A candy company has 201 kg of chocolate covered nuts and 117 kg of chocolate cov
ID: 3196135 • Letter: A
Question
A candy company has 201 kg of chocolate covered nuts and 117 kg of chocolate covered raisans to be dols 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. Complete parts a and b.
a. How many kg of each mix should the company prepare for the maximum revenue? Find the max revenue.
The company should prepare ____kg of the first mix and ___kg of the second mix for a maximum revenue of $_______.
b. The company raises the price of the second mix to $11 per kg. Now how many kg of each mix should the company prepare for the maximum revenue? Find the maximum revenue.
The company should prepare ____kg of the first mix and ____kg of the second mix for a maximum revenue of $____.
Explanation / Answer
Nuts =201 kg and Raisans = y= 117 kg;
Mix 1 (x) = x/2 + x/2
Mix 2 (y) = 3y/4 + y/4;
a)
We know that 2x/4 + 3y/4 = 201 or 2x+3y <= 804;
and 2x/4 + y/4 = 117 or 2x+y <= 468;
Revenue function = 7x + 9.5y;
The 2 constraints meet at 2y = 336 so y= 168; and x = 150;
The points of feasible region are (0,0) , (234,0) , (150,168) and (0,268)
The maximum revenue will occur at the point (150,168)
Revenue max = 7 (150) + 9.5 (168) = 2646$;
Thus, company should prepare 150 kg of first mix and 168 kg of second mix for maximum revenue of 2646$ ;
b)
Now the second mix is of 11$ per kg so objective function is Revenue = 7.5x + 11y;
Here we see that the maximum revenue occurs at (0,268) point of the feasible region given by
7.5 (0) + 11 (268) = 2948 $ ( vs a revenue of 2898 at (150,168) )
Thus, company should prepare 0 kg of first mix and 268 kg of second mix for a maximum revenue of 2948$
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.