To solve this problem, we need to analyze the constraints and identify the feasible region for the problem. Let's break it down step by step.

Problem Breakdown:

1. Objective:

   • Maximize profit from selling bread and cake.
   • Bread sells for $2, cake for $3.

2. Constraints:

   • Time: 5 minutes per bread, 4 minutes per cake.
   • Dough: 0.15 pounds per bread, 0.6 pounds per cake.
   • Available time: $6 \frac{2}{3}$ hours = 400 minutes.
   • Available dough: 43.2 pounds.

3. Inequalities:

   • Time: $5x + 4y \leq 400$
   • Dough: $0.15x + 0.6y \leq 43.2$
   • Non-negativity: $x \geq 0, y \geq 0$

Step 1: Analyze the Inequalities

1. Time Constraint: $5x + 4y \leq 400$

2. Dough Constraint: $0.15x + 0.6y \leq 43.2$ This can be simplified by multiplying the entire inequality by 10: $1.5x + 6y \leq 432$

Step 2: Find the Feasible Region

We need to find the region bounded by these inequalities, considering that $x \geq 0$ and $y \geq 0$.

1. Time Constraint:

   • Intercept at $x$-axis: $x = \frac{400}{5} = 80$
   • Intercept at $y$-axis: $y = \frac{400}{4} = 100$

2. Dough Constraint:

   • Intercept at $x$-axis: $x = \frac{432}{1.5} = 288$
   • Intercept at $y$-axis: $y = \frac{432}{6} = 72$

3. Feasible Region: The feasible region is bounded by the above constraints.

Step 3: Identify the Corners of the Feasible Region

The corners of the feasible region are where the boundary lines intersect.

1. Intersection of Time Constraint with $x$- and $y$-axes:

   • (0, 100)
   • (80, 0)

2. Intersection of Dough Constraint with $x$- and $y$-axes:

   • (0, 72)
   • (288, 0) (but this point is beyond $x = 80$ so not part of feasible region)

3. Intersection of the Two Constraints: To find the intersection of $5x + 4y = 400$ and $1.5x + 6y = 432$: Solve simultaneously: $\text{Multiply the first equation by 1.5: } 7.5x + 6y = 600$ Subtract the second equation: $7.5x + 6y - (1.5x + 6y) = 600 - 432 \Rightarrow 6x = 168 \Rightarrow x = 28$ Substitute $x = 28$ in the first equation: $5(28) + 4y = 400 \Rightarrow 140 + 4y = 400 \Rightarrow 4y = 260 \Rightarrow y = 65$ Intersection point: (28, 65)

Step 4: Evaluate the Given Options

Given the potential options, let's evaluate:

1. (0, 0)
2. (0, 100)
3. (0, 72)
4. (28, 65)
5. (80, 0)

From these, the corners of the feasible region should include points that satisfy all constraints.

Conclusion:

The correct set of corners for the feasible region that satisfies all constraints and is within bounds are:

• (0, 0)
• (0, 72)
• (28, 65)
• (80, 0)

So, the correct answer is:

(0, 0), (0, 72), (28, 65), and (80, 0)

This corresponds to the third option:

(0, 0), (0, 72), (28, 65), and (80, 0)

Would you like more details or have any questions?

Here are 5 related questions you might be interested in:

1. How do you graphically represent the feasible region for a linear programming problem?
2. What is the significance of the corners of the feasible region in linear programming?
3. How can you determine the optimal solution in a linear programming problem?
4. How does changing one of the constraints affect the feasible region?
5. What happens if two constraints in a linear programming problem are parallel?

Tip: Always verify the intersection points of constraints, as these are crucial in identifying the feasible region in linear programming.