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Sagot :
### Solution:
### 1. Find the vertices of the feasible region.
To find the vertices, we need to determine where the constraints intersect. The constraints are:
1. [tex]\( 2x + 3y \leq 300 \)[/tex]
2. [tex]\( 4x + y \leq 400 \)[/tex]
3. [tex]\( y \leq x \)[/tex]
4. [tex]\( x \geq 0 \)[/tex]
5. [tex]\( y \geq 0 \)[/tex]
#### a. Find the coordinates of vertex 1.
This vertex is where the constraint [tex]\( x \geq 0 \)[/tex] intersects [tex]\( y \leq x \)[/tex] and [tex]\( 4x + y = 400 \)[/tex].
1. Set [tex]\( x = 0 \)[/tex] and solve for [tex]\( y \)[/tex].
2. [tex]\( 4(0) + y = 400 \implies y = 400 \)[/tex].
Since [tex]\( y \leq x \)[/tex], this intersection does not satisfy the constraint [tex]\( y \leq x \)[/tex]. Instead, we check where [tex]\( y = x \)[/tex]:
3. Substitute [tex]\( y = x \)[/tex] in [tex]\( 4x + y = 400 \)[/tex]: [tex]\( 4x + x = 400 \implies 5x = 400 \implies x = 80 \)[/tex], thus [tex]\( y = 80 \)[/tex].
So the coordinates of vertex 1 are [tex]\((80, 80)\)[/tex].
#### b. Find the coordinates of vertex 2.
This vertex is where the constraint [tex]\( y = 0 \)[/tex] intersects [tex]\( 4x + y = 400 \)[/tex].
1. Set [tex]\( y = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
2. [tex]\( 4x + 0 = 400 \implies 4x = 400 \implies x = 100 \)[/tex].
So the coordinates of vertex 2 are [tex]\((100, 0)\)[/tex].
#### c. Find the coordinates of vertex 3.
This vertex is where the constraints [tex]\( 2x + 3y = 300 \)[/tex] and [tex]\( 4x + y = 400 \)[/tex] intersect.
1. Set up the two equations:
[tex]\( 2x + 3y = 300 \)[/tex]
[tex]\( 4x + y = 400 \)[/tex]
2. Solve the first equation for [tex]\( y \)[/tex]:
[tex]\( y = \frac{400 - 4x}{1} = 400 - 4x \)[/tex].
3. Substitute [tex]\( y = \frac{300 - 2x}{3} \)[/tex] into the second equation:
[tex]\( 4x + \frac{300 - 2x}{3} = 400 \)[/tex]
4. Multiply through by 3 to clear the fraction:
[tex]\( 12x + 300 - 2x = 1200 \)[/tex]
[tex]\( 10x = 900 \)[/tex]
[tex]\( x = 90 \)[/tex]
5. Substitute [tex]\( x = 90 \)[/tex] back into [tex]\( y = \frac{300 - 2x}{3} \)[/tex]:
[tex]\( y = \frac{300 - 2(90)}{3} = \frac{300 - 180}{3} = \frac{120}{3} = 40 \)[/tex].
So the coordinates of vertex 3 are [tex]\((90, 40)\)[/tex].
#### d. Find the coordinates of vertex 4.
This vertex is where the constraint [tex]\( y = 0 \)[/tex] intersects [tex]\( 2x + 3y = 300 \)[/tex].
1. Set [tex]\( y = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
2. [tex]\( 2x + 0 = 300 \implies 2x = 300 \implies x = 150 \)[/tex].
So the coordinates of vertex 4 are [tex]\((150, 0)\)[/tex].
