Question 1123346: A candy company has 114kg of chocolate-covered nuts and 72kg 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 three fourths nuts and one fourth raisins and will sell for $9.50 per kg.The company should prepare how many kg of the first mix and how many kg of the second mix for a maximum revenue of how much. 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.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x = the number of kilograms of the first mix.
let y = the number of kilograms of the second mix.
the first mix contains 50% nuts and 50% raisins.
the second mix contains 75% nuts and 25% raisins.
the price for the first mix will be 7 per kilogram.
the price for the second mix will be 9.5 per kilogram.
the constraints are that maximum amount of nuts is 112 kilograms and maximum amount of raisins is 72 kilograms.
for the nuts, your constraint equation becomes .5x + .75y <= 114
that's because the first mix has 50% nuts and the second mix has 75% nuts.
for the raisins, your constraint equation becomes .5x + .25y <= 72
that's because the first mix has 50% raisins and the second mix has 25% raisins.
those are your constraint inequalities along with x >= 0 and y >= 0 because the quantities of nuts or raisins can't be negative.
using the desmos.com calculator, you would graph the opposite of these constraint inequalities.
the area of the graph that is not shaded is your region of feasibility.
the corner points of this region are where your maximum revenue will reside.
here's what the graph looks like.
the corner points of the feasible region of this graph are:
(0,152)
(102,84)
(144,0)
when the revenue for x is 7 and the revenue for y is 9.5, you get the following revenues.
(0,152) = 0*7 + 152*9.5 = 1444
(102,84) = 102*7 + 84*9.5 = 1512 ***** maximum revenue
(144,0) = 144*7 + 0*9.5 = 1008
1512 is the most revenue when the price of the first mix is 7 per kilogram and the price of the second mix is 9.5 per kilogram.
when the revenue for x is 7 and the revenue for y is 11, you get the following revenues.
(0,152) = 0*7 + 152*11 = 1672 ***** maximum revenue
(102,84) = 102*7 + 84*11 = 1638
(144,0) = 144*7 + 0*11 = 1008
1672 is the most revenue when the price of the first mix is 7 per kilogram and the price of the second mix is 11 per kilogram.
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