP problem has three ingredients:

Decision variables — the quantities you control (how many chairs, how many tables)
An objective — what you want to maximise (profit) or minimise (cost)
Constraints — the limits you must respect (wood, labour, non-negativity)

The "linear" part means everything — the profit, the resource usage — adds up in straight-line proportions. Double the chairs, double the wood needed. No fancy curves, no discounts for bulk.

The Precise Statement

A Linear Programming problem is an optimisation problem of the following form:

\text{Maximise (or Minimise) } Z = c_1 x_1 + c_2 x_2 + \dots + c_n x_n

Let's decode this piece by piece.

Decision Variables

$x_1, x_2, \dots, x_n$ are the variables you control. In our factory: let $x_1$ = number of chairs, $x_2$ = number of tables.

Objective Function

$Z = c_1 x_1 + c_2 x_2 + \dots + c_n x_n$ is what you want to optimise. The $c_j$ are coefficients — profit per unit or cost per unit. For our factory: $Z = 200x_1 + 300x_2$ (profit to maximise).

Constraints

Each constraint is a linear inequality (or equality) that limits the variables. The $a_{ij}$ are the resource usage per unit of product $j$ for resource $i$ . The $b_i$ are the available amounts of each resource.

For our factory:

Wood: $2x_1 + 4x_2 \leq 240$
Labour: $1x_1 + 3x_2 \leq 100$

Non-negativity

$x_j \geq 0$ for all $j$ . You can't produce negative chairs. This is almost always present in real problems.

Watch out

A common mistake is forgetting the non-negativity constraints. Without them, the solver might "produce" negative quantities — which is nonsense in the real world.

The Complete Formulation for Our Example

\text{Maximise } Z = 200x_1 + 300x_2

subject to:

2x_1 + 4x_2 \leq 240

x_1 + 3x_2 \leq 100

x_1, x_2 \geq 0

That's it. This is a complete LP formulation. Every LP problem, no matter how complex, follows this same skeleton.

Why "Linear"?

The objective and every constraint are linear functions — they involve only first powers of variables, no $x^2$ , no $\sin x$ , no $x_1 x_2$ . This linearity is what makes LP problems solvable efficiently (using the Simplex method or interior-point methods). If you have products of variables or powers, it's not linear programming anymore — it becomes nonlinear programming, which is much harder. …

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We convert the minimization problem into a maximization of $-Z$ and use the iso-cost line method. The feasible region is bounded, and the minimum value of $Z$ occurs at the corner point $(0, 14)$ , giving $Z_{\min} = -28$ .

1. Understanding the problem and the iso-cost method

We are asked to minimize $Z = 4x - 2y$ subject to linear constraints. The iso-cost method works by drawing the feasible region, then sliding a line of constant $Z$ (the objective function) in the direction that reduces $Z$ until it just touches the feasible region. For minimization, we move the line opposite to the direction of the gradient of $Z$ .

The gradient of $Z$ is $(4, -2)$ , which points in the direction of increasing $Z$ . So to minimize, we slide the line in the direction of $(-4, 2)$ — that is, decreasing $x$ and increasing $y$ .

2. Plotting the constraints

We first rewrite each constraint as an equation to draw the boundary lines.

$x + y = 14$

Intercepts: $(14, 0)$ and $(0, 14)$ . The inequality is $\leq$ , so the region is below this line.
$2x + y = 24$

Intercepts: $(12, 0)$ and $(0, 24)$ . The inequality is $\leq$ , so the region is below this line.
$3x + 2y = 14$

Intercepts: $(14/3, 0) \approx (4.67, 0)$ and $(0, 7)$ . The inequality is $\geq$ , so the region is above this line.
$x \geq 0$ , $y \geq 0$ — the first quadrant.

Watch out

A common mistake is to treat the third constraint ( $\geq$ ) as a "below" region. Always check the inequality sign: $3x + 2y \geq 14$ means the half-plane containing the origin is not feasible (since $0 \geq 14$ is false). So the feasible region lies away from the origin relative to this line.

