Question

A company makes three types of candy and packages them in three assortments. assortment i contains 44 cherrycherry​, 44 lemonlemon​, and 1212 limelime ​candies, and sells for a profit of ​$4.004.00. assortment ii contains 1212 cherrycherry​, 44 lemonlemon​, and 44 limelime ​candies, and sells for a profit of ​$3.003.00. assortment iii contains 88 cherrycherry​, 88 lemonlemon​, and 88 limelime ​candies, and sells for a profit of $ 5.00$5.00. they can make 4 comma 8004,800 cherrycherry​, 4 comma 0004,000 lemonlemon​, and 6 comma 0006,000 limelime candies weekly. how many boxes of each type should the company produce each week in order to maximize its profit​ (assuming that all boxes produced can be​ sold)? what is the maximum​ profit?

Answer 1

Let x be the number of assortment I items sold
Let y be the number of assortment II items sold
Let z be the number of assortment III items sold
The cost for making assortment I is .20(4) + .25(4) + .30(12) = $5.40
The cost for making assortment II is .20(12) + .25(4) + .30(4) = $4.60
The cost for making assortment III is .20(8) + .25(8) + .30(8) = $6.00
The equation for total cost is: cost = 5.4x + 4.6y +6z
The equation for total income is: income = 9.4x + 7.6y + 11z
The equation for total profit is: profit = (9.4-5.4)x + (7.6-4.6)y + (11-6)z --> 4x + 3y + 5z
The total profit equation is the objective function for this example.
The constraint equations are listed below:
4x + 12y + 8z ≤ 5000 (sour candy)
4x + 4y + 8z ≤ 3800 (lemon candy)
12x + 4y + 8z ≤ 5400 (lime candy)
x ≥ 0
y ≥ 0
z ≥ 0
This system can be solved using the Simplex method:initial simplex tableau:
x y z s1 s2 s3 n
4 12 8 1 0 0 5000
4 4 8 0 1 0 3800
12 4 8 0 0 1 5400
-4 -3 -5 0 0 0 0
The solution to this is listed below:
x y z s1 s2 s3 n
0 1 0 .125 -.125 0 150
0 0 1 -.0625 .25 -.0625 300
1 0 0 0 -.125 .125 200
0 0 0 .0625 .375 .1875 2750
The maximum profit of $2750 is obtained by selling 200 units of assortment I, 150 units of assortment II, and 300 units of assortment III.