Engineering Optimization: Theory and Practice, Fourth Edition

204 Linear Programming II: Additional Topics and Extensions

FertilizerAshould not contain more than 60% of ammonia andBshould contain

at least 50% of ammonia. On the average, the plant can sell up to 1000 lb/hr and due

to limitations on the production facilities, not more than 600 lb of fertilizerAcan be

produced per hour. The availability of chemicalC 1 is restricted to 500 lb/hr. Assuming

that the production costs are same for bothAandB, determine the quantities ofA

andBto be produced per hour for maximum return if the plant sellsAandBat the

rates of $6 and $7 per pound, respectively.

SOLUTION Letx 1 andx 2 indicate the amounts of chemicalsC 1 andC 2 used in

fertilizerA, andy 1 andy 2 in fertilizerBper hour. Thus the total amounts ofAand

Bproduced per hour are given byx 1 +x 2 andy 1 +y 2 , respectively. The objective

function to be maximized is given by

f=selling price− cost of chemicalC 1 andC 2

= 6 (x 1 +x 2 )+ 7 (y 1 +y 2 ) − 5 (x 1 +y 1 ) − 4 (x 2 +y 2 )

The constraints are given by

(x 1 +x 2 ) +(y 1 +y 2 ) ≤ 1000 (amount that can be sold)

x 1 +y 1 ≤ 005 (availability ofC 1 )

x 1 +x 2 ≤ 006 (production limitations on A)

7

10 x^1 +

4

10 x^2 ≤

6

10 (x^1 +x^2 ) (A^ should not contain more

than 60% of ammonia)

7

10 y^1 +

4

10 y^2 ≥

5

10 (y^1 +y^2 ) (B^ should contain at least

50% of ammonia)

Thus the problem can be restated as

Maximizef=x 1 + 2 x 2 + 2 y 1 + 3 y 2 (E 1 )

subjectto

x 1 +x 2 +y 1 +y 2

x 1 +y 1

≤ 1000

≤ 500 (E^2 )

x 1 +x 2

x 1 − 2 x 2

≤ 600

≤ 0

(E 3 )

− 2 y 1 +y 2 ≤ (^0) (E 4 )
xi≥ 0 , yi≥ 0 , i= 1 , 2
This problem can also be stated in matrix notation as follows:
Maximizef (X)=cT 1 X 1 +cT 2 X 2