Also, we must have *20. y20. Therefore, the above problem can be fomulated as follows: find x and y that maximize - = 2x + 1.25y subject to the constraints: y < 130 -x 3* 2 y < 170 2 0 y 2 0 +

Advanced Engineering Mathematics
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ISBN:9780470458365
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Chapter2: Second-order Linear Odes
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A candy manufacturer has 130 pounds of chocolate-covered chenies and 170 pounds of
chocolate-covered mints in stock. He decides to sell them in the form of two different
mixtures. One mixture will contain half cheries and half mints by weight and will sell for
$2.00 per pound. The other mixture will contain one-third cheries and two-thirds mints by
weight and will sell for $1.25 per pound. How many pounds of each mixture should the
candy manufacturer prepare in order to maximize his sales revenue?
let us call A the mixture of half cheies and half mints, and B the mixture which is one-
third chenries and two-thirds mints. Let x be the number of pounds of A to be prepared and y
the number of pounds of B to be prepared. The revenue function can then be written as
: = 2x + 1.25 y
Since each pound of A contains one-half pound of chenies and each pound of B contains
one-third pound of chenries, the total number of pounds of chenies used in both mixtures is
1
x +
1
Similarly, the total number of pounds of mints used in both mixtures is:
1
Now, since the manufacturer can use at most 130 pounds of chenies and 170 pounds of
mints, we have the constraints:
-y
< 130
1
< 170
+
+
Transcribed Image Text:A candy manufacturer has 130 pounds of chocolate-covered chenies and 170 pounds of chocolate-covered mints in stock. He decides to sell them in the form of two different mixtures. One mixture will contain half cheries and half mints by weight and will sell for $2.00 per pound. The other mixture will contain one-third cheries and two-thirds mints by weight and will sell for $1.25 per pound. How many pounds of each mixture should the candy manufacturer prepare in order to maximize his sales revenue? let us call A the mixture of half cheies and half mints, and B the mixture which is one- third chenries and two-thirds mints. Let x be the number of pounds of A to be prepared and y the number of pounds of B to be prepared. The revenue function can then be written as : = 2x + 1.25 y Since each pound of A contains one-half pound of chenies and each pound of B contains one-third pound of chenries, the total number of pounds of chenies used in both mixtures is 1 x + 1 Similarly, the total number of pounds of mints used in both mixtures is: 1 Now, since the manufacturer can use at most 130 pounds of chenies and 170 pounds of mints, we have the constraints: -y < 130 1 < 170 + +
Also, we must have *20, y 20. Therefore, the above problem can be formulated as
= 2x + 1.25y subject to the constraints:
follows: find x and y that maximize
1
< 130
-y
3
< 170
2 0
y 2 0
Use the technique of linear programming and find feasible region of the problem and locate
our extreme points.
Q 6.
(B)
Make a linear programming graph from the following LP model and find out the most
profitable solution.
Maximize CM = $25A + $40B
Subject to: 2A + 4B < 100 hours
3A + 2B < 90
A2 0, B20
Transcribed Image Text:Also, we must have *20, y 20. Therefore, the above problem can be formulated as = 2x + 1.25y subject to the constraints: follows: find x and y that maximize 1 < 130 -y 3 < 170 2 0 y 2 0 Use the technique of linear programming and find feasible region of the problem and locate our extreme points. Q 6. (B) Make a linear programming graph from the following LP model and find out the most profitable solution. Maximize CM = $25A + $40B Subject to: 2A + 4B < 100 hours 3A + 2B < 90 A2 0, B20
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