### 2. Find the value of the revenue function [tex]\( R \)[/tex] at each vertex.
Revenue Function: [tex]\( R(x, y) = 1.95x + 2.25y \)[/tex]
#### a. Value of [tex]\( R \)[/tex] at vertex 1:
Vertex 1 coordinates: [tex]\((80, 80)\)[/tex]
[tex]\[ R(80, 80) = 1.95(80) + 2.25(80) = 1.95 \cdot 80 + 2.25 \cdot 80 = 156 + 180 = 336. \][/tex]
#### b. Value of [tex]\( R \)[/tex] at vertex 2:
Vertex 2 coordinates: [tex]\((100, 0)\)[/tex]
[tex]\[ R(100, 0) = 1.95(100) + 2.25(0) = 1.95 \cdot 100 + 0 = 195 \][/tex]
#### c. Value of [tex]\( R \)[/tex] at vertex 3:
Vertex 3 coordinates: [tex]\((90, 40)\)[/tex]
[tex]\[ R(90, 40) = 1.95(90) + 2.25(40) = 1.95 \cdot 90 + 2.25 \cdot 40 = 175.5 + 90 = 265.5 \][/tex]
#### d. Value of [tex]\( R \)[/tex] at vertex 4:
Vertex 4 coordinates: [tex]\((150, 0)\)[/tex]
[tex]\[ R(150, 0) = 1.95(150) + 2.25(0) = 1.95 \cdot 150 + 0 = 292.5 \][/tex]
### 3. How many standard-mix packages and how many deluxe-mix packages will the confectioner need to produce to maximize her revenue?
To maximize revenue, we look for the highest value of [tex]\( R \)[/tex]:
- [tex]\( R = 336 \)[/tex] at vertex [tex]\((80, 80)\)[/tex]
- [tex]\( R = 195 \)[/tex] at vertex [tex]\((100, 0)\)[/tex]
- [tex]\( R = 265.5 \)[/tex] at vertex [tex]\((90, 40)\)[/tex]
- [tex]\( R = 292.5 \)[/tex] at vertex [tex]\((150, 0)\)[/tex]
The highest value of [tex]\( R \)[/tex] is 336 at vertex [tex]\((80, 80)\)[/tex].
Thus, to maximize revenue, the confectioner should produce 80 standard-mix packages and 80 deluxe-mix packages.
### 1. Find the vertices of the feasible region.
To find the vertices, we need to determine where the constraints intersect. The constraints are:
1. [tex]\( 2x + 3y \leq 300 \)[/tex]
2. [tex]\( 4x + y \leq 400 \)[/tex]
3. [tex]\( y \leq x \)[/tex]
4. [tex]\( x \geq 0 \)[/tex]
5. [tex]\( y \geq 0 \)[/tex]
#### a. Find the coordinates of vertex 1.
This vertex is where the constraint [tex]\( x \geq 0 \)[/tex] intersects [tex]\( y \leq x \)[/tex] and [tex]\( 4x + y = 400 \)[/tex].
1. Set [tex]\( x = 0 \)[/tex] and solve for [tex]\( y \)[/tex].
2. [tex]\( 4(0) + y = 400 \implies y = 400 \)[/tex].
Since [tex]\( y \leq x \)[/tex], this intersection does not satisfy the constraint [tex]\( y \leq x \)[/tex]. Instead, we check where [tex]\( y = x \)[/tex]:
3. Substitute [tex]\( y = x \)[/tex] in [tex]\( 4x + y = 400 \)[/tex]: [tex]\( 4x + x = 400 \implies 5x = 400 \implies x = 80 \)[/tex], thus [tex]\( y = 80 \)[/tex].
So the coordinates of vertex 1 are [tex]\((80, 80)\)[/tex].
#### b. Find the coordinates of vertex 2.
This vertex is where the constraint [tex]\( y = 0 \)[/tex] intersects [tex]\( 4x + y = 400 \)[/tex].
1. Set [tex]\( y = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
2. [tex]\( 4x + 0 = 400 \implies 4x = 400 \implies x = 100 \)[/tex].
So the coordinates of vertex 2 are [tex]\((100, 0)\)[/tex].
#### c. Find the coordinates of vertex 3.
This vertex is where the constraints [tex]\( 2x + 3y = 300 \)[/tex] and [tex]\( 4x + y = 400 \)[/tex] intersect.
