A manufacturer has a maximum of 240, 360, and 180 kilograms of wood, plastic and steel available. The company produces two products, A and B. Each unit of A requires 1, 3 and 2 kilograms of wood, plastic and steel respectively; each unit of B requires 3, 4 and 1 kilograms of wood, plastic and steel respectively. The profit per unit of A and B is $4.00 and $6.00 respectively.

a) Write out the objective profit function.
b) Write out the constraints this company faces.
c) List all corner points and determine which combination of items will produce the maximum profit.

This is really linear programming, not statistics.

p = 4 a + 6 b
That is the objective function we wish to maximize

1 a + 3 b </=240 wood constraint
3 a + 4 b </=360 plastic constraint
2 a + 1 b </=180 steel constraint

graph those three constraint lines with a on x axis and b on y axis
for w
intercepts to make that easy
wood (0,80) and (240,0)
plastic (0,90) and (120,0)
steel (0,180) and (90,0)
so
wood y = -x/3 + 80
plastic y = -3x/4 + 90
steel y = - 2x + 180

so corners (except 0,0 )
wood-plastic
-x/3 + 80 = -3x/4 + 90
5 x/12 = 10
x = 24 and y = 72
what is p there ?
p = 3*24 + 6*72

also
do that for the wood - steel
and for the plastic - steel
pick the biggest p

Answer for this question.

a) The objective profit function can be written as follows:

Profit = Profit(A) + Profit(B) = 4A + 6B

b) The constraints the company faces are as follows:
Wood constraint: 1A + 3B ≤ 240
Plastic constraint: 3A + 4B ≤ 360
Steel constraint: 2A + B ≤ 180
Non-negativity constraint: A ≥ 0, B ≥ 0

c) To find the corner points, we need to solve the system of equations formed by the constraints.

Let's solve each constraint equation separately to find the feasible region:

Wood constraint:
1A + 3B = 240
3B = 240 - 1A
B = (240 - 1A)/3

Plastic constraint:
3A + 4B = 360
4B = 360 - 3A
B = (360 - 3A)/4

Steel constraint:
2A + B = 180
B = 180 - 2A

Now, let's substitute the expression for B from the wood constraint into the expression for B from the plastic constraint:

(240 - 1A)/3 = (360 - 3A)/4

Solving this equation gives us A = 60

Substituting the value of A into B = (240 - 1A)/3, we get B = 60

Therefore, the first corner point is (A, B) = (60, 60)

Now, let's substitute the expression for B from the wood constraint into the expression for B from the steel constraint:

(240 - 1A)/3 = 180 - 2A

Solving this equation gives us A = 60

Substituting the value of A into B = 180 - 2A, we get B = 60

Therefore, the second corner point is (A, B) = (60, 60)

The third corner point is (A, B) = (0, 120)

The fourth corner point is (A, B) = (80, 0)

Now, let's calculate the profit for each corner point:

Profit(A, B) = 4A + 6B

Profit(60, 60) = 4(60) + 6(60) = 600

Profit(60, 60) = 4(60) + 6(60) = 600

Profit(0, 120) = 4(0) + 6(120) = 720

Profit(80, 0) = 4(80) + 6(0) = 320

The combination of items that will produce the maximum profit is at the corner point (0, 120) with a profit of $720.

a) The objective profit function can be written as:

Profit (P) = 4A + 6B

where A represents the number of units of product A and B represents the number of units of product B.

b) The constraints this company faces are:

Wood constraint: 1A + 3B ≤ 240 kg
Plastic constraint: 3A + 4B ≤ 360 kg
Steel constraint: 2A + B ≤ 180 kg

These constraints represent the maximum amount of wood, plastic, and steel that can be used by the company.

c) To find the corner points and determine which combination of items will produce the maximum profit, we need to solve the system of equations created by the constraints.

Let's solve these equations one by one:

Wood constraint: 1A + 3B ≤ 240
If we set A = 0, we have 3B ≤ 240, which gives B ≤ 80.
If we set B = 0, we have 1A ≤ 240, which gives A ≤ 240.
So, the corner points for this constraint are (0,0), (240,0), and (0,80).

Plastic constraint: 3A + 4B ≤ 360
If we set A = 0, we have 4B ≤ 360, which gives B ≤ 90.
If we set B = 0, we have 3A ≤ 360, which gives A ≤ 120.
So, the corner points for this constraint are (0,0), (120,0), and (0,90).

Steel constraint: 2A + B ≤ 180
If we set A = 0, we have B ≤ 180.
If we set B = 0, we have 2A ≤ 180, which gives A ≤ 90.
So, the corner points for this constraint are (0,0) and (90,0).

Now, let's find the profit (P) for each corner point:

P(0,0) = 4(0) + 6(0) = 0
P(240,0) = 4(240) + 6(0) = 960
P(0,80) = 4(0) + 6(80) = 480
P(120,0) = 4(120) + 6(0) = 480
P(0,90) = 4(0) + 6(90) = 540
P(90,0) = 4(90) + 6(0) = 360

The maximum profit is achieved at the combination (240,0) with a profit of $960.

a) The objective profit function is the equation that represents the total profit the company will make. In this case, the profit per unit of A is $4.00, and the profit per unit of B is $6.00. Let's use variables x and y to represent the number of units produced for A and B, respectively.

The profit from product A is 4x, and the profit from product B is 6y. Therefore, the objective profit function can be written as:

Profit = 4x + 6y

b) The constraints are the restrictions that the company faces in terms of the available resources. In this case, the constraints are based on the maximum amounts of wood, plastic, and steel that the manufacturer has.

Wood constraint: The maximum amount of wood available is 240 kilograms. Each unit of A requires 1 kilogram, and each unit of B requires 3 kilograms. Therefore, the wood constraint can be written as:

1x + 3y ≤ 240

Plastic constraint: The maximum amount of plastic available is 360 kilograms. Each unit of A requires 3 kilograms, and each unit of B requires 4 kilograms. Therefore, the plastic constraint can be written as:

3x + 4y ≤ 360

Steel constraint: The maximum amount of steel available is 180 kilograms. Each unit of A requires 2 kilograms, and each unit of B requires 1 kilogram. Therefore, the steel constraint can be written as:

2x + 1y ≤ 180

Non-negativity constraint: The number of units produced cannot be negative:

x, y ≥ 0

c) To find all the corner points, we need to graph the feasible region defined by the constraints and find the intersection points.

Graphing the constraints on a coordinate plane, we can determine the points of intersection.

The corner points are:

Point 1: (60, 60)
Point 2: (80, 30)
Point 3: (120, 0)

We can substitute these points into the objective profit function to find the profit at each corner point.

For Point 1: Profit = 4(60) + 6(60) = $600
For Point 2: Profit = 4(80) + 6(30) = $520
For Point 3: Profit = 4(120) + 6(0) = $480

Therefore, the combination of items that will produce the maximum profit is at Point 1, where 60 units of A and 60 units of B should be produced. The maximum profit achievable in this case is $600.