To solve this problem, we need to formulate it as a linear programming problem. Let's denote:

• $x$ as the number of ounces of chicken.
• $y$ as the number of ounces of grain.

Objective: Minimize the cost $C$, which is given by:

$C = 9x + y$

Constraints:

1. Protein Constraint: The total protein content must be at least 230 grams. $10x + 2y \geq 230$

2. Fat Constraint: The total fat content must be at least 180 grams. $5x + 2y \geq 180$

3. Non-negativity Constraints: $x \geq 0, \quad y \geq 0$

Step 1: Convert Inequalities to Equalities (Finding the Feasible Region)

For each constraint, we'll convert the inequalities to equalities to find the boundary lines.

1. $10x + 2y = 230$
   Divide by 2: $5x + y = 115$ So, the line equation is $y = 115 - 5x$.

2. $5x + 2y = 180$
   So, the line equation is $y = 90 - \frac{5}{2}x$.

Step 2: Identify the Intersection Points (Vertices of the Feasible Region)

We find the intersection points of the boundary lines and the axes.

1. Intersection of $y = 115 - 5x$ and $y = 90 - \frac{5}{2}x$:

   Set $115 - 5x = 90 - \frac{5}{2}x$: $115 - 90 = 5x - \frac{5}{2}x$ $25 = \frac{10x - 5x}{2}$ $25 = \frac{5x}{2}$ $x = 10$ Substitute $x = 10$ into $y = 115 - 5(10)$: $y = 115 - 50 = 65$ So, the intersection point is $(10, 65)$.

2. Intersection with the y-axis ($x = 0$):

   From $y = 115 - 5x$: $y = 115 \quad \text{(Intersection: (0, 115))}$

   From $y = 90 - \frac{5}{2}x$: $y = 90 \quad \text{(Intersection: (0, 90))}$

3. Intersection with the x-axis ($y = 0$):

   From $y = 115 - 5x = 0$: $x = 23 \quad \text{(Intersection: (23, 0))}$

   From $y = 90 - \frac{5}{2}x = 0$: $x = 36 \quad \text{(Intersection: (36, 0))}$

Step 3: Evaluate the Objective Function at the Vertices

Now we evaluate the cost function $C = 9x + y$ at the vertices found:

1. $C(10, 65) = 9(10) + 65 = 90 + 65 = 155$
2. $C(0, 115) = 9(0) + 115 = 115$
3. $C(36, 0) = 9(36) + 0 = 324$
4. $C(23, 0) = 9(23) + 0 = 207$

Step 4: Identify the Minimum Cost

The minimum cost occurs at $C = 115$ when $x = 0$ and $y = 115$.

Answer:

• Grain: $y = 115$ ounces
• Chicken: $x = 0$ ounces

This is the optimal solution to minimize the cost.

Would you like any further details, or do you have any questions?

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Relative Questions:

1. How would the cost change if the price of chicken increased to 12¢ per ounce?
2. What if the minimum fat requirement increased to 200 grams? How would the solution change?
3. How would adding a minimum calorie requirement affect the problem?
4. Can you solve this problem graphically to confirm the algebraic solution?
5. What is the impact on the cost if the protein content in grain increases by 1 gram per ounce?

Tip: When solving linear programming problems, always verify the feasible region before optimizing the objective function.