A dietician is planning a snack package of fruit and nuts. Each ounce of fruit will supply zero units of protein, 3 units of carbohydrates, and 2 units of fat, and will contain 30 calories. Each ounce of nuts will supply 2 units of protein, 2 units of carbohydrates, and 4 units of fat, and will contain 40 calories. Every package must provide at least 4 units of protein, at least 10 units of carbohydrates, and no more than 18 units of fat. Find the number of ounces of fruit and number of ounces of nuts that will meet the requirement with the least number of calories. What is the least number of calories?
you want to minimize c=30f+40n subject to
0f+2n >= 4
3d+2n >= 10
2f+4n <= 18
plot the lines and evaluate c at the vertices.
Let's assume the number of ounces of fruit to be x, and the number of ounces of nuts to be y.
Given:
1 ounce of fruit supplies 0 units of protein, 3 units of carbohydrates, and 2 units of fat, with 30 calories.
1 ounce of nuts supplies 2 units of protein, 2 units of carbohydrates, and 4 units of fat, with 40 calories.
We need to find x and y such that:
- The total number of units of protein is at least 4.
- The total number of units of carbohydrates is at least 10.
- The total number of units of fat is at most 18.
Now, let's calculate the total units of protein, carbohydrates, and fat in terms of x and y:
Total units of protein = 0x + 2y
Total units of carbohydrates = 3x + 2y
Total units of fat = 2x + 4y
We need to satisfy the following conditions:
0x + 2y ≥ 4 (equation 1) [Total units of protein condition]
3x + 2y ≥ 10 (equation 2) [Total units of carbohydrates condition]
2x + 4y ≤ 18 (equation 3) [Total units of fat condition]
We can solve this system of linear inequalities to find the minimum number of calories:
To minimize the number of calories, we need to minimize 30x + 40y.
Let's solve the system of inequalities using the Simplex method:
Step 1: Convert the inequalities into equations by adding slack variables:
0x + 2y + s1 = 4
3x + 2y + s2 = 10
2x + 4y - s3 = 18
Step 2: Convert the equations into an objective function:
Z = 30x + 40y
Step 3: Create the initial tableau:
| 0 2 1 0 0 4 |
| 3 2 0 1 0 10 |
| 2 4 0 0 -1 18 |
| 30 40 0 0 0 0 |
Step 4: Use the simplex method to optimize the objective function.
- Applying the pivot operation to equation 2 (column 2, row 2):
| 1 -2/3 0 1/3 0 10/3 |
| 0 8/3 0 -1/3 0 2/3 |
| 0 4/3 0 -2/3 -1 14/3 |
| 0 10/3 0 -10/3 0 34/3 |
- Applying the pivot operation to equation 1 (column 1, row 3):
| 1 0 0 0 2/3 14/3 |
| 0 0 0 1/6 -4/3 4/3 |
| 0 1 0 -1/2 -3/4 21/4 |
| 0 0 0 -10/3 -10/3 10 |
- Applying the pivot operation to equation 4 (column 2, row 4):
| 1 0 0 0 2/3 14/3 |
| 0 0 0 1/6 -4/3 4/3 |
| 0 1 0 -1/2 -3/4 21/4 |
| 0 0 0 0 0 12 |
The tableau is in its final form. The minimum value of the objective function Z is 12, which corresponds to 14/3 ounces of fruit and 21/4 ounces of nuts.
Therefore, the least number of calories required is (30 * 14/3) + (40 * 21/4) = 420/3 + 840/4 = 140 + 210 = 350.
To find the number of ounces of fruit and nuts that will meet the requirement with the least number of calories, we can use a system of linear inequalities.
Let x represent the number of ounces of fruit and y represent the number of ounces of nuts.
The constraints are as follows:
- Protein constraint: 0 units of protein from fruit + 2 units of protein from nuts ≥ 4 units of protein
This can be written as 2y ≥ 4, which simplifies to y ≥ 2.
- Carbohydrates constraint: 3 units of carbohydrates from fruit + 2 units of carbohydrates from nuts ≥ 10 units of carbohydrates
This can be written as 3x + 2y ≥ 10.
- Fat constraint: 2 units of fat from fruit + 4 units of fat from nuts ≤ 18 units of fat
This can be written as 2x + 4y ≤ 18.
Additionally, we want to minimize the total number of calories, which can be expressed as the objective function:
Total calories = 30x + 40y
Now we have the following system of inequalities:
2y ≥ 4
3x + 2y ≥ 10
2x + 4y ≤ 18
To solve this linear programming problem, we need to graph the feasible region for these inequalities and determine the corner point that minimizes the objective function.
Let's plot the inequalities on a graph to visualize the feasible region.
Graph:
1. Plot the line y = 2 (proteins constraint).
2. Plot the line 3x + 2y = 10 (carbohydrates constraint).
3. Plot the line 2x + 4y = 18 (fat constraint).
4. Shade the region above the line y = 2.
5. Shade the region above the line 3x + 2y = 10.
6. Shade the region below the line 2x + 4y = 18.
Now, let's find the coordinates of the corner point that minimizes the objective function.
The feasible corner points for individual ounces (x, y) in this region are:
A(2, 2)
B(2, 6)
C(3, 2)
D(4.5, 1)
Now we can calculate the total calories for each feasible corner point using the objective function: Total calories = 30x + 40y.
Calories at A = 30(2) + 40(2) = 120
Calories at B = 30(2) + 40(6) = 300
Calories at C = 30(3) + 40(2) = 150
Calories at D = 30(4.5) + 40(1) = 195
So, the least number of calories is 120, which corresponds to point A (2 ounces of fruit and 2 ounces of nuts).
Therefore, the dietician should include 2 ounces of fruit and 2 ounces of nuts in the snack package, which will provide the least number of calories (120).