Answer :
To determine the maximum value of the objective function [tex]\( P = 10x + 50y \)[/tex] subject to the given constraints:
1. The system of constraints is:
[tex]\[ \begin{cases} x + y \leq 18 \\ x + 3y \leq 30 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
2. Let's first analyze the feasible region given by these inequalities. The feasible region is the area where all these inequalities overlap:
- The line [tex]\( x + y = 18 \)[/tex] represents one boundary of the feasible region.
- The line [tex]\( x + 3y = 30 \)[/tex] represents another boundary.
3. We need to find the vertices of the feasible region, where the objective function might reach its maximum. These vertices can be found by solving the system of linear equations formed by the boundaries of these constraints.
4. The intersection points are:
- Intersection of [tex]\( x + y = 18 \)[/tex] and [tex]\( x+3y=30 \)[/tex]:
[tex]\[ \begin{cases} x + y = 18 \\ x + 3y = 30 \end{cases} \][/tex]
To solve for these equations, we can use substitution or elimination method.
Subtract the first equation from the second:
[tex]\[ (x + 3y) - (x + y) = 30 - 18 \implies 2y = 12 \implies y = 6 \][/tex]
Now substitute [tex]\( y = 6 \)[/tex] back into [tex]\( x + y = 18 \)[/tex]:
[tex]\[ x + 6 = 18 \implies x = 12 \][/tex]
So intersection point is [tex]\( (12, 6) \)[/tex].
- Intersection of [tex]\( x + y = 18 \)[/tex] with [tex]\( y \)[/tex]-axis:
[tex]\[ x = 0 \implies 0 + y = 18 \implies y = 18 \][/tex]
So intersection point is [tex]\( (0, 18) \)[/tex].
- Intersection of [tex]\( x + 3y = 30 \)[/tex] with [tex]\( y \)[/tex]-axis:
[tex]\[ x = 0 \implies 0 + 3y = 30 \implies y = 10 \][/tex]
So intersection point is [tex]\( (0, 10) \)[/tex].
- Additionally, intersections with [tex]\( x \)[/tex]-axis:
For [tex]\( x + y = 18 \rightarrow y = 0, x = 18 \)[/tex]
For [tex]\( x + 3y = 30 \rightarrow y = 0, x = 30 \)[/tex]
5. We now evaluate the objective function [tex]\( P = 10x + 50y \)[/tex] at each of these intersection points to find the maximum.
- At [tex]\( (12, 6) \)[/tex]:
[tex]\[ P = 10(12) + 50(6) = 120 + 300 = 420 \][/tex]
- At [tex]\( (0, 18) \)[/tex]:
[tex]\[ P = 10(0) + 50(18) = 0 + 900 = 900 \quad \text{(not feasible as it exceeds other boundary)} \][/tex]
- At [tex]\( (0, 10) \)[/tex]:
[tex]\[ P = 10(0) + 50(10) = 0 + 500 = 500 \][/tex]
- At [tex]\( (18, 0) \)[/tex]:
[tex]\[ P = 10(18) + 50(0) = 180 + 0 = 180 \][/tex]
- At [tex]\( (30, 0) \)[/tex]:
[tex]\[ P = 10(30) + 50(0) = 300 + 0 = 300 \quad \text{(not feasible as it exceeds other boundary)} \][/tex]
6. Comparing these values, the maximum value of the objective function [tex]\( P = 10x + 50y \)[/tex] subject to the given constraints occurs at the point [tex]\( (0, 10) \)[/tex] with a value of 500.
Thus, the maximum value of the objective function is [tex]\( \boxed{500} \)[/tex].
1. The system of constraints is:
[tex]\[ \begin{cases} x + y \leq 18 \\ x + 3y \leq 30 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
2. Let's first analyze the feasible region given by these inequalities. The feasible region is the area where all these inequalities overlap:
- The line [tex]\( x + y = 18 \)[/tex] represents one boundary of the feasible region.
- The line [tex]\( x + 3y = 30 \)[/tex] represents another boundary.
3. We need to find the vertices of the feasible region, where the objective function might reach its maximum. These vertices can be found by solving the system of linear equations formed by the boundaries of these constraints.
4. The intersection points are:
- Intersection of [tex]\( x + y = 18 \)[/tex] and [tex]\( x+3y=30 \)[/tex]:
[tex]\[ \begin{cases} x + y = 18 \\ x + 3y = 30 \end{cases} \][/tex]
To solve for these equations, we can use substitution or elimination method.
Subtract the first equation from the second:
[tex]\[ (x + 3y) - (x + y) = 30 - 18 \implies 2y = 12 \implies y = 6 \][/tex]
Now substitute [tex]\( y = 6 \)[/tex] back into [tex]\( x + y = 18 \)[/tex]:
[tex]\[ x + 6 = 18 \implies x = 12 \][/tex]
So intersection point is [tex]\( (12, 6) \)[/tex].
- Intersection of [tex]\( x + y = 18 \)[/tex] with [tex]\( y \)[/tex]-axis:
[tex]\[ x = 0 \implies 0 + y = 18 \implies y = 18 \][/tex]
So intersection point is [tex]\( (0, 18) \)[/tex].
- Intersection of [tex]\( x + 3y = 30 \)[/tex] with [tex]\( y \)[/tex]-axis:
[tex]\[ x = 0 \implies 0 + 3y = 30 \implies y = 10 \][/tex]
So intersection point is [tex]\( (0, 10) \)[/tex].
- Additionally, intersections with [tex]\( x \)[/tex]-axis:
For [tex]\( x + y = 18 \rightarrow y = 0, x = 18 \)[/tex]
For [tex]\( x + 3y = 30 \rightarrow y = 0, x = 30 \)[/tex]
5. We now evaluate the objective function [tex]\( P = 10x + 50y \)[/tex] at each of these intersection points to find the maximum.
- At [tex]\( (12, 6) \)[/tex]:
[tex]\[ P = 10(12) + 50(6) = 120 + 300 = 420 \][/tex]
- At [tex]\( (0, 18) \)[/tex]:
[tex]\[ P = 10(0) + 50(18) = 0 + 900 = 900 \quad \text{(not feasible as it exceeds other boundary)} \][/tex]
- At [tex]\( (0, 10) \)[/tex]:
[tex]\[ P = 10(0) + 50(10) = 0 + 500 = 500 \][/tex]
- At [tex]\( (18, 0) \)[/tex]:
[tex]\[ P = 10(18) + 50(0) = 180 + 0 = 180 \][/tex]
- At [tex]\( (30, 0) \)[/tex]:
[tex]\[ P = 10(30) + 50(0) = 300 + 0 = 300 \quad \text{(not feasible as it exceeds other boundary)} \][/tex]
6. Comparing these values, the maximum value of the objective function [tex]\( P = 10x + 50y \)[/tex] subject to the given constraints occurs at the point [tex]\( (0, 10) \)[/tex] with a value of 500.
Thus, the maximum value of the objective function is [tex]\( \boxed{500} \)[/tex].
