A FOOD PROCESSING COMPANY PRODUCES REGULAR AND DELUXE ICE CREAM AT 3 PLANTS. PER

Question

A FOOD PROCESSING COMPANY PRODUCES REGULAR AND DELUXE ICE CREAM AT 3 PLANTS. PER HOUR OF OPERATION, THE PLANT IN CEDARBURG PRODUCES 20 GALLONS OF REGULAR ICE CREAM AND 10 GALLONS OF DELUXE ICE CREAM. GRAFTON PLANT PRODUCES 10 GALLONS OF REGULAR ICE CREAM AND 20 GALLONS OF DELUXE ICE CREAM, AND THE WEST BEND PLANT PRODUCES 20 GALLONS OF REGULAR ICE CREAM AND 20 GALLONS OF DELUXE ICE CREAM.

IT COST $70 PER HOUR TO OPERATE CEDARBURG PLANT, $75 PER HOUR TO OPERATE GRAFTON PLANT, AND $90 PER HOUR TO OPERATE WEST BEND PLANT.

THE COMPANY NEEDS ATLEAST 300 GALLONS OF REGULAR ICE CREAM AND ATLEAST 200 GALLONS OF DELUXE ICE CREAM EACH DAY.

HOW MANY HOURS PER DAY SHOULD EACH PLANT OPERATE IN ORDER TO PRODUCE THE REQUIRED AMOUNTS OF ICE CREAM AND MINIMIZE THE COST OF PRODUCTION?

WHAT IS THE MINIMUM PRODUCTION COST?

Explanation / Answer

CEDARBURG PRODUCES 20 GALLONS OF REGULAR ICE CREAM AND 10 GALLONS OF DELUXE ICE CREAM.(20,10)

GRAFTON PRODUCES 10 GALLONS OF REGULAR ICE CREAM AND 20 GALLONS OF DELUXE ICE CREAM.

(10,20)

THE WEST BEND PRODUCES 20 GALLONS OF REGULAR ICE CREAM AND 20 GALLONS OF DELUXE ICE CREAM.(20,20)

CEDARBURG= $70 per hour to operate

GRAFTON= $75

WEST BEND= $90

let the minimum number of hours for each company is x,y,z

total number gallons produced per day=x(20,10)+y(10,20)+z(20,20)

regulaar ice cream per day in gallons=20x+10y+20z

deluxe ice cream per day in gallons=10x+20y+20z

given

20x+10y+20z>=300 and 10x+20y+20z>=200

so10x+10x+10y+20z>=300 and10x+10y+10y+20z>=200

20x+10y+20z>=10x+20y+20z+100

10x-10y>=100

x-y>=10

x>=10+y

so x minimum value is 10+y

substitute x in 10x+20y+20z>=200

10(10+y)+20y+20z>=200

3y+2z>=10 this is ABC

we also know that x,y,z<24(total hours of day)

total production cost=$70*x+$75*y+$90*z

it has to be minimum

70(10+y)+75y+90z

=700+70y+75y+90z

=700+145y+90z

=700+45(3y+2z)+10y from ABC

=700+45(10)+10y

=700+450+10y

=1150$+10y is the minimum cost of production

3y+2z=10

2z=10-3y

z=5-3y/2

z>=0

so

5-3y/2>=0

y<=10/3

x=10+y<=24

y<=14

y<=10/3

so y can take value from 0,1,2,3

corresponding x and z values are

for y=0

x=10 and z=5

for y=1

x=11 and z=7/2

for y=2

x=12 and z=2

for y=3

x=13 and z=1/2

among these the integer pairs are

x=10,y=0 and z=5

and

x=12,y=2 and z=2

the minimum cost of production 1150+10y so y should be minimum so take it as zero

=1150+10*0

=1150$

and x=10,y=0 and z=5

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