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Part 2: Solve Linear Programming Problems Graphically

Problem 2

Problem Statement: A wholesaler contains 100 grams of cashew and 200 grams of peanuts and sells for \[tex]$1.95. The deluxe-mix contains 150 grams of cashew and 50 grams of peanuts and sells for \$[/tex]2.25. The confectioner has 15 kilograms of cashews and 20 kilograms of peanuts available. Based on past sales, the confectioner needs to have at least as many standard as deluxe packages available. How many bags of each nut mixture should she package to maximize revenue?

Note: If we let [tex]\( x \)[/tex] represent the number of standard-mix packages and [tex]\( y \)[/tex] represent the number of deluxe-mix packages sold, then we know the revenue produced by selling the packages is given by

[tex]\[ f(x, y) = 1.95x + 2.25y \][/tex]

Clearly, to maximize revenue, the confectioner would simply sell as many packages as possible. However, it is not as simple as that since there are some "constraints" that must be satisfied.

Constraint 1: Deluxe-mix uses 150 grams of cashews, we need to ensure that
[tex]\[ 100x + 150y \leq 15000 \Rightarrow 2x + 3y \leq 300 \][/tex]

Constraint 2: There are 20,000 grams of peanuts available. Since the standard-mix uses 200 grams and the deluxe-mix uses 50 grams, we need to ensure that
[tex]\[ 200x + 50y \leq 20000 \Rightarrow 4x + y \leq 400 \][/tex]

Constraint 3: The confectioner needs to have at least as many standard-mix packages as she has deluxe-mix packages, so we also require that
[tex]\[ x \geq y \Rightarrow y \leq x \][/tex]

Constraints 4 and 5: Since the confectioner cannot produce a negative number of packages, it must be that
[tex]\[ x \geq 0 \text{ and } y \geq 0 \][/tex]

If we plot the inequalities given by the constraints, the ordered pair [tex]\((x, y)\)[/tex] that maximizes the revenue function will be a vertex of the feasible region. The feasible region is the region in the [tex]\( xy \)[/tex]-plane defined by the constraints and can be seen in the figure here.

Questions

1. Find the vertices of the feasible region.
a. Find the coordinates of vertex 1.
b. Find the coordinates of vertex 2.
c. Find the coordinates of vertex 3.
d. Find the coordinates of vertex 4.

2. Find the value of the revenue function [tex]\( R \)[/tex] at each vertex.
a. Value of [tex]\( R \)[/tex] at vertex 1.
b. Value of [tex]\( R \)[/tex] at vertex 2.
c. Value of [tex]\( R \)[/tex] at vertex 3.
d. Value of [tex]\( R \)[/tex] at vertex 4.

3. How many standard-mix packages and how many deluxe-mix packages will the confectioner need to maximize her revenue?



Answer :

GinnyAnswer
### 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.

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