Question 295942: Please can anyone help me understand this word problem.
A manufacturer makes three different drinks. Drink 1 contains 20% juice and 80% carbonated water. Drink 2 contains 50% juice and 50% carbonated water. Drink 3 contains 80% juice and 20% carbonated water. The manufacturer currently has available 100 liters of juice and 120 liters of carbonated water. Drink 1 sells for $1 per liter, Drink 2 sells for $2 per liter, and Drink 3 sells for $3.80 per liter. How many liters of each should the manufacturer make to maximize revenues?
A) 96.7 liters of Drink 1, 60 liters of Drink 2, 70.0 liters of Drink 3
B)66.0 liters of Drink 1, 20 liters of Drink 2, 83.3 liters of Drink 3
C)126.7 liters of Drink 1, 0 liters of Drink 2, 93.3 liters of Drink 3
D)106.7liters of Drink 1, 40 liters of Drink 2, 80.0 liters of drink 3
This is what I got so far, but no understand what to do next.
Drink 1: .2j .8c and sells for 1
Drink 2: .5j .5c and sells for 2
Drink 3: .8j .2c and sells for 3.8
Maximize: z = x1 + 2x2 + 3.8x3
Subject to: .2x1 + .5x2 + .8x3 = 100
.8x1 + .5x2 + .2x3 = 120
Thanks to anyone who can help.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! I'm not sure if I did this correctly, but I did determine that the answer is selection C.
Here's how:
x = number of liters of drink 1
y = number of liters of drink 2
z = number of liters of drink 3
number of liters of juice totals 100
number of liters of carbonated water totals 120
r = x + 2y + 3.8z (equation 1)
Equation 1 is the equation you want to maximize.
number of liters of juice in drink 1 = .2*x
number of liters of juice in drink 2 = .5*y
number of liters of juice in drink 3 = .8*z
.2*x + .5*y + .8*z <= 100 (equation 2)
Equation 2 is the equation for the amount of soda.
number of liters of carbonated water in drink 1 = .8*x
number of liters of carbonated water in drink 2 = .5*x
number of liters of carbonated water in drink 3 = .2*x
.8*x + .5*y + .2*z <= 120 (equation 3)
Equation 3 is the equation for the amount of carbonated water.
Your constraints are:
x >= 0
y >= 0
z >= 0
Equation 2
Equation 3
If you could graph in 3 dimensions, then you could graph these equations and pick one of the intersection points which would then yield you the answer you were looking for.
Since we can't do that here (don't have 3 dimension graphing capability), then we'll need to determine the intersection points without graphing.
To find the intersection points, we set all equations equal to what they are.
x = 0
y = 0
z = 0
equation 2 as an equality rather than an inequality (replace <= with =).
equation 3 as an equality rather than an inequality (replace <= with =).
The intersection of equations 2 and 3 is a line, not a point. This is because we have 2 equations in 3 unknowns which means the planes represented by those equations intersect in a line rather than a point. In 3 dimensions, it takes 3 equations in 3 unknowns to find an intersection point.
Full utilization of the resources is not required for maximum revenue, but it's certainly something to look at.
With 2 equations in 3 unknowns, there is not 1 choice for full utilization of resources, but a range of choices.
From equation 2 and 3, you can, however, determine that x has to be equal to z + 33 and 1/3 in order for there to be full utilization of the resources (soda + water).
You do this by solving those 2 equations simultaneously.
The 2 equations are:
.2*x + .5*y + .8*z = 100 (equation 2)
.8*x + .5*y + .2*z = 120 (equation 3)
You subtract equation 3 from equation 2 to get:
.6*x - .6*z = 20
You solve for x to get:
x = (.6*z + 20)/.6 which becomes:
x = z + 33 and 1/3.
To get full utilization of the resources, x and y could be anything as long as x = z + 33 and 1/3 and the total of the resources equals 100 for soda and 120 for water.
As an example, if we set z equal to 1, then x has to be 33 and 1/3 and y will need to be solved for as follows:
The 2 equations are:
.2*x + .5*y + .8*z = 100 (equation 2)
.8*x + .5*y + .2*z = 120 (equation 3)
Substitute 1 for z and 1 + 33 and 1/3 for x to get:
.2*(34 and 1/3) + .5*y + .8*1 = 100
.8*(34 and 1/3) + .5*y + .2*1 = 120
This becomes:
.5*y = 92 and 1/3
.5*y = 92 and 1/3
As long as x = z + 33 and 1/3, we have automatically solved for y.
