Question

A candy company has 99 kg of chocolate-covered nuts and 63 kg of chocolate-covered raisins to be sold as two different mixes. One mix will contain half nuts and half raisins and will sell for $7 per kg. The other mix will containnuts and raisins and will sell for $9.50 per kg. Complete parts a. and b a. How many kilograms of each mix should the company prepare for the maximum revenue? Find the maximum revenue The company should prepare kg of the firast mik and kg of the second mix for a maximum revenue of s b. The company raises the price of the second mix to $11 per kg. Now how many kilograms of each mix should the company prepare for the maximum revenue? Find the maximum revenue. The company should preparekg of the first mix and kg of the second mix for a maximum revenue of

Answer 1

make a table from given data

	chocolate nuts	chocolate raisins	selling price per kg
one mixture = x1	1/2=0.5	1/2=0.5	7
second mixture = x2	3/4=0.75	1/4=0.25	9.5
total we have	99	63

Max Z = 72.1 + 9.52.2

subject to

0.5.r1 0.752 < 99

0.5r1 0.25r2 < 63

After introducing slack variables

$Max\:\:Z=7x_1+9.5x_2+0S_1+0S_2$

subject to

0.52.1 0.752.2 Si 99

0.52.1 + 0.25-r-+ S-= 63

Iteration-1		Cj	7	9.5	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XB/x2
S1	0	99	0.5	(0.75)	1	0	99/0.75=132→
S2	0	63	0.5	0.25	0	1	63/0.25=252
Z=0		Zj	0	0	0	0
		Zj-Cj	-7	-9.5↑	0	0

Negative minimum Zj-Cj is -9.5 and its column index is 2

Minimum ratio is 132 and its row index is 1.

The pivot element is 0.75.

Entering =x2, Departing =S1,

R, ← R1 * 1.333

R2 ← R2-0.25 RI

Iteration-2		Cj	7	9.5	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XB/x1
x2	9.5	132	0.6667	1	1.3333	0	132/0.6667=198
S2	0	30	(0.3333)	0	-0.3333	1	30/0.3333=90→
Z=1254		Zj	6.3333	9.5	12.6667	0
		Zj-Cj	-0.6667↑	0	12.6667	0

Negative minimum Zj-Cj is -0.6667 and its column index is 1

Minimum ratio is 90 and its row index is 2.

The pivot element is 0.3333.

Entering =x1, Departing =S2

R2 ← R) * 3

R1 ← R1-0.6667 R2

Iteration-3		Cj	7	9.5	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio
x2	9.5	72	0	1	2	-2
x1	7	90	1	0	-1	3
Z=1314		Zj	7	9.5	12	2
		Zj-Cj	0	0	12	2

Since all  $Zj-Cj \geq 0$

Hence, optimal solution is arrived

${\color{Red} x_1=90,\:\:x_2=72}$

Max Z1314

.

.

.

the company should prepare 90 kg of first mix and 72 kg of the second mix for a maximum revenue of $1314

.

.

.

.

.

.

when second mix price is $11

$Max\:\:Z=7x_1+11x_2+0S_1+0S_2$

subject to

0.52.1 0.752.2 Si 99

0.52.1 + 0.25-r-+ S-= 63

Iteration-1		Cj	7	11	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio XB/x2
S1	0	99	0.5	(0.75)	1	0	99/0.75=132→
S2	0	63	0.5	0.25	0	1	63/0.25=252
Z=0		Zj	0	0	0	0
		Zj-Cj	-7	-11↑	0	0

Negative minimum Zj-Cj is -11 and its column index is 2

Minimum ratio is 132 and its row index is 1.

The pivot element is 0.75.

Entering =x2, Departing =S1,

$R_1 \leftarrow R_1*1.3333$

$R_2 \leftarrow R_2-0.25R_1$

Iteration-2		Cj	7	11	0	0
B	CB	XB	x1	x2	S1	S2	MinRatio
x2	11	132	0.6667	1	1.3333	0
S2	0	30	0.3333	0	-0.3333	1
Z=1452		Zj	7.3333	11	14.6667	0
		Zj-Cj	0.3333	0	14.6667	0

Since all $Zj-Cj \geq 0$

Hence, optimal solution is arrive

${\color{Red} x_1= 0,\:\:x_2=132}$

${\color{Red} Max \:\:Z=1452}$

.

.

the company should prepare 0 kg of first mix and 132 kg of the second mix for a maximum revenue of $1452