|
Question 1165321: Bob, a nutritionist who works for the University Medical Center, has been asked to prepare special diets for two patients, Susan and Tom. Bob has decided that Susan's meals should contain at least 330 mg of calcium, 18 mg of iron, and 44 mg of vitamin C, whereas Tom's meals should contain at least 280 mg of calcium, 13 mg of iron, and 34 mg of vitamin C. Bob has also decided that the meals are to be prepared from three basic foods: Food A, Food B, and Food C. The special nutritional contents of these foods are summarized in the accompanying table. Find how many ounces of each type of food should be used in a meal so that the minimum requirements of calcium, iron, and vitamin C are met for each patient's meals.
Contents (mg/oz)
Calcium Iron Vitamin C
Food A 30 1 2
Food B 25 1 5
Food C 20 2 4
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! This is a system of linear inequalities problem, but since the problem asks to "find how many ounces of each type of food should be used... so that the minimum requirements... are met," without an objective function (like minimizing cost), we are looking for the **feasible region** that satisfies all constraints for both patients simultaneously.
Let:
* $A$ = ounces of **Food A** used in the meal.
* $B$ = ounces of **Food B** used in the meal.
* $C$ = ounces of **Food C** used in the meal.
We need to set up the nutritional constraints for Susan and Tom.
### 1. Set Up Constraints
The contents table summarizes the nutrients per ounce:
| Food | Calcium (mg/oz) | Iron (mg/oz) | Vitamin C (mg/oz) |
| :---: | :---: | :---: | :---: |
| A | 30 | 1 | 2 |
| B | 25 | 1 | 5 |
| C | 20 | 2 | 4 |
#### A. Susan's Constraints
| Nutrient | Requirement | Inequality |
| :---: | :---: | :---: |
| **Calcium** | $\ge 330$ mg | $30A + 25B + 20C \ge 330$ |
| **Iron** | $\ge 18$ mg | $1A + 1B + 2C \ge 18$ |
| **Vitamin C** | $\ge 44$ mg | $2A + 5B + 4C \ge 44$ |
#### B. Tom's Constraints
| Nutrient | Requirement | Inequality |
| :---: | :---: | :---: |
| **Calcium** | $\ge 280$ mg | $30A + 25B + 20C \ge 280$ |
| **Iron** | $\ge 13$ mg | $1A + 1B + 2C \ge 13$ |
| **Vitamin C** | $\ge 34$ mg | $2A + 5B + 4C \ge 34$ |
#### C. Non-Negativity Constraints
$$A \ge 0, \quad B \ge 0, \quad C \ge 0$$
### 2. Identify the Overall Feasible Region
Since the meal must meet **both** patients' requirements, the overall system is the *union* of the more restrictive constraints from Susan and Tom.
For example, if Susan needs $\ge 330$ mg of Calcium and Tom needs $\ge 280$ mg, the final meal must satisfy the more restrictive condition: $\ge 330$ mg.
| Nutrient | Susan's Minimum | Tom's Minimum | **Governing Constraint** |
| :---: | :---: | :---: | :---: |
| **Calcium** | $330$ | $280$ | $30A + 25B + 20C \ge 330$ |
| **Iron** | $18$ | $13$ | $A + B + 2C \ge 18$ |
| **Vitamin C** | $44$ | $34$ | $2A + 5B + 4C \ge 44$ |
The meal must satisfy the following system of linear inequalities:
1. $$30A + 25B + 20C \ge 330$$
2. $$A + B + 2C \ge 18$$
3. $$2A + 5B + 4C \ge 44$$
4. $$A, B, C \ge 0$$
### 3. Finding a Specific Solution (Example)
Since no objective function (like cost) was given to minimize, the solution is any set of non-negative values for $(A, B, C)$ that satisfies all three governing inequalities. The feasible region is a three-dimensional region (a polyhedron).
To provide a concrete answer, we can look for corner points (where the inequalities become equalities), as these often represent efficient solutions in optimization problems. However, without minimization, we'll look for a simple integer solution that works.
Let's test $A=5, B=5, C=5$:
1. $30(5) + 25(5) + 20(5) = 150 + 125 + 100 = 375 \ge 330$. (OK)
2. $5 + 5 + 2(5) = 5 + 5 + 10 = 20 \ge 18$. (OK)
3. $2(5) + 5(5) + 4(5) = 10 + 25 + 20 = 55 \ge 44$. (OK)
The meal mixture of **5 ounces of Food A, 5 ounces of Food B, and 5 ounces of Food C** meets both patients' minimum requirements.
**Final Answer:**
The feasible region is defined by the following system of inequalities:
$$\begin{cases} 30A + 25B + 20C \ge 330 \\ A + B + 2C \ge 18 \\ 2A + 5B + 4C \ge 44 \\ A \ge 0, B \ge 0, C \ge 0 \end{cases}$$
One possible solution is to use:
$$\mathbf{A = 5 \text{ ounces}, B = 5 \text{ ounces}, C = 5 \text{ ounces}}$$
|
|
|
| |
