Food and clothing are shipped to victims of a natural disaster. Each carton of food will feed 13 people, while each carton of clothing will help 6 people. Each 20-cubic-foot box of food weighs 40 pounds and each 5-cubic-foot box of clothing weighs 25 pounds. The commercial carriers transporting food and clothing are bound by the following constraints:- The total weight per carrier cannot exceed 21 000 pounds.- The total volume must be no more than 7000 cubic feet.How many cartons of food and clothing should be sent with each plane shipment to maximize the number of people who can be helped?

This linear programming problem can be formulated as two inequalities (in addition to the usual constraints that the variables be non-negative). One of these expresses the constraint on weight. Let f and c represent numbers of food and clothing containers, respectively.

40f +25c ≤ 21000

The other expresses the limit on volume.

20f + 5c ≤ 7000

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Feasible Region vertex

We can subtract the boundary line equation of the first inequality from that of 5 times the second to find f:

5(20f +5c) -(40f +25c) = 5(7000) -21000

60f = 14000

f = 233 1/3

The second boundary line equation can be rearranged to find c:

c = 1400 -4f = 466 2/3

The nearest integer numbers to these values are ...

(f, c) = (233, 467)

The other vertices of the feasible region are associated with one or the other variable being zero: (f, c) = (0, 840) or (350, 0).

Check of Integer Solution

Trying these in the constraint inequalities gives ...

40·233 +25·467 = 20,995 < 21000
20·233 +5·467 = 6995 < 7000

Selection of the Answer

The answer to the question will be the feasible region vertex that maximizes the number of people helped. That is, we want to maximize ...

p = 13f + 6c

The values of p at the vertices are ...

p = 13·233 + 6·467 = 5831

p = 13·0 + 6·840 = 5040

p = 13·350 + 6·0 = 2100

The most people are helped when the plane is filled with 233 food cartons and 467 clothing cartons.