A candy company has 120 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 the company prepare for the maximum revenue? Find the maximum revenue.
Let x represent the number of kilograms of the 1/2:1/2 mix, and y the number of kilograms of the 3/4:1/4 mix. (Here the ratio is nuts to raisins.)
The amount of profit can be represented by P=7x+9.5y.
There are some constraints -- namely the amount of raw materials available. The constraints can be represented by a system of inequalities:
`x>=0,y>=0` are the natural constraints.
`1/2 x+3/4y<=120` represents the amount of nuts available. Each kg of mix x uses 1/2 kg of nuts, while each kg mix y uses 3/4 kg.
`1/2x+1/4y<=90` represents the amount of raisins available.
Graphing the system of inequalities yields a feasible region (in this case a bounded polygon) where every point in the region satisfies all of the inequalities. By the corner point principle, there is a maximum and a minimum for the objective function (the profit function in this case) and they occur at the corner points of the feasible region.
The points are (0,0),(0,160),(180,0) and (150,60). These points represent, in order, making no product, only producing 160kg of product y, only producing 180kg of product x, and finally producing 150kg of x and 60kg of y.
Substituting the values into the objective function we get:
Thus your maximum profit occurs when you produce 150kg of x, and 60kg of y.
(Note that 150kg of x uses 75kg nuts and 75kg of raisins. 60kg of y uses 45kg of nuts and 15kg of raisins. Thus we use a total of 120kg of nuts and 90kg of raisins.)
The candy company has 120 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.
If the company sells only the first mix at $7 per kg, it can make 90 kg of the same. The revenue when this is sold is 7*90 = $630.
If the company sells only the second mix at $9.5 per kg it can make 120*(3/4) = 160 kg and the total revenue is 160*9.5 = $1520
A third option would be to sell both the mixes. If the company makes x kg of mix one and y kg of the second mix, it needs a total of 0.5*x + 0.75*y of nuts and 0.5*x + 0.25*y of raisins. It has 120 kg of nuts in stock and 90 kg of raisins in stock.
0.5*x + 0.75*y = 120 ...(1)
0.5*x + 0.25*y = 90 ...(2)
(1) - (2) gives
0.5y = 30
=> y = 60
x = 120 - 0.75*60 = 75
The revenue when 75 kg of the first mix and 60 kg of the second mix is sold is
75*7 + 60*9.5 = $1095
This shows that revenue is maximized when only the second mix is sold and the maximum revenue is equal to $1520.
For making one kilogram of the first mix, the candy maker needs 0.5 kg of nuts and 0.5 kg of raisins and for the second mix it needs 0.75 kg of nuts and 0.25 kg of raisins.
First we can assume that to get maximum revenue all the ingredients should be used. If the candy maker sells x kg of the first mix and y kg of the second mix it uses 0.5*x + 0.75*y of nuts and 0.5*x + 0.25*y of raisins.
0.5*x + 0.75*y = 120
0.5*x + 0.25*y = 90
Subtracting the two equations 0.5y = 30, y = 60 and x = 75
Mix one can be sold at $7 and mix 2 can be sold at $9.5. The total revenue is $1095.
If the candy maker decides on selling only the second mix as it sells at $9.5, it can make 160 kg and the total revenue is $1520. This is greater than $1095 though it still has 50 kg of raisin remaining.
The candy maker should prepare 160 kg of the second mix and none of the first mix. This gives a revenue of $1520.