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 unitss of fat, and will contain 20 calories. Each ounce of nuts will supply 2 units of protein, 11 unitnothing of carbohydratenothing, and 4 units of fat, and will contain 40 calories. Every package must provide at least 4 units of protein, at least 17 units of carbohydrates, and no more than 20 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?
20X+40Y=Z
What are you measuring?
assign variables for the quantities of interest.
What are the constraints?
This doesn’t give me the answer but ok🤷🏽♀️
To find the number of ounces of fruit and nuts that will meet the requirements with the least number of calories, we need to set up a linear programming problem.
Let X be the number of ounces of fruit and Y be the number of ounces of nuts.
Our objective is to minimize the number of calories, so our objective function is Z = 20X + 40Y.
We have the following constraints:
- Protein constraint: 0X + 2Y ≥ 4 (at least 4 units of protein)
- Carbohydrate constraint: 3X + 0Y ≥ 17 (at least 17 units of carbohydrates)
- Fat constraint: 2X + 4Y ≤ 20 (no more than 20 units of fat)
- Non-negativity constraint: X ≥ 0, Y ≥ 0 (number of ounces cannot be negative)
Let's solve this linear programming problem to find the least number of calories.
First, let's plot the constraints on a graph to visualize the feasible region.
Protein constraint: 0X + 2Y ≥ 4
This constraint represents the region above the line 2Y = 4, which simplifies to Y ≥ 2.
Carbohydrate constraint: 3X + 0Y ≥ 17
This constraint represents the region to the right of the line 3X = 17, which simplifies to X ≥ 17/3.
Fat constraint: 2X + 4Y ≤ 20
This constraint represents the region below the line 2X + 4Y = 20, which simplifies to X + 2Y ≤ 10.
Based on the feasible region formed by these constraints, we can evaluate the objective function at the corner points of the feasible region to find the least number of calories.
The corner points are:
1. (17/3, 2)
2. (20, 0)
3. (10, 0)
4. (10, 5)
Now, we substitute these corner points into the objective function Z = 20X + 40Y and find the corresponding value of Z.
1. Z = 20(17/3) + 40(2) ≈ 340
2. Z = 20(20) + 40(0) = 400
3. Z = 20(10) + 40(0) = 200
4. Z = 20(10) + 40(5) = 600
Among these values, the least number of calories is 200, which occurs at the point (10, 0).
Therefore, the least number of calories is 200, and it can be achieved by using 10 ounces of fruit and 0 ounces of nuts.
To find the number of ounces of fruit and nuts that will meet the requirement with the least number of calories, we need to set up a system of linear inequalities.
Let's use variables X and Y to represent the number of ounces of fruit and nuts, respectively.
Given the requirements:
1. Protein: We need at least 4 units of protein. Each ounce of fruit provides zero units, while each ounce of nuts provides 2 units. So, the protein requirement can be expressed as: 0X + 2Y ≥ 4.
2. Carbohydrates: We need at least 17 units of carbohydrates. Each ounce of fruit provides 3 units, while each ounce of nuts provides zero units. So, the carbohydrate requirement can be expressed as: 3X + 0Y ≥ 17.
3. Fat: We need no more than 20 units of fat. Each ounce of fruit provides 2 units, while each ounce of nuts provides 4 units. So, the fat requirement can be expressed as: 2X + 4Y ≤ 20.
4. Non-negativity: Both X and Y should be non-negative, meaning they cannot be less than zero.
Now, we need to minimize the number of calories, given that each ounce of fruit has 20 calories, and each ounce of nuts has 40 calories.
We can express the total number of calories as: 20X + 40Y = Z, where Z represents the total number of calories.
To solve this optimization problem, we need to graph the feasible region determined by the inequalities and find the point at which the least number of calories is achieved.
The graphing and optimization process can be performed using linear programming techniques, which involve solving systems of linear inequalities.