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: $4x = 11 \Rightarrow x = 2.75$ [