Question 158560This question is from textbook
: Mixture Problem. A store sells almonds for $6 per pound, cashews for $5 per pound, and peanuts for $2 per pound. One week the manager decides to prepare 100 16-ounce packages of nuts by mixing the peanuts, almonds, and cashews. Each package will be sold for $4. The mixture is to produce the same revenue as selling the nuts separately. Prepare a table that shows some of the possible ways the manager can prepare the mixture.
This question is from textbook
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! SEVERAL DAYS AGO, I STARTED WORKING ON THIS PROBLEM AND WAS CALLED AWAY. I AM NOW BACK AND SEE THAT A SOLUTION HAS NOT BEEN PROVIDED. YOU PROBABLY HAVE ALREADY FOUND A SOLUTION BUT I THINK IT'S A GOOD PROBLEM, SO I'LL GIVE YOU MY TWO BITS:
This is yet another example of more unknowns than equations. Many times, multiple answers are possible.
First, lets deal in ounces:
Let x=amount of almonds @ $6 per lb or $3/8 per ounce
Let y=amount of cashews @ $5 per lb or $5/16 per ounce
Let z=amount of peanuts @ $2 per lb or $1/8 per ounce
Now we are told that:
x+y+z=16------------------------eq1
and ($ are understood)
(3/8)x+(5/16)y+(1/8)z=4 multiply each term by 16
6x+5y+2z=64-----------------------------------------eq2
Multiply eq1 by 6 and subtract eq2 from it and we get:
y+4z=32 solve for y
y=32-4z-------------------------------eq1a
Now we know a couple of things about this problem:
(1) We cannot have negative values for x, y or z
(2) We cannot have non-zero values for x, y, or z (I'm assuming that there has to be some of each in every packet)
Given the above, we see that in eq1a, 32-4z has to be greater than zero, other wise we will have either negative or a zero value for y, so:
32-4z>0 subtract 32 from each side:
-4z>-32 divide each side by -4 (note: the inequality sign changes)
z<8 in order for y to be positive and non-zero
Now lets plug y=32-4z from eq1a into eq1
x+32-4z+z=16 or
x+32-3z=16 subtract 32 from and add 3z to each side
x=3z-16 now in order for x to be positive and non-zero, 3z-16 must be greater than zero:
3z-16>0 add 16 to each side
3z>16 divide each side by 3
z>5 1/3 in order for x to be positive and non-zero
WE HAVE NOW PLACED CONSTRAINTS ON Z:
(5 1/3)< z <(8). In order for x and y to be positive and non-zero
Now we can start building the table:
Choose ANY value (INFINITE NUMBER DEPENDING ON THE ACCURACY OF THE SCALES) for z within the above constraints, plug the value into eq1a and solve for y, then plug both values into eq1 and solve for x. Lets try an example:
Let z be 7, then from eq1a, y=4, then from eq1 x+4+7=16
x+16-11; x=5
CK
5+4+7=16
6*5+5*4+7*2=64
30+20+14=64
64=64
Niche problem!!!
Hope this helps---ptaylor
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