Answer :
To solve the problem, let's write out the constraints, determine the vertices of the feasible region, write the optimization equation, and find the maximum number of trucks that can be built during the shift.
### Constraints:
1. Total candy canes:
[tex]\(40x + 64y \leq 480\)[/tex]
This constraint ensures that the total payment to apprentice and senior elves does not exceed 480 candy canes.
2. Total number of elves:
[tex]\(x + y \leq 9\)[/tex]
This constraint ensures that the total number of elves working the shift does not exceed 9.
3. Non-negative number of apprentice elves:
[tex]\(x \geq 0\)[/tex]
4. Non-negative number of senior elves:
[tex]\(y \geq 0\)[/tex]
### Vertices of the Feasible Region:
The feasible region is bounded by the above constraints. The vertices (corner points) of the feasible region can be found by solving the equations formed by these constraints:
1. Intersection of [tex]\(40x + 64y = 480\)[/tex] and [tex]\(x + y = 9\)[/tex]:
- Multiply the second equation by 40 to align it for elimination with the first equation:
[tex]\[ 40(x + y) = 40 \cdot 9 \][/tex]
[tex]\[ 40x + 40y = 360 \][/tex]
- Subtract this from the first equation:
[tex]\[ 40x + 64y - (40x + 40y) = 480 - 360 \][/tex]
[tex]\[ 24y = 120 \implies y = 5 \][/tex]
- Substitute [tex]\( y = 5 \)[/tex] into [tex]\( x + y = 9 \)[/tex]:
[tex]\[ x + 5 = 9 \implies x = 4 \][/tex]
- Intersection point is [tex]\((4, 5)\)[/tex].
2. Intersection of [tex]\(40x + 64y = 480\)[/tex] and [tex]\(x = 0\)[/tex]:
- Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ 40 \cdot 0 + 64y = 480 \implies 64y = 480 \implies y = 7.5 \][/tex]
- Intersection point is [tex]\((0, 7.5)\)[/tex].
3. Intersection of [tex]\(x + y = 9\)[/tex] and [tex]\(y = 0\)[/tex]:
- Substitute [tex]\( y = 0 \)[/tex]:
[tex]\[ x + 0 = 9 \implies x = 9 \][/tex]
- Intersection point is [tex]\((9, 0)\)[/tex].
4. Intersection of [tex]\(40x + 64y = 480\)[/tex] and [tex]\(y = 0\)[/tex]:
- Substitute [tex]\( y = 0 \)[/tex]:
[tex]\[ 40x + 64 \cdot 0 = 480 \implies 40x = 480 \implies x = 12 \][/tex]
- However, [tex]\( x = 12 \)[/tex] exceeds the constraint [tex]\( x + y \leq 9 \)[/tex], so this point is outside the feasible region.
Thus, the vertices of the feasible region are [tex]\((0, 0)\)[/tex], [tex]\((4, 5)\)[/tex], [tex]\((0, 7.5)\)[/tex], and [tex]\((9, 0)\)[/tex].
### Optimization Equation:
The number of trucks [tex]\( T \)[/tex] is given by:
[tex]\[ T = 32x + 48y \][/tex]
### Maximum Value:
Evaluating the objective function at each vertex:
1. [tex]\((0, 0)\)[/tex]:
[tex]\[ T = 32(0) + 48(0) = 0 \][/tex]
2. [tex]\((4, 5)\)[/tex]:
[tex]\[ T = 32(4) + 48(5) = 128 + 240 = 368 \][/tex]
3. [tex]\((0, 7.5)\)[/tex]:
[tex]\[ T = 32(0) + 48(7.5) = 0 + 360 = 360 \][/tex]
4. [tex]\((9, 0)\)[/tex]:
[tex]\[ T = 32(9) + 48(0) = 288 + 0 = 288 \][/tex]
The maximum number of trucks [tex]\( \textbf{368} \)[/tex] is achieved at the vertex [tex]\( \textbf{(4, 5)} \)[/tex]. Therefore, Santa should schedule [tex]\( \textbf{4} \)[/tex] apprentice elves and [tex]\( \textbf{5} \)[/tex] senior elves.
