To solve this linear programming problem by the method of corners, we need to follow these steps:

1. Identify the inequalities and plot them.
2. Find the feasible region where all inequalities intersect, considering the constraints $x \geq 0$ and $y \geq 0$.
3. Determine the corner points (vertices) of the feasible region.
4. Evaluate the objective function $P = 5x + 2y$ at each corner point to find the maximum and minimum values.

Step 1: Write down the inequalities

The inequalities given are:

1. $3x + 5y \geq 20$ (or equivalently $5y \geq 20 - 3x$)
2. $3x + y \leq 16$
3. $-2x + y \leq 3$
4. $x \geq 0$
5. $y \geq 0$

Step 2: Find the intersection points (corners of the feasible region)

To find the intersection points, we solve pairs of equations:

1. Intersection of $3x + 5y = 20$ and $3x + y = 16$:

$3x + 5y = 20$ $3x + y = 16$ Subtracting the second equation from the first: $4y = 4 \Rightarrow y = 1$ Substitute $y = 1$ into $3x + y = 16$: $3x + 1 = 16 \Rightarrow 3x = 15 \Rightarrow x = 5$ So, one corner point is $(5, 1)$.

2. Intersection of $3x + 5y = 20$ and $-2x + y = 3$:

$3x + 5y = 20$ $-2x + y = 3 \Rightarrow y = 3 + 2x$ Substitute $y = 3 + 2x$ into $3x + 5y = 20$: $3x + 5(3 + 2x) = 20$ $3x + 15 + 10x = 20$ $13x = 5 \Rightarrow x = \frac{5}{13}$ Substitute $x = \frac{5}{13}$ into $y = 3 + 2x$: $y = 3 + 2\left(\frac{5}{13}\right) = 3 + \frac{10}{13} = \frac{49}{13}$ So, another corner point is $\left(\frac{5}{13}, \frac{49}{13}\right)$.

3. Intersection of $3x + y = 16$ and $-2x + y = 3$:

$3x + y = 16$ $-2x + y = 3$ Subtract the second equation from the first: $5x = 13 \Rightarrow x = \frac{13}{5}$ Substitute $x = \frac{13}{5}$ into $3x + y = 16$: $3\left(\frac{13}{5}\right) + y = 16$ $\frac{39}{5} + y = 16 \Rightarrow y = 16 - \frac{39}{5} = \frac{41}{5}$ So, another corner point is $\left(\frac{13}{5}, \frac{41}{5}\right)$.

Step 3: Evaluate $P = 5x + 2y$ at each corner point

Let's calculate $P$ at each corner point:

1. At $(5, 1)$: $P = 5(5) + 2(1) = 25 + 2 = 27$

2. At $\left(\frac{5}{13}, \frac{49}{13}\right)$: $P = 5\left(\frac{5}{13}\right) + 2\left(\frac{49}{13}\right) = \frac{25}{13} + \frac{98}{13} = \frac{123}{13} \approx 9.46$

3. At $\left(\frac{13}{5}, \frac{41}{5}\right)$: $P = 5\left(\frac{13}{5}\right) + 2\left(\frac{41}{5}\right) = 13 + \frac{82}{5} = 13 + 16.4 = 29.4$

Step 4: Conclusion

The minimum and maximum values of $P$ are:

• Minimum $P \approx 9.46$ at $\left(\frac{5}{13}, \frac{49}{13}\right)$.
• Maximum $P = 29.4$ at $\left(\frac{13}{5}, \frac{41}{5}\right)$.

Would you like a further breakdown of any step?

Related Questions

1. How can the graphical method of linear programming help visualize constraints?
2. What are the practical applications of the corner-point method?
3. How would the solution change if we wanted to maximize a different objective function?
4. What is the role of slack variables in linear programming?
5. How can we interpret dual variables in this problem?

Tip: Always re-check the feasible region to ensure all inequalities are correctly considered before solving.