Answer :
To find the maximum value of [tex]\( P = 3x + 2y \)[/tex] given the constraints:
[tex]\[ \begin{cases} x + 3y \leq 15 \\ 4x + y \leq 16 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
we should approach it step-by-step by using graphical or algebraic methods. Here, we will consider the points where the constraints intersect and evaluate [tex]\( P \)[/tex] at these points. These intersection points occur at the boundaries of the feasible region defined by the constraints.
### Finding Intersection Points
1. Intersection of [tex]\( x + 3y = 15 \)[/tex] and [tex]\( 4x + y = 16 \)[/tex]:
To find this intersection, solve the system of linear equations:
[tex]\[ \begin{cases} x + 3y = 15 \\ 4x + y = 16 \end{cases} \][/tex]
Multiply the second equation by 3 to align the [tex]\( y \)[/tex]-terms:
[tex]\[ 4x + y = 16 \implies 12x + 3y = 48 \][/tex]
Now, subtract the first equation from this result:
[tex]\[ 12x + 3y - (x + 3y) = 48 - 15 \\ 11x = 33 \\ x = 3 \][/tex]
Substitute [tex]\( x = 3 \)[/tex] back into [tex]\( x + 3y = 15 \)[/tex]:
[tex]\[ 3 + 3y = 15 \\ 3y = 12 \\ y = 4 \][/tex]
So, the intersection point here is [tex]\( (3, 4) \)[/tex].
2. Intersection with Axes:
- For [tex]\( x + 3y = 15 \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[ 3y = 15 \\ y = 5 \][/tex]
Hence, point [tex]\( (0, 5) \)[/tex].
- For [tex]\( x + 3y = 15 \)[/tex] when [tex]\( y = 0 \)[/tex]:
[tex]\[ x = 15 \][/tex]
Hence, point [tex]\( (15, 0) \)[/tex].
- For [tex]\( 4x + y = 16 \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 16 \][/tex]
Hence, point [tex]\( (0, 16) \)[/tex].
- For [tex]\( 4x + y = 16 \)[/tex] when [tex]\( y = 0 \)[/tex]:
[tex]\[ 4x = 16 \\ x = 4 \][/tex]
Hence, point [tex]\( (4, 0) \)[/tex].
### Evaluating [tex]\( P \)[/tex] at the Intersection Points
Let's evaluate [tex]\( P = 3x + 2y \)[/tex] at each significant point within the constraints:
- At [tex]\( (0, 5) \)[/tex], [tex]\( P = 3(0) + 2(5) = 10 \)[/tex]
- At [tex]\( (3, 4) \)[/tex], [tex]\( P = 3(3) + 2(4) = 9 + 8 = 17 \)[/tex]
- At [tex]\( (4, 0) \)[/tex], [tex]\( P = 3(4) + 2(0) = 12 \)[/tex]
### Conclusion
In conclusion, the maximum value of [tex]\( P \)[/tex] given the constraints is:
[tex]\[ P = 3(3) + 2(4) = 9 + 8 = 17 \][/tex]
So, the maximum value of [tex]\( P \)[/tex] is [tex]\( 17 \)[/tex]. This maximum occurs at the point [tex]\( (3, 4) \)[/tex].
[tex]\[ \begin{cases} x + 3y \leq 15 \\ 4x + y \leq 16 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
we should approach it step-by-step by using graphical or algebraic methods. Here, we will consider the points where the constraints intersect and evaluate [tex]\( P \)[/tex] at these points. These intersection points occur at the boundaries of the feasible region defined by the constraints.
### Finding Intersection Points
1. Intersection of [tex]\( x + 3y = 15 \)[/tex] and [tex]\( 4x + y = 16 \)[/tex]:
To find this intersection, solve the system of linear equations:
[tex]\[ \begin{cases} x + 3y = 15 \\ 4x + y = 16 \end{cases} \][/tex]
Multiply the second equation by 3 to align the [tex]\( y \)[/tex]-terms:
[tex]\[ 4x + y = 16 \implies 12x + 3y = 48 \][/tex]
Now, subtract the first equation from this result:
[tex]\[ 12x + 3y - (x + 3y) = 48 - 15 \\ 11x = 33 \\ x = 3 \][/tex]
Substitute [tex]\( x = 3 \)[/tex] back into [tex]\( x + 3y = 15 \)[/tex]:
[tex]\[ 3 + 3y = 15 \\ 3y = 12 \\ y = 4 \][/tex]
So, the intersection point here is [tex]\( (3, 4) \)[/tex].
2. Intersection with Axes:
- For [tex]\( x + 3y = 15 \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[ 3y = 15 \\ y = 5 \][/tex]
Hence, point [tex]\( (0, 5) \)[/tex].
- For [tex]\( x + 3y = 15 \)[/tex] when [tex]\( y = 0 \)[/tex]:
[tex]\[ x = 15 \][/tex]
Hence, point [tex]\( (15, 0) \)[/tex].
- For [tex]\( 4x + y = 16 \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 16 \][/tex]
Hence, point [tex]\( (0, 16) \)[/tex].
- For [tex]\( 4x + y = 16 \)[/tex] when [tex]\( y = 0 \)[/tex]:
[tex]\[ 4x = 16 \\ x = 4 \][/tex]
Hence, point [tex]\( (4, 0) \)[/tex].
### Evaluating [tex]\( P \)[/tex] at the Intersection Points
Let's evaluate [tex]\( P = 3x + 2y \)[/tex] at each significant point within the constraints:
- At [tex]\( (0, 5) \)[/tex], [tex]\( P = 3(0) + 2(5) = 10 \)[/tex]
- At [tex]\( (3, 4) \)[/tex], [tex]\( P = 3(3) + 2(4) = 9 + 8 = 17 \)[/tex]
- At [tex]\( (4, 0) \)[/tex], [tex]\( P = 3(4) + 2(0) = 12 \)[/tex]
### Conclusion
In conclusion, the maximum value of [tex]\( P \)[/tex] given the constraints is:
[tex]\[ P = 3(3) + 2(4) = 9 + 8 = 17 \][/tex]
So, the maximum value of [tex]\( P \)[/tex] is [tex]\( 17 \)[/tex]. This maximum occurs at the point [tex]\( (3, 4) \)[/tex].
