A veterinarian has been asked to prepare a diet for a group of dogs to be used in a nutrition study at the School of Animal Science. It has been stipulated that each serving should be no larger than 8 oz and must contain at least 29 units of Nutrient I and 20 units of Nutrient II. The vet has decided that the diet may be prepared from two brands of dog food: Brand A and Brand B. Each ounce of Brand A contains 3 units of Nutrient I and 4 units of Nutrient II. Each ounce of Brand B contains 5 units of Nutrient I and 2 units of Nutrient II. Brand A costs 4 cents/oz and Brand B costs 5 cents/oz. Determine how many ounces of each brand of dog food should be used per serving to meet the given requirements at a minimum cost.
Brand A
oz
Brand B
oz

What is the minimum cost per serving?
cents

To solve this problem, we’ll set up a linear programming model.

### Define the variables:
Let:
- \( x \) = ounces of Brand A used per serving.
- \( y \) = ounces of Brand B used per serving.

### Objective:
Minimize the cost of the serving. The cost function is:
\[
\text{Cost} = 4x + 5y
\]

### Constraints:
1. Each serving should be no larger than 8 oz:
   \[
   x + y \leq 8
   \]

2. The serving must contain at least 29 units of Nutrient I:
   \[
   3x + 5y \geq 29
   \]

3. The serving must contain at least 20 units of Nutrient II:
   \[
   4x + 2y \geq 20
   \]

4. Non-negativity constraints:
   \[
   x \geq 0, \quad y \geq 0
   \]

### Solving the System:

To find the optimal values of \( x \) and \( y \) that satisfy all these constraints and minimize the cost, we can use graphical methods or linear programming tools. For simplicity here, I'll outline the solution:

### Steps:

1. **Identify the feasible region** defined by the constraints.
2. **Determine the corner points** of the feasible region (intersection points of the constraints).
3. **Evaluate the cost function** at each corner point to find the minimum cost.

Let's go through the steps.

### Step 1: Solve the inequalities for intersections

#### Intersection of \( x + y = 8 \) and \( 3x + 5y = 29 \):

Solving these two equations:
1. \( x + y = 8 \Rightarrow y = 8 - x \).
2. Substitute into \( 3x + 5y = 29 \):
   \[
   3x + 5(8 - x) = 29
   \]
   \[
   3x + 40 - 5x = 29
   \]
   \[
   -2x = -11 \Rightarrow x = 5.5
   \]
   \[
   y = 8 - 5.5 = 2.5
   \]
   So, one intersection point is \( (5.5, 2.5) \).

#### Intersection of \( x + y = 8 \) and \( 4x + 2y = 20 \):

1. \( x + y = 8 \Rightarrow y = 8 - x \).
2. Substitute into \( 4x + 2y = 20 \):
   \[
   4x + 2(8 - x) = 20
   \]
   \[
   4x + 16 - 2x = 20
   \]
   \[
   2x = 4 \Rightarrow x = 2
   \]
   \[
   y = 8 - 2 = 6
   \]
   So, another intersection point is \( (2, 6) \).

#### Intersection of \( 3x + 5y = 29 \) and \( 4x + 2y = 20 \):

1. Solve these equations by elimination or substitution:
   - Multiply the second equation by 2 to align \( y \)-terms:
     \[
     3x + 5y = 29
     \]
     \[
     8x + 4y = 40
     \]
   - Solve by elimination:
     \[
     4x = 11 \Rightarrow x = 2.75
     \]
   \[


A veterinarian has been asked to prepare a diet for a group of dogs to be used in a nutrition study at the School of Animal Science. It has been stipulated that each serving should be no larger than 8 oz and must contain at least 29 units of Nutrient I and 20 units of Nutrient II. The vet has decided that the diet may be prepared from two brands of dog food: Brand A and Brand B. Each ounce of Brand A contains 3 units of Nutrient I and 4 units of Nutrient II. Each ounce of Brand B contains 5 units of Nutrient I and 2 units of Nutrient II. Brand A costs 4 cents/oz and Brand B costs 5 cents/oz. Determine how many ounces of each brand of dog food should be used per serving to meet the given requirements at a minimum cost.

Determine the optimal amounts of Brand A and Brand B dog foods to minimize cost while meeting nutritional requirements for a dog diet. Learn about linear programming, constraints in optimization, and cost calculations in this advanced problem for Grades 11-12 and college-level students.

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