Answer :
To find the maximum value of \( P = 3x + 2y \) subject to the given constraints, we can use a systematic approach.
1. Identify the constraints and the objective function:
The constraints are:
[tex]\[ \begin{cases} x + 3y \leq 15 \\ 4x + y \leq 16 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
The objective function we want to maximize is \( P = 3x + 2y \).
2. Graph the inequalities:
- First, convert each inequality into an equation to find the boundary lines.
[tex]\[ x + 3y = 15 \][/tex]
[tex]\[ 4x + y = 16 \][/tex]
- Find the intercepts for each line:
- For \( x + 3y = 15 \):
- When \( x = 0 \), \( y = 5 \)
- When \( y = 0 \), \( x = 15 \)
- For \( 4x + y = 16 \):
- When \( x = 0 \), \( y = 16 \)
- When \( y = 0 \), \( x = 4 \)
3. Find the feasible region:
The feasible region is the intersection area where all inequalities are satisfied.
4. Determine the vertices of the feasible region:
The vertices of the feasible region can be found at the intersections of the lines and the x/y-axes. These include:
- The intersection of \( x + 3y = 15 \) and \( 4x + y = 16 \)
- The x-intercept and y-intercept of each line within the constraints \( x \geq 0 \) and \( y \geq 0 \)
5. Solve for the intersection point:
Solve the system of equations:
[tex]\[ \begin{cases} x + 3y = 15 \\ 4x + y = 16 \end{cases} \][/tex]
Multiply the second equation by 3 to facilitate elimination:
[tex]\[ 3(4x + y) = 3 \times 16 \implies 12x + 3y = 48 \][/tex]
Now subtract the first equation from this result:
[tex]\[ (12x + 3y) - (x + 3y) = 48 - 15 \implies 11x = 33 \implies x = 3 \][/tex]
Substitute \( x = 3 \) back into one of the original equations:
[tex]\[ 4(3) + y = 16 \implies 12 + y = 16 \implies y = 4 \][/tex]
Thus, the intersection point is \( (3, 4) \).
6. Evaluate the objective function at the vertices:
The vertices of the feasible region are \( (0, 0) \), \( (15, 0) \), \( (0, 5) \), and \( (3, 4) \).
- At \( (0, 0) \):
[tex]\[ P = 3(0) + 2(0) = 0 \][/tex]
- At \( (15, 0) \):
[tex]\[ P = 3(15) + 2(0) = 45 \][/tex]
However, \((15, 0)\) does not lie within the feasible region because it does not satisfy all constraints.
- At \( (0, 5) \):
[tex]\[ P = 3(0) + 2(5) = 10 \][/tex]
- At \( (3, 4) \):
[tex]\[ P = 3(3) + 2(4) = 9 + 8 = 17 \][/tex]
7. Determine the maximum value:
The maximum value of \( P = 3x + 2y \) within the feasible region is therefore 17, and it occurs at \( (3, 4) \).
Thus, the maximum value of [tex]\( P \)[/tex] is [tex]\( \boxed{17} \)[/tex].
1. Identify the constraints and the objective function:
The constraints are:
[tex]\[ \begin{cases} x + 3y \leq 15 \\ 4x + y \leq 16 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
The objective function we want to maximize is \( P = 3x + 2y \).
2. Graph the inequalities:
- First, convert each inequality into an equation to find the boundary lines.
[tex]\[ x + 3y = 15 \][/tex]
[tex]\[ 4x + y = 16 \][/tex]
- Find the intercepts for each line:
- For \( x + 3y = 15 \):
- When \( x = 0 \), \( y = 5 \)
- When \( y = 0 \), \( x = 15 \)
- For \( 4x + y = 16 \):
- When \( x = 0 \), \( y = 16 \)
- When \( y = 0 \), \( x = 4 \)
3. Find the feasible region:
The feasible region is the intersection area where all inequalities are satisfied.
4. Determine the vertices of the feasible region:
The vertices of the feasible region can be found at the intersections of the lines and the x/y-axes. These include:
- The intersection of \( x + 3y = 15 \) and \( 4x + y = 16 \)
- The x-intercept and y-intercept of each line within the constraints \( x \geq 0 \) and \( y \geq 0 \)
5. Solve for the intersection point:
Solve the system of equations:
[tex]\[ \begin{cases} x + 3y = 15 \\ 4x + y = 16 \end{cases} \][/tex]
Multiply the second equation by 3 to facilitate elimination:
[tex]\[ 3(4x + y) = 3 \times 16 \implies 12x + 3y = 48 \][/tex]
Now subtract the first equation from this result:
[tex]\[ (12x + 3y) - (x + 3y) = 48 - 15 \implies 11x = 33 \implies x = 3 \][/tex]
Substitute \( x = 3 \) back into one of the original equations:
[tex]\[ 4(3) + y = 16 \implies 12 + y = 16 \implies y = 4 \][/tex]
Thus, the intersection point is \( (3, 4) \).
6. Evaluate the objective function at the vertices:
The vertices of the feasible region are \( (0, 0) \), \( (15, 0) \), \( (0, 5) \), and \( (3, 4) \).
- At \( (0, 0) \):
[tex]\[ P = 3(0) + 2(0) = 0 \][/tex]
- At \( (15, 0) \):
[tex]\[ P = 3(15) + 2(0) = 45 \][/tex]
However, \((15, 0)\) does not lie within the feasible region because it does not satisfy all constraints.
- At \( (0, 5) \):
[tex]\[ P = 3(0) + 2(5) = 10 \][/tex]
- At \( (3, 4) \):
[tex]\[ P = 3(3) + 2(4) = 9 + 8 = 17 \][/tex]
7. Determine the maximum value:
The maximum value of \( P = 3x + 2y \) within the feasible region is therefore 17, and it occurs at \( (3, 4) \).
Thus, the maximum value of [tex]\( P \)[/tex] is [tex]\( \boxed{17} \)[/tex].