1. Set up the two equations:
[tex]\( 2x + 3y = 300 \)[/tex]
[tex]\( 4x + y = 400 \)[/tex]
2. Solve the first equation for [tex]\( y \)[/tex]:
[tex]\( y = \frac{400 - 4x}{1} = 400 - 4x \)[/tex].
3. Substitute [tex]\( y = \frac{300 - 2x}{3} \)[/tex] into the second equation:
[tex]\( 4x + \frac{300 - 2x}{3} = 400 \)[/tex]
4. Multiply through by 3 to clear the fraction:
[tex]\( 12x + 300 - 2x = 1200 \)[/tex]
[tex]\( 10x = 900 \)[/tex]
[tex]\( x = 90 \)[/tex]
5. Substitute [tex]\( x = 90 \)[/tex] back into [tex]\( y = \frac{300 - 2x}{3} \)[/tex]:
[tex]\( y = \frac{300 - 2(90)}{3} = \frac{300 - 180}{3} = \frac{120}{3} = 40 \)[/tex].
So the coordinates of vertex 3 are [tex]\((90, 40)\)[/tex].
#### d. Find the coordinates of vertex 4.
This vertex is where the constraint [tex]\( y = 0 \)[/tex] intersects [tex]\( 2x + 3y = 300 \)[/tex].
1. Set [tex]\( y = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
2. [tex]\( 2x + 0 = 300 \implies 2x = 300 \implies x = 150 \)[/tex].
So the coordinates of vertex 4 are [tex]\((150, 0)\)[/tex].
### 2. Find the value of the revenue function [tex]\( R \)[/tex] at each vertex.
Revenue Function: [tex]\( R(x, y) = 1.95x + 2.25y \)[/tex]
#### a. Value of [tex]\( R \)[/tex] at vertex 1:
Vertex 1 coordinates: [tex]\((80, 80)\)[/tex]
[tex]\[ R(80, 80) = 1.95(80) + 2.25(80) = 1.95 \cdot 80 + 2.25 \cdot 80 = 156 + 180 = 336. \][/tex]
#### b. Value of [tex]\( R \)[/tex] at vertex 2:
Vertex 2 coordinates: [tex]\((100, 0)\)[/tex]
[tex]\[ R(100, 0) = 1.95(100) + 2.25(0) = 1.95 \cdot 100 + 0 = 195 \][/tex]
#### c. Value of [tex]\( R \)[/tex] at vertex 3:
Vertex 3 coordinates: [tex]\((90, 40)\)[/tex]
[tex]\[ R(90, 40) = 1.95(90) + 2.25(40) = 1.95 \cdot 90 + 2.25 \cdot 40 = 175.5 + 90 = 265.5 \][/tex]
#### d. Value of [tex]\( R \)[/tex] at vertex 4:
Vertex 4 coordinates: [tex]\((150, 0)\)[/tex]
[tex]\[ R(150, 0) = 1.95(150) + 2.25(0) = 1.95 \cdot 150 + 0 = 292.5 \][/tex]
### 3. How many standard-mix packages and how many deluxe-mix packages will the confectioner need to produce to maximize her revenue?
To maximize revenue, we look for the highest value of [tex]\( R \)[/tex]:
- [tex]\( R = 336 \)[/tex] at vertex [tex]\((80, 80)\)[/tex]
- [tex]\( R = 195 \)[/tex] at vertex [tex]\((100, 0)\)[/tex]
- [tex]\( R = 265.5 \)[/tex] at vertex [tex]\((90, 40)\)[/tex]
- [tex]\( R = 292.5 \)[/tex] at vertex [tex]\((150, 0)\)[/tex]
The highest value of [tex]\( R \)[/tex] is 336 at vertex [tex]\((80, 80)\)[/tex].
Thus, to maximize revenue, the confectioner should produce 80 standard-mix packages and 80 deluxe-mix packages.
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