danithename said:
i think this is a variation problem but i need help.
A coffee company has 1440 pounds of sumatran coffee and 700 pounds of kona coffee a batch of hawaiian blend requires 8 pounds of kona and 12 pounds of sumatran and yeilds a profit of $90. A batch of classic blend requires 4 pounds of kona and 16 pounds of sumatran and yeilds a $55 profit. How many batches of each kind should be made in order to maximize profit? What is the Maximum profit?
I have not tried to much of this problem just because i just dont do well with word problems if you could offer any help i would apperciate it thank you!
Ok....it looks from your work that you might need a "push" in the right direction.
The first thing you should do is define the variables you are going to use. I'll define them this way:
let x = number of batches of Hawaiian blend produced
let y = number of batches of Classic blend produced
Now, the profit earned on 1 batch of Hawaiian blend is $90, so if x batches are produced, the profit will be 90x. The profit earned on 1 batch of Classic blend is $55, so if y batches are produced, the profit will be 55y.
total profit = profit on Hawaiian blend + profit on Classic blend
total profit = 90x + 55y
<-----this is the "objective function"....the quantity you want to maximize
There are some constraints that must be considered:
First, there are 1440 pounds of Sumatran coffee. Each batch of Hawaiian blend takes 12 pounds of Sumatran coffee, so "x" batches will use up 12x pounds. And each batch of Classic coffee takes 16 pounds of Sumatran coffee, so "y" batches will require 16y pounds of Sumatran coffee. Since there are only 1440 pounds of Sumatran coffee available,
12x + 16y
< 1440
Second, there are 700 pounds of Kona coffee available. Each batch of Hawaiian blend requires 8 pounds of Kona coffee, so it will require 8x pounds of Kona to produce "x" batches of Hawaiian blend. And each batch of Classic blend uses 4 pounds of Kona coffee; to produce "y" batches of Classic blend they will need 4y pounds of Kona coffee. The total amount of Kona coffee used cannot exceed the 700 available pounds, so
8x + 4y
< 700
Third, they cannot produce a negative number of batches, so
x
> 0
y
> 0
Graph the region which satisfies all of the constraints....this is known as the "feasible region." Any point within this region should satisfy all of the constraints, but you'll find the maximum and minimum values of the objective function at a vertex, or corner, of the region. So...graph the feasible region, and test the coordinates of each corner in the objective function to find which gives the maximum.
If this is still a mystery for you, I think you'll need to talk to your teacher, as we can't provide instruction here.
