Romans Food Market, located in Saratoga, New York, carries a variety of specialty foods from around the world. Two of the store's leading products use the Romans Food Market name: Romans Regular Coffee and Romans DeCaf Coffee. These coffees are blends of Brazilian Natural and Colombian Mild coffee beans, which are purchased from a distributor located in New York City. Because Romans purchases large quantities, the coffee beans may be purchased on an as-needed basis for a price 10% higher than the market price the distributor pays for the beans. The current market price is $0.47 per pound for Brazilian Natural and $0.62 per pound for Colombian Mild. The compositions of each coffee blend are as follows: Blend Bean Regular DeCaf
Brazilian Natural 75% 40%
Columbian Mild 25% 60%
Romans sells the regular blend for $3.60 per pound and the Decaf blend for $4.40 per pound. Romans would like to place an order for the Brazilian and Columbian coffee beans that will enable the production of 1000 pounds of Romans Regular coffee and 500 pound of Romans DeCaf coffee. The production cost is $0.80 per pound for the Regular blend. Because of the extra steps required to produce DeCaf, the production cost for the DeCaf blend is $1.05 per pound. Packaging costs for both products are $0.25 per pound. Formulate a linear programming model that can be used to determine the pounds of Brazilian Natural and Columbian Mild that will maximize the total contribution to profit. What is the optimal solution and what is the contribution profit?

z(max) =  2996.13  $

x₁  = 968      x₂  =  430       ( quantities of regular and Decaf coffee respectevely)

Total quantity of BN = 898 pounds

Total quantity of CM =  500 pounds

Step-by-step explanation:

Cost of the beans

Brazilian natural   =  Price market + 10 %    =  0.47 + 0.047  

BN  Cost   = 0.517 $/lb

Clombian Mild      =  Price market + 10 %    =  0.62  +  0.062

CM Cost   =  0.682  $/lb

Composition of the coffee blend

Regular coffee     0.75 BN  +  0.25 CM

De Caf   coffee     0.40 BN + 0.60  CM

PRICES

Regular Roman  =  3.60  $

Decaf                   =  4.40  $

Production costs:

Regular Roman  =  0.80  $/lb

Decaf                   =  1.05  $/lb

Packaging costs:   0.25  $/Lb       both

Profit  =  Price  -  cost

Profit  of regular coffee  =  3.60 - 0.80 - 0.25 -Cost of bean

for regular coffee cost of BN + CM

BN   is :   0.75*BN cost  = 0.75*0.517  =  0.38775       and

CM  is : 0.25*0.682   =   0.1705

Profit  of regular coffee  =  1.99175  $

Profit for Decaf coffee  =  4.4 - 1.05 - 0.25 - ( 0.517*0.4 + 0.6*0.682)

Profit for Decaf coffee  =  4.4  - 1.30 - 0.616

Profit for Decaf coffee  =  2.484  $

Let´s call  x₁  pounds of regular coffee   and  x₂   pounds of Decaf coffee then:

 Objective Function is:  

z  =  1.99175*x₁  +  2.484*x₂      to maximize

Subject to:

Availability of beans for 1000 pounds of Regular coffee means:

750 pounds of BN  +  250 pounds of CM

Availability of beans for 500 pounds of Decaf coffee means

200 pounds of BN  +   300 pounds of CM

Then   750 + 200  =  900 pounds of BN

And     250 + 300 =  550  pounds of CM

Availability of beans for 1000 pounds of Decaf  coffee correspond to

0.75 *x₁  +  0.40*x₂  ≤ 900

Availability of beans for 500 pounds of Regular  coffee correspond to

0.25*x₁  +  0.60*x₂  ≤ 500

Then the model is:

z  =  1.99175*x₁  +  2.484*x₂      to maximize

Subject to:

0.75 *x₁  +  0.40*x₂  ≤ 900

0.25*x₁  +  0.60*x₂  ≤ 500

General constraints  x₁ ≥ 0      x₂ ≥   0  both integers

After 6 iterations optimal solution  ( maximum z) is

z(max) =  2996.13  $

x₁  = 968      x₂  =  430  

x₁   and  x₂  are quantities of  Regular and Decaf coffee respectively, to find out quantities of Brazilian Natural and Colombian Mild

we proceed as follows

Regular coffee is :  0.75*968 =   726 pounds of BN

Decaf coffee is :      0.40*430 =   172  pounds of BN

Total quantity of BN = 898 pounds

Regular coffee is :  0.25*968   =  242 pounds of CM

Decaf coffee is :  0.6*430  =  258 pounds of CM

Total quantity of CM =  500 pounds

idk lol

step-by-step explanation:

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