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Chemco produces two chemicals A and B. These chemicals are produced via two manufacturing processes. Process...

Chemco produces two chemicals A and B. These chemicals are produced via two manufacturing processes. Process 1 requires 2 hours of labor and 1 lb of raw materials to produce 2 oz of A and 1 oz of B. Process 2 requires 3 hours of labor and 2 lbs of raw material to produce 3 oz of A and 2 oz of B. Sixty hours of labor and 40 lbs of raw material are available. Demand for Chemical A is unlimited, but only 20 oz of B can be sold. Chemical A sells for $16/oz, and B sells for $14/oz. Any chemical B that is unsold must be disposed of at a cost of $2/oz. Formulate an LP model to maximize Chemicos revenue and solve.

1. Maximum Revenue

2. Amount of chemical A produced (oz)

3. Amount of chemical B produced and sold (oz)

4. Amunt of chemical B destroyed (oz)

Solutions

Expert Solution

Solution:-

Consider Variable as:-

X₁ = Process 1

_X2= Process 2

A = Chemical A produced

B = Chemical B produced

B₂= Chemical unsold

Now,

To maximize chemicos revenue,

Z= ( Cost of A) (Units of process 1) + (cost of B) ( units of process 2)- (cost of B₂)(units of B unsold)

Z = 16 (2x₁ + 3x₂) + 14 (x₁ + 2x₂ - B₂) - 2B₂

Maximize are at the Z value

Z = 16 (2x₁ + 3x₂) + 14 (x₁ + 2x₂ - B₂) - 2B₂

Let the equation are:-

Constraint 1:- At most, 60 hours of labor are available

[( hour of labor) (Units of chemical produced by process 1) + (hour of labor) (Units of a chemical produced by process 2)]

= 2x₁ + 3x₂

Constraint 2:- At most, 40 lbs of raw materials are available

[(pounds of raw materials ) (Units of a chemical produced by process 1) + (pounds of materials) (Units of a chemical produced by process 2)]
= x₁ + 2x₂

Constraint 3:- At most, 20 ounces of chemical B can be sold.

[(ounces of a chemical produced by process 1) + (ounces of a chemical produced by process 2) - (ounces of a chemical unsold)]

x₁ + 2x₂ - B₂

Thus, the formulation of linear programme model is shown as

Maximize Z = 16 (2x₁ + 3x₂) + 14 (x₁ + 2x₂ - B₂) - 2B₂

Z= 32x₁ + 48x₂ + 14x₁ + 28x₂ - 14 B₂ - 2B₂

Mamimize Z= 46x₁ + 76x₂ - 16B₂

Depending on the constraint

2x₁ + 3x₂

x₁ + 2x₂

x₁ + 2x₂ - B₂

Thus all variables x_1, x_2, B₂ are

keosha answered 1 year ago

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