Question

Linear Programming A candy company makes three types of candy, solid-center, fruit-filled, and cream-filled, and packages these candies in three different assortments. A box of assortment I contains 4 solid-center, 4 fruit-filled, and 12 cream-filled candies, and sells for $17.95. A box of assortment II contains 12 solid-center, 4 fruit-filled, and 4 cream-filled candies, and sells for $18.45. A box of assortment III contains 8 solid-center, 8 fruit-filled, and 8 cream-filled candies, and sells for $20.85. The manufacturing costs per piece of candy are $0.01 for solid-center, $0.02 for fruit-filled, and $0.03 for cream-filled. The company can manufacture 4,800 solid-center, 4,000 fruit-filled, and 5,600 cream-filled candies weekly. How many boxes of each type should the company produce each week in order to maximize their profits? What is the maximum profit? *Will thumbs up for correct answer, thank you*

Answer 1

By given conditions, profit in selling each box of assortment I is = $[17.95-{(4*0.01)+(4*0.02)+(12*0.03)}] = $17.47

Profit in selling each box of assortment II is = $[18.45-{(12*0.01)+(4*0.02)+(4*0.03)}] = $18.13

Profit in selling each box of assortment III is = $[20.85-{(8*0.01)+(8*0.02)+(8*0.03)}] = $20.37

Let, the company should produce x boxes of assortment I, y boxes of assortment II, z boxes of assortment III each week.

Then the problem becomes,

Maximize A = 17.47x+18.13y+20.37z

Subject to 4x+12y+8z 4800

4x+4y+8z 4000

12x+4y+8z 5600

x,y,z 0

After introducing slack variables a, b and c, we rewrite the problem in the standard form as,

Maximize A = 17.47x+18.13y+20.37z+0a+0b+0c

Subject to 4x+12y+8z+1a+0b+0c = 4800

4x+4y+8z+0a+1b+0c = 4000

12x+4y+8z+0a+0b+1c = 5600

x,y,z,a,b,c 0

		cj	17.47	18.13	20.37	0	0	0
cB	B	b	a1	a2	a3	a4	a5	a6
0	a4	4800	4	12	8	1	0	0
0	a5	4000	4	4	8	0	1	0
0	a6	5600	12	4	8	0	0	1
			-17.47	-18.13	-20.37	0	0	0
0	a4	800	0	8	0	1	-1	0
20.37	a3	500	1/2	1/2	1	0	1/8	0
0	a6	1600	8	0	0	0	-1	1
			-7.285	-7.945	0	0	2.546	0
18.13	a2	100	0	1	0	1/8	-1/8	0
20.37	a3	450	1/2	0	1	-1/16	3/16	0
0	a6	1600	8	0	0	0	-1	1
			-7.285	0	0	0.993	1.553	0
18.13	a2	100	0	1	0	1/8	-1/8	0
20.37	a3	350	0	0	1	-1/16	1/4	-1/16
17.47	a1	200	1	0	0	0	-1/8	1/8
			0	0	0	0.993	0.642	0.911

In the first table a₄,a₅ and a₆ are the basis vectors and the largest negative (z_j-c_j) is (z₃-c₃) and hence a₃ is the entering vector. To find the leaving vector, we have Min = ; hence a₅ is the leaving vector, 8 being the key number.

We then transform the first table by simplex formula and get the second table.

In the second table a₄,a₃ and a₆ are the basis vectors and the largest negative (z_j-c_j) is (z₂-c₂) and hence a₂ is the entering vector. To find the leaving vector, we have Min = ; hence a₄ is the leaving vector, 8 being the key number.

We then transform the table following usual rule.

In the third table a₂,a₃ and a₆ are the basis vectors and the largest negative (z_j-c_j) is (z₁-c₁) and hence a₁ is the entering vector. To find the leaving vector, we have Min = [Since ]; hence a₆ is the leaving vector, 8 being the key number.

We then transform the table following usual rule.

In the next table all z_j-c_j 0. Hence, this table gives the optimal solution.

The optimal solution is x = 200, y = 100, z = 350 and A_max = 12436.5.

Therefore, the company should produce 200 boxes of assortment I, 100 boxes of assortment II, 350 boxes of assortment III each week.

And, the maximum profit is $12,436.50