3. Finding the corner points of the feasible region

The feasible region is the intersection of all half-planes. Let's find where the boundary lines intersect.

Intersection of $x + y = 14$ and $2x + y = 24$ :

Subtract first from second: $(2x + y) - (x + y) = 24 - 14 \Rightarrow x = 10$ . Then $y = 4$ .

So point $A = (10, 4)$ .

Intersection of $x + y = 14$ and $3x + 2y = 14$ :

From $x + y = 14$ , we have $y = 14 - x$ . Substitute:

$3x + 2(14 - x) = 14 \Rightarrow 3x + 28 - 2x = 14 \Rightarrow x = -14$ .

That gives $x = -14$ , which is not in the first quadrant. So these two lines do not meet in the feasible region.

Intersection of $2x + y = 24$ and $3x + 2y = 14$ :

From $2x + y = 24$ , we have $y = 24 - 2x$ . Substitute:

$3x + 2(24 - 2x) = 14 \Rightarrow 3x + 48 - 4x = 14 \Rightarrow -x = -34 \Rightarrow x = 34$ .

Then $y = 24 - 68 = -44$ , which is not feasible. So no intersection in the first quadrant.

Intersection with axes:

$x + y = 14$ meets $x=0$ at $(0, 14)$ and $y=0$ at $(14, 0)$ .
$2x + y = 24$ meets $x=0$ at $(0, 24)$ and $y=0$ at $(12, 0)$ .
$3x + 2y = 14$ meets $x=0$ at $(0, 7)$ and $y=0$ at $(14/3, 0) \approx (4.67, 0)$ .

Now we need to find which of these are actually in the feasible region by checking all constraints.

4. Identifying the feasible corner points

Let's list candidate points and test them.

Point	Coordinates	Satisfies $x+y \leq 14$ ?	Satisfies $2x+y \leq 24$ ?	Satisfies $3x+2y \geq 14$ ?	Feasible?
$P_1$	$(0, 7)$	$0+7=7 \leq 14$ ✓	$0+7=7 \leq 24$ ✓	$0+14=14 \geq 14$ ✓	✓
$P_2$	$(0, 14)$	$0+14=14 \leq 14$ ✓	$0+14=14 \leq 24$ ✓	$0+28=28 \geq 14$ ✓	✓
$P_3$	$(4.67, 0)$	$4.67+0=4.67 \leq 14$ ✓	$9.34+0=9.34 \leq 24$ ✓	$14+0=14 \geq 14$ ✓	✓
$P_4$	$(12, 0)$	$12+0=12 \leq 14$ ✓	$24+0=24 \leq 24$ ✓	$36+0=36 \geq 14$ ✓	✓
$P_5$	$(14, 0)$	$14+0=14 \leq 14$ ✓	$28+0=28 \leq 24$ ✗	—	✗
$P_6$	$(0, 24)$	$0+24=24 \leq 14$ ✗	—	—	✗
$P_7$	$(10, 4)$	$10+4=14 \leq 14$ ✓	$20+4=24 \leq 24$ ✓	$30+8=38 \geq 14$ ✓	✓

So the feasible corner points are: $(0, 7)$ , $(0, 14)$ , $(4.67, 0)$ , $(12, 0)$ , and $(10, 4)$ . …

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Q.Solve the following Linear Programming Problem graphically by using Iso-cost method: Minimize Z=4x−2yZ = 4x - 2yZ=4x−2y Subject to the constraints: x+y≤14x + y \leq 14x+y≤14 2x+y≤242x + y \leq 242x+y≤24 3x+2y≥143x + 2y \geq 143x+2y≥14 and x≥0,y≥0x \geq 0, y \geq 0x≥0,y≥0

1. Understanding the problem and the iso-cost method

2. Plotting the constraints

3. Finding the corner points of the feasible region

4. Identifying the feasible corner points

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Q.Solve the following Linear Programming Problem graphically by using Iso-cost method: Minimize $Z = 4x - 2y$ Subject to the constraints: $x + y \leq 14$ $2x + y \leq 24$ $3x + 2y \geq 14$ and $x \geq 0, y \geq 0$

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