A packet of plain biscuits costs ₹ 6 and that of choloate biscuits costs ₹ 9 . A house-wife has ₹ 72 and wants to buy at least three packets of plain biscuits and at least four of choloate biscuits.How many of each type should she buy so that she can have maximum number of packets?Make it as an LPP and solve it graphically.

Answer

To solve the problem, we will formulate it as a Linear Programming Problem (LPP) and then solve it graphically.

### Step 1: Define the Variables
Let:
- \( x \) = number of packets of plain biscuits
- \( y \) = number of packets of chocolate biscuits

### Step 2: Formulate the Objective Function
The objective is to maximize the total number of packets, which can be expressed as:
\[ Z = x + y \]

### Step 3: Formulate the Constraints
From the problem, we have the following constraints:
1. The cost of plain biscuits is ₹6 and chocolate biscuits is ₹9. The total cost must not exceed ₹72:
\[ 6x + 9y \leq 72 \]
2. The housewife wants at least 3 packets of plain biscuits:
\[ x \geq 3 \]
3. The housewife wants at least 4 packets of chocolate biscuits:
\[ y \geq 4 \]
4. Non-negativity constraints:
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Simplify the Cost Constraint
We can simplify the cost constraint:
\[ 6x + 9y \leq 72 \]
Dividing the entire equation by 3 gives:
\[ 2x + 3y \leq 24 \]

### Step 5: Graph the Constraints
To graph the constraints, we will find the intercepts for the equation \( 2x + 3y = 24 \):
- For \( x \)-intercept (set \( y = 0 \)):
\[ 2x = 24 \Rightarrow x = 12 \]
- For \( y \)-intercept (set \( x = 0 \)):
\[ 3y = 24 \Rightarrow y = 8 \]

Now we can plot the line \( 2x + 3y = 24 \) on a graph.

### Step 6: Identify the Feasible Region
The feasible region is bounded by the lines:
1. \( x = 3 \) (vertical line)
2. \( y = 4 \) (horizontal line)
3. \( 2x + 3y = 24 \) (line we just plotted)

The feasible region will be the area that satisfies all constraints, which is the area where all inequalities overlap.

### Step 7: Find the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( x = 3 \) and \( y = 4 \):
\[ (3, 4) \]
2. Intersection of \( x = 3 \) and \( 2x + 3y = 24 \):
\[ 2(3) + 3y = 24 \Rightarrow 6 + 3y = 24 \Rightarrow 3y = 18 \Rightarrow y = 6 \]
So, the point is \( (3, 6) \).
3. Intersection of \( y = 4 \) and \( 2x + 3y = 24 \):
\[ 2x + 3(4) = 24 \Rightarrow 2x + 12 = 24 \Rightarrow 2x = 12 \Rightarrow x = 6 \]
So, the point is \( (6, 4) \).
4. Intersection of \( 2x + 3y = 24 \) with axes:
- \( (12, 0) \) (not in feasible region)
- \( (0, 8) \) (not in feasible region)

### Step 8: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = x + y \) at each of the corner points:
1. At \( (3, 4) \):
\[ Z = 3 + 4 = 7 \]
2. At \( (3, 6) \):
\[ Z = 3 + 6 = 9 \]
3. At \( (6, 4) \):
\[ Z = 6 + 4 = 10 \]

### Step 9: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( (6, 4) \) where \( Z = 10 \).

### Conclusion
The housewife should buy:
- 6 packets of plain biscuits
- 4 packets of chocolate biscuits

This gives her a total of 10 packets.