At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
