To determine the optimal daily quantities of Nutri Chomp and Healthy Bites that the store should sell to maximize revenue while meeting the protein and fat constraints, we need to approach this problem using linear programming.

Step-by-Step Process:

1. Define the Variables:

• Let $x$ be the number of pounds of Nutri Chomp sold.
• Let $y$ be the number of pounds of Healthy Bites sold.

2. Formulate the Objective Function (Revenue):

The revenue the store generates is: $R = 12x + 16y$ where $12 is the price per pound of Nutri Chomp and $16 is the price per pound of Healthy Bites. The goal is to maximize this revenue function.

3. Set Up the Constraints:

We are given the following nutritional constraints:

• Protein Constraint: Nutri Chomp contains 10% protein, and Healthy Bites contains 5% protein. The blend must contain at least 8% protein. This gives the inequality: $0.10x + 0.05y \geq 0.08(x + y)$ Simplifying this inequality: $0.02x \geq 0.03y$ or equivalently: $2x \geq 3y$ which can also be written as: $x \geq \frac{3}{2}y$

• Fat Constraint: Nutri Chomp contains 5% fat, and Healthy Bites contains 10% fat. The blend must contain at least 9% fat. This gives the inequality: $0.05x + 0.10y \geq 0.09(x + y)$ Simplifying this inequality: $0.01y \geq 0.04x$ or equivalently: $y \geq 4x$

• Non-negativity Constraints: The pounds of both dog foods sold cannot be negative: $x \geq 0 \quad \text{and} \quad y \geq 0$

4. Graphical Solution:

To solve this graphically, plot the two inequalities $x \geq \frac{3}{2}y$ and $y \geq 4x$ in the first quadrant, where $x \geq 0$ and $y \geq 0$.

• Plotting $x = \frac{3}{2}y$: This line has a slope of $\frac{3}{2}$ and passes through the origin.

• Plotting $y = 4x$: This line has a slope of 4 and passes through the origin.

The feasible region will be the area bounded by these two lines and the axes. The points where the lines intersect the axes are critical.

5. Determine the Feasible Region:

The feasible region will be the area that satisfies both inequalities. Graphically, it will be a triangular region, and the vertices of this region will represent potential candidates for the optimal solution.

6. Identify the Vertices:

The vertices of the feasible region can be found by solving the system of equations formed by the two constraints:

1. $x = \frac{3}{2}y$
2. $y = 4x$

Solve these two equations simultaneously to find the intersection point of the lines.

Substituting $y = 4x$ into $x = \frac{3}{2}y$: $x = \frac{3}{2}(4x)$ $x = 6x$ Since this leads to a contradiction, it means the lines do not intersect in the first quadrant. Therefore, the feasible region is bounded by the axes and the two lines.

The vertices are:

1. $(0, 0)$
2. Intersection of $y = 4x$ with the $y$-axis: $(0, 0)$
3. Intersection of $x = \frac{3}{2}y$ with the $y$-axis: (