A candy company has 150 kg of chocolate-covered nuts and 90 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 contain 3/4 nuts and 1/4 raisins and will sell for $9.50 per kg. How many kilograms of each mix should be company prepare for the maximum revenue? Find the maximum revenue.

1. 7x + 9.5y

2. 1980

Step-by-step explanation:

In Mix A, there is an equal amount of nuts and raisins.

Let x = kg of nuts = kg of raisins.

In Mix B, there is 3 times as much nuts as raisins.

Let y = kg of raisins, then 3y = kg of nuts.

We have this information:

           Nuts    Raisins   Price  

mix A    x   +   x     =  $ 7.00

mix B    3y  +  y     =  $  9.50

Total:   150   90

Hence, the number of kilograms of each mix should company prepare for the maximum revenue is = 7x + 9.5y

We have these inequalties:

{x≥0y≥0x+3y≤150x+y≤90} (equation 1)

There will be

x+x=2x kg of Mix A at $7.00 per kg.

The revenue from Mix A is:

7(2x)=14x dollars.

There will be

3y+y=4y kg of Mix B at $9.50 per kg.

The revnue from Mix B is:

9.5(4y)=38y dollars.

Hence, the total revenue is:

R=14x+38y (equation 2).

Graph the region determined by [1] and locate its vertices.

You should get:

(0,0),(90,0),(0,50),(60,30)

Test the vertices into [2] and see which vertex produces maximum revenue

(0,0): R=14(0)+38(0)=0 Minimum revenue

(90,0):R=14(90)+38(0)=1260

(0,50): R=14(0)+38(50)=1900

(60,30):R=14(60)+38(30)=1980 Maximum revenue.

answer: a

step-by-step explanation:

10.2 units