Given the function g(x) = 6(4)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.
1 answer:
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Answer:
233 cartons of food; 467 cartons of clothing
Step-by-step explanation:
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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<u>Feasible Region vertex</u>
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).
<u>Check of Integer Solution</u>
Trying these in the constraint inequalities gives ...
- 40·233 +25·467 = 20,995 < 21000
- 20·233 +5·467 = 6995 < 7000
<u>Selection of the Answer</u>
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.
