To find the maximum value of the objective function [tex]\( P = 4x + 2y \)[/tex] given the constraints, we can solve the linear programming problem step-by-step:
### Step 1: Define the Objective Function
We want to maximize:
[tex]\[ P = 4x + 2y \][/tex]
### Step 2: Write Down the Constraints
The problem is subject to the following constraints:
1. [tex]\( 2x + 2y \leq 10 \)[/tex]
2. [tex]\( 3x + y \leq 9 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
### Step 3: Identify the Feasible Region
To find the feasible region, we need to plot the following lines on a graph:
1. [tex]\( 2x + 2y = 10 \)[/tex] implies [tex]\( x + y = 5 \)[/tex]
2. [tex]\( 3x + y = 9 \)[/tex]
Then, we determine the region that satisfies both inequality constraints and the non-negativity constraints [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex].
### Step 4: Find the Intersection Points
Next, we need to find the intersection points of these lines, which will give us the vertices that potentially maximize [tex]\( P \)[/tex].
1. Intersection of [tex]\( x + y = 5 \)[/tex] and [tex]\( 3x + y = 9 \)[/tex]:
- Solving these equations simultaneously:
[tex]\[
\begin{cases}
x + y = 5 \\
3x + y = 9
\end{cases}
\][/tex]
Subtracting the first equation from the second results in:
[tex]\[
2x = 4 \implies x = 2
\][/tex]
Substituting [tex]\( x = 2 \)[/tex] back into [tex]\( x + y = 5 \)[/tex] gives:
[tex]\[
2 + y = 5 \implies y = 3
\][/tex]
So, the intersection is [tex]\( (2, 3) \)[/tex].
### Step 5: Evaluate the Objective Function at Vertices
Evaluate [tex]\( P \)[/tex] at the found intersection (and any other vertices defined by the constraints, but here focusing on the critical intersection):
1. At [tex]\( (2, 3) \)[/tex]:
[tex]\[
P = 4(2) + 2(3) = 8 + 6 = 14
\][/tex]
### Conclusion
The maximum value of [tex]\( P \)[/tex] within the feasible region defined by the constraints is:
[tex]\[ \boxed{P = 14} \][/tex]
and it occurs at the point [tex]\( (x, y) = (2, 3) \)[/tex].
