1. ## linear programming question

A fruit drink is made from 3 ingredients.

let j be the fraction of the drink that is juice
s be the fraction of the drink that is sugar and flavouring
w be the fraction of the drink that is water.

the cost of the drink per liter is $12j+3s+w, and it can sold at$2+s+j/litre. choose j and s to maximize the profit, for marketing reasons j>=1/20

1. express profit per litre as a function of j and s.
How do express profit, say i let z=profit in dollars per litre. z= price sold-cost=2-11j-2s????

2. Express problem to maximize profit, subject to constraints.
well my constraints are j>=1/20, s,j,w>=0... what are my others?

2. Originally Posted by fredrick08
A fruit drink is made from 3 ingredients.

let j be the fraction of the drink that is juice
s be the fraction of the drink that is sugar and flavouring
w be the fraction of the drink that is water.

the cost of the drink per liter is $12j+3s+w, and it can sold at$2+s+j/litre. choose j and s to maximize the profit, for marketing reasons j>=1/20

1. express profit per litre as a function of j and s.
How do express profit, say i let z=profit in dollars per litre. z= price sold-cost=2-11j-2s????

2. Express problem to maximize profit, subject to constraints.
well my constraints are j>=1/20, s,j,w>=0... what are my others?
1. $\text{Profit per litre} = \text{revenue - cost} = (2+j+s) - (12j+3s+w)=2-11j-2s-w$

2. As $s,j,w$ are fractions by volume you also have $s+j+w=1$

3. ok i drew it out, and done it in solver, but its says that there is no water?? Is this true? and my optimal solution is at (0.95,0,05) profit=1.55??

4. i did the math and subbed j=1/20 and s=19/20, then profit=-9/20??? im really confused now, how did solver get 1.55??

5. Originally Posted by fredrick08
ok i drew it out, and done it in solver, but its says that there is no water?? Is this true? and my optimal solution is at (0.95,0,05) profit=1.55??
You should be maximising profits (looks like you have minimised?)

By inspection it looks like maximum profit is returned with j=0.05, w=0.95, s=0. When the profit is \$0.5

CB

6. yes, that is what I thought, but my lecturer told me to use j+s+w=1, to eliminate w.... i got no idea what that means. im jet gunna leave it is profit=0.5

7. Originally Posted by fredrick08
yes, that is what I thought, but my lecturer told me to use j+s+w=1, to eliminate w.... i got no idea what that means. im jet gunna leave it is profit=0.5
In the profit expression replace w by 1-j-s

CB

8. i see, but how am i meant to know when i draw the feasible region that s=0?

9. Originally Posted by fredrick08
i see, but how am i meant to know when i draw the feasible region that s=0?
Draw the feasible region, it is defined by the constraints $s\ge 0$, $j\ge 0.05$, $s+j\le 1$.

Now the optimum (maximum profit) occurs at a vertex of the feasible region, so check them and keep the one with maximum profit.

CB