this problem is given me very much difficulty because of the set up and what to do. can someone assit me if you have a few minutes. thanks.
Income. A baker has 250 units of flour, 90 of sugar, and 150 of raisins. A loaf of raisin bread requires 1 unit of flour, 1 of sugar, and 2 of raisins, while a raisin cake needs 5,2,and 1 unit, respectively.
a) If raisin bread sells for $1.75 a loaf and raisin cake for $4 each , how many of each should be baked so that gross income is maximized?
b) what is the maximum gross income?
c) Does it require all of the available units of flour, sugar, and raisins to produce the number of loaves of raisins bread and raisin cakes that produce the maximum profit? If not, how much of each ingredient isleft over?
I suggest you make a basic chart similar to the one I did for your previous question. Since the cakes provide more income, how many cakes can the baker make with these ingredients? How many loaves of bread? Can you juggle these numbers to use more of the ingredients to produce more income?
Look at the example of the Simplex method here:
http://www.markschulze.net/LinearProgramming.pdf
Why did the baker go to school?
To improve his dough-cumentation skills!
But don't worry, I'm here to assist you with your problem. Let's get baking!
To solve this problem, we can use a linear programming approach. First, let's define our variables:
Let x be the number of loaves of raisin bread.
Let y be the number of raisin cakes.
a) To maximize the gross income, we need to maximize the objective function. The objective function in this case is the total income, which can be calculated as follows:
Total Income = 1.75x (income from raisin bread) + 4y (income from raisin cakes)
But we also have some constraints:
Flour constraint: 1x + 5y <= 250
Sugar constraint: 1x + 2y <= 90
Raisin constraint: 2x + y <= 150
b) In order to find the maximum gross income, we need to solve this linear programming problem. If you plug in these constraints into the Simplex method, you should be able to find the optimal values of x and y.
c) As for whether all the available units of flour, sugar, and raisins will be used, it depends on the optimal values of x and y. If the optimal solution requires using all the available units, then there won't be any leftovers. But if the optimal solution allows for some slack in the constraints, then there might be some ingredients left over.
To solve this problem, we can use the concept of linear programming to determine the optimal solution.
a) To maximize the gross income, we need to determine the number of loaves of raisin bread (x) and raisin cakes (y) that should be baked.
We can set up the following equations based on the given information:
1x + 5y ≤ 250 (equation for flour)
1x + 2y ≤ 90 (equation for sugar)
2x + 1y ≤ 150 (equation for raisins)
We also have the constraints that both x and y should be non-negative.
Now, let's use the simplex method to solve this linear programming problem.
1. Start with the initial tableau:
```
| z | x | y | s1 | s2 | s3 | RHS |
------|---|---|---|----|----|----|-----|
z | 0 | 1 | 1 | 0 | 0 | 0 | |
------|---|---|---|----|----|----|-----|
s1 | | 1 | 5 | 1 | 0 | 0 | 250 |
s2 | | 1 | 2 | 0 | 1 | 0 | 90 |
s3 | | 2 | 1 | 0 | 0 | 1 | 150 |
```
2. Choose the entering variable (the one with the most negative coefficient in the z-row). In this case, x has the most negative coefficient (-1), so it will be the entering variable.
3. Choose the leaving variable (the one with the smallest positive ratio of RHS to the entering variable's coefficient in its column). In this case, s2 has the smallest positive ratio of 90/2 = 45, so it will be the leaving variable.
4. Perform row operations to make the pivot element (in this case, the coefficient of x in the s2 row) equal to 1 and all other coefficients in the column zero. After row operations:
```
| z | x | y | s1 | s2 | s3 | RHS |
------|----|-----|-----|-----|-----|-----|-----|
z | 0 | 0 | 2/5 | 1/5 | 0 | 0 | 10 |
------|----|-----|-----|-----|-----|-----|-----|
s1 | | 7/2 | 5/2 | 1 | 0 | 0 | 375 |
x | | 1/2 | 1 | 0 | 1/2 | 0 | 45 |
s3 | | 0 | -1 | -2 | 0 | 1 | -90 |
```
5. Repeat steps 2-4 until all coefficients in the z-row are non-negative.
Next, apply the simplex method:
6. The entering variable is y because it has the most negative coefficient (-2) in the z-row.
7. The leaving variable is s3 because it has the smallest positive ratio of -90/-1 = 90 in its column.
8. Perform row operations to make the pivot element (in this case, the coefficient of y in the s3 row) equal to 1 and all other coefficients in the column zero. After row operations:
```
| z | x | y | s1 | s2 | s3 | RHS |
------|-----|------|------|-----|-----|-----|-----|
z | 1/3 | 0 | 2/15 | 1/15| 0 | -2/3| 100 |
------|-----|------|------|-----|-----|-----|-----|
s1 | | 7/2 | 5/2 | 1 | 0 | 0 | 375 |
x | | 1/2 | 1 | 0 | 1/2 | 0 | 45 |
y | | -1/2 | 0 | 2 | 0 | -1 | 90 |
```
9. Repeat steps 6-8 until all coefficients in the z-row are non-negative.
10. The final tableau shows that the maximum value of the objective function (gross income) is $100. To achieve this, the baker should bake 45 loaves of raisin bread and 90 raisin cakes.
b) The maximum gross income is $100.
c) No, it does not require all the available units of flour, sugar, and raisins. After producing the maximum profit, there will be some leftover ingredients. To determine the amounts, we can check the slack columns (s1, s2, s3) in the final tableau:
- s1: 375 units of flour are left
- s2: 0 units of sugar are left (fully used)
- s3: 0 units of raisins are left (fully used)
Therefore, 375 units of flour are left over after producing the maximum profit.
To solve this problem, we can use linear programming to find the optimal solution. Linear programming involves setting up a mathematical model to represent the problem and then using optimization techniques to find the best solution.
Let's start by setting up the variables:
Let x - the number of raisin bread loaves to be baked
Let y - the number of raisin cakes to be baked
Now, let's set up the objective function and constraints for each part of the problem:
a) To maximize the gross income:
Objective function: z = 1.75x + 4y
Constraints:
1 unit of flour is required for each loaf of raisin bread, and 5 units for each cake.
Thus, the constraint for flour is: x + 5y <= 250
1 unit of sugar is required for each loaf of raisin bread, and 2 units for each cake.
The constraint for sugar is: x + 2y <= 90
2 units of raisins are required for each loaf of raisin bread, and 1 unit for each cake.
The constraint for raisins is: 2x + y <= 150
All the variables (x and y) should be non-negative: x >= 0, y >= 0
b) To determine the maximum gross income, we need to solve the linear programming problem using optimization techniques. You can use methods such as the Simplex method or graphical method to find the optimal solution. The link you provided (http://www.markschulze.net/LinearProgramming.pdf) explains the Simplex method in detail.
c) After finding the optimal solution (values of x and y that maximize the gross income), you can evaluate whether all the available units of flour, sugar, and raisins are used to produce the maximum profit. To do this, compare the optimal values of x and y to the constraints for flour, sugar, and raisins. If any of the constraints are not fully utilized, then there will be some ingredients left over. You can calculate the remaining amount by subtracting the used amount from the available amount for each ingredient.
Hope this helps in understanding the problem and how to approach it!