Need help in F5 question (minimising cost)

[Jocelyn F]

A manufacturer has signed a contract to supply total minimum 260units of component X and Y to customer next month. At least half of the units must be component X.
Variable cost
X= $120 per unit
Y=$100 per unit

X Y
Labour Hrs/unit hrs/unit hrs available for next mth
class 1 4 6 1200
class 2 4 2 800

Labour hours can be increased by hiring casual workers at short notice. the manufacturer wish to minimise cost, whats the product mix to meet the contract requirements?

My answer:
Obj: min cost
min 120x + 100y = Total cost

Conditions:
x + y >= 260
x >= y
Class 1 lab 4x + 6y <= 1200
Class 2 lab 4x + 2y <= 800
Non negativity x,y >= 0

I did until the iso contribution line. How do I know which equation to choose to get total cost? And how do I continue from here?

Please help, thanks! :)

[Jocelyn F]

Hey there, can someone help me? :(

[Sonal]

Hi,


Considering the total minimum requirement of 260 units and half of it must be of the X component, 130 units of X component must be produced.

Inequalities:
X > 130
X + Y > 260
4X + 6Y > 1200
4X + 2Y > 800

By eliminating the last two inequalities, X = 150 and Y = 100 units, which does not satisfy the second inequality.

Available hours of class 1 and class 2 are 1200 and 800 respectively whereas required hours are 2080.

Assuming the 130 units of component X and 130 units of component Y

Class 1: 4 (130) + 6 (130) = 520 + 780 = 1300 hours
Class 2: 4 (130) + 2 (130) = 520 + 260 = 780 hours

As mentioned in the example, labour hours can be increased by hiring casual workers at short notice, required addition 100 hours of class 1 can be hired for the production of component Y. (This is because, for minimising costs, it is advisable to produce Y component as its variable cost is lower than that of component Y)

Even if its is possible to produce 150 units of X, only 130 units should be produced and remaining hours (20 units x 4 = 80) should be diverted to the production of Y.

Additional production of component Y = (80 hours + 100 hours)/ 6 hour per unit = 30 units

Hence,
X = 150 -20 = 130 units
Y = 100 + 30 = 130 units

Minimum cost: 120X + 100Y
120 (130) + 100 (130) = 15600 + 13000 = 28600