### Constraints:
1. Total candy canes:
[tex]\(40x + 64y \leq 480\)[/tex]
This constraint ensures that the total payment to apprentice and senior elves does not exceed 480 candy canes.
2. Total number of elves:
[tex]\(x + y \leq 9\)[/tex]
This constraint ensures that the total number of elves working the shift does not exceed 9.
3. Non-negative number of apprentice elves:
[tex]\(x \geq 0\)[/tex]
4. Non-negative number of senior elves:
[tex]\(y \geq 0\)[/tex]
### Vertices of the Feasible Region:
The feasible region is bounded by the above constraints. The vertices (corner points) of the feasible region can be found by solving the equations formed by these constraints:
1. Intersection of [tex]\(40x + 64y = 480\)[/tex] and [tex]\(x + y = 9\)[/tex]:
- Multiply the second equation by 40 to align it for elimination with the first equation:
[tex]\[ 40(x + y) = 40 \cdot 9 \][/tex]
[tex]\[ 40x + 40y = 360 \][/tex]
- Subtract this from the first equation:
[tex]\[ 40x + 64y - (40x + 40y) = 480 - 360 \][/tex]
[tex]\[ 24y = 120 \implies y = 5 \][/tex]
- Substitute [tex]\( y = 5 \)[/tex] into [tex]\( x + y = 9 \)[/tex]:
[tex]\[ x + 5 = 9 \implies x = 4 \][/tex]
- Intersection point is [tex]\((4, 5)\)[/tex].
2. Intersection of [tex]\(40x + 64y = 480\)[/tex] and [tex]\(x = 0\)[/tex]:
- Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ 40 \cdot 0 + 64y = 480 \implies 64y = 480 \implies y = 7.5 \][/tex]
- Intersection point is [tex]\((0, 7.5)\)[/tex].
3. Intersection of [tex]\(x + y = 9\)[/tex] and [tex]\(y = 0\)[/tex]:
- Substitute [tex]\( y = 0 \)[/tex]:
[tex]\[ x + 0 = 9 \implies x = 9 \][/tex]
- Intersection point is [tex]\((9, 0)\)[/tex].
4. Intersection of [tex]\(40x + 64y = 480\)[/tex] and [tex]\(y = 0\)[/tex]:
- Substitute [tex]\( y = 0 \)[/tex]:
[tex]\[ 40x + 64 \cdot 0 = 480 \implies 40x = 480 \implies x = 12 \][/tex]
- However, [tex]\( x = 12 \)[/tex] exceeds the constraint [tex]\( x + y \leq 9 \)[/tex], so this point is outside the feasible region.
Thus, the vertices of the feasible region are [tex]\((0, 0)\)[/tex], [tex]\((4, 5)\)[/tex], [tex]\((0, 7.5)\)[/tex], and [tex]\((9, 0)\)[/tex].
### Optimization Equation:
The number of trucks [tex]\( T \)[/tex] is given by:
[tex]\[ T = 32x + 48y \][/tex]
### Maximum Value:
Evaluating the objective function at each vertex:
1. [tex]\((0, 0)\)[/tex]:
[tex]\[ T = 32(0) + 48(0) = 0 \][/tex]
2. [tex]\((4, 5)\)[/tex]:
[tex]\[ T = 32(4) + 48(5) = 128 + 240 = 368 \][/tex]
3. [tex]\((0, 7.5)\)[/tex]:
[tex]\[ T = 32(0) + 48(7.5) = 0 + 360 = 360 \][/tex]
4. [tex]\((9, 0)\)[/tex]:
[tex]\[ T = 32(9) + 48(0) = 288 + 0 = 288 \][/tex]
The maximum number of trucks [tex]\( \textbf{368} \)[/tex] is achieved at the vertex [tex]\( \textbf{(4, 5)} \)[/tex]. Therefore, Santa should schedule [tex]\( \textbf{4} \)[/tex] apprentice elves and [tex]\( \textbf{5} \)[/tex] senior elves.