If we let z = 60, then x = 60 + 33 and 1/3 = 93 and 1/3.
Our equations are:
.2*x + .5*y + .8*z = 100 (equation 2)
.8*x + .5*y + .2*z = 120 (equation 3)
Substitute 60 for z and 93 and 1/3 for x to get:
.2*(93 and 1/3) + .5*y + .8*60 = 100
.8*(93 and 1/3) + .5*y + .2*60 = 120
This becomes:
.5*y = 33 and 1/3.
.5*y = 33 and 1/3.
As long as x = z + 33 and 1/3, we have automatically solved for y.
So, to get full utilization of the resources, x has to be equal to z + 33 and 1/3.
Our constraints are therefore still:
x >= 0
y >= 0
z >= 0
.2x + .5y + .8z <= 100
.8x + .5y + .2z <= 120
But, to get full utilization of the resources, x has to be equal to z + 33 and 1/3.
You could go through all the possible scenarios and calculate the revenue for each, but since they already gave you the 4 possible answers to look at, you can simply look through each of the possible answers to see which gives you the most revenue.
Looking at all the possible answers, the only one where x = z + 33 and 1/3 was selection c.
That means that selection c was the only answer where full utilization of the resources was achieved.
That doesn't mean that selection c provided the most revenue, but in this case it did as you will soon see.
It was necessary to calculate the revenue for all the possible selections to determine that.
If you had determined what the possible selections were by yourself, you would still have had to calculate the revenue for each, so this is not cheating.
In addition, it was necessary to determine that the constraints were met.
the revenue calculated for each selection was based on:
$1 per drink for drink 1 * x
$2 per drink for drink 2 * y
$3.80 per drink for drink 3 * z
selection a: $482.7
selection b: $422.54
selection c: $481.24
selection d: $490.7
From a pure revenue standpoint, selection d looks like the winner.
We have to see, however, if selection d met all the constraints.
Selection d is:
D)106.7liters of Drink 1, 40 liters of Drink 2, 80.0 liters of drink 3
Total resources used by selection d were 106.7 + 40 + 80 = 226.7 liters which is more than the total of 220 liters available.
Selection d used more soda and water than was available so selection d cannot be one of the choices even though it had the highest revenue.
Since selection a had the next highest revenue, then check selection a to see if it met the resource constraints.
Selection a is:
A) 96.7 liters of Drink 1, 60 liters of Drink 2, 70.0 liters of Drink 3
Selection a used a total of 226.7 liters which is more than the 220 liters that were available.
Selection a is discounted because it did not meet the resource constraints, even though the revenue was higher than selection c.
Since selection c had the next highest revenue, then check selection c to see if it met the resource constraints:
C)126.7 liters of Drink 1, 0 liters of Drink 2, 93.3 liters of Drink 3
Selection c used 126.7 + 0 + 93.3 = 220 liters of resources.
Since 220 liters of resource were available, selection c meets the resource constraint requirements.
Since selection b did not have as much revenue as selection c, we don't need to check selection b.
Your answer is selection c since selection c gave the maximum revenue while keeping within the constraints of the resources available.
As I said before, I don't know if this is the way they wanted you to solve this problem, and it certainly took a lot longer than I expected it to, but I know the answer is correct, even if it was a chore to find it.
It's also worth noting, that even if we did not go through the process of determining all the intersections points, that the solution of selection c was at one of them.
Selection c had an intersection point with y = 0.
When y = 0, the 2 equations of:
.2x + .5y + .8z <= 100
.8x + .5y + .2z <= 120
became equal to:
.2x + .8z = 100
.8x + .2z = 120
Multiply the first equation by 4 to get:
.8x + 3.2z = 400
.8x + .2z = 120
Subtract the second equation from the first equation to get:
3z = 280
z = 93.333333333 which is the same as 93 and 1/3.
Solving for x when y = 93 and 1/3 yielded x = 126 and 2/3 in both equations.
Your intersection point for selection c was:
x = 126 and 2/3
y = 0
z = 93 and 1/3
The theory states that the maximum / minimum solution will be at one of the intersection points of the constraint equations.
That theory holds for this problem since the solution is at one of the intersection points of the constraint equations.
My answer was a lot more long winded than necessary because I was actually trying to solve the problem.
Since you were already given the possible solutions to consider, the simplest solution for this problem would have been to calculate the revenue from the possible solutions and calculate the resources used from the possible solutions and then determine which solution had the highest revenue while still meeting the resource constraints.
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