To solve this linear programming problem using the method of corners, we need to:

1. Graph the inequalities to find the feasible region.
2. Determine the corner points (vertices) of the feasible region.
3. Evaluate the objective function $C = 3x + 5y$ at each corner point to find the minimum.

Step 1: Convert Inequalities to Equations

The constraints are:

1. $4x + y \geq 40$
2. $2x + y \geq 30$
3. $x + 3y \geq 30$
4. $x \geq 0$
5. $y \geq 0$

We'll turn these inequalities into equalities to find where they intersect.

Step 2: Find Points of Intersection (Corner Points)

1. Intersection of $4x + y = 40$ and $2x + y = 30$:

   • Subtract the second equation from the first: $(4x + y) - (2x + y) = 40 - 30$ $2x = 10 \Rightarrow x = 5$
   • Substitute $x = 5$ into $2x + y = 30$: $2(5) + y = 30$ $10 + y = 30 \Rightarrow y = 20$
   • Point: $(5, 20)$

2. Intersection of $4x + y = 40$ and $x + 3y = 30$:

   • Solve $4x + y = 40$ for $y$: $y = 40 - 4x$
   • Substitute into $x + 3y = 30$: $x + 3(40 - 4x) = 30$ $x + 120 - 12x = 30$ $-11x = -90 \Rightarrow x = \frac{90}{11} \approx 8.18$
   • Substitute $x = \frac{90}{11}$ back into $y = 40 - 4x$: $y = 40 - 4 \left(\frac{90}{11}\right)$ Simplifying gives $y \approx 4.55$.
   • Point: $\left(\frac{90}{11}, \frac{50}{11}\right)$.

3. Intersection of $2x + y = 30$ and $x + 3y = 30$:

   • Solve $2x + y = 30$ for $y$: $y = 30 - 2x$
   • Substitute into $x + 3y = 30$: $x + 3(30 - 2x) = 30$ $x + 90 - 6x = 30$ $-5x = -60 \Rightarrow x = 12$
   • Substitute $x = 12$ back into $y = 30 - 2x$: $y = 30 - 2(12) = 6$
   • Point: $(12, 6)$

Step 3: Evaluate the Objective Function $C = 3x + 5y$ at Each Corner Point

1. At $(5, 20)$: $C = 3(5) + 5(20) = 15 + 100 = 115$

2. At $\left(\frac{90}{11}, \frac{50}{11}\right)$: $C = 3\left(\frac{90}{11}\right) + 5\left(\frac{50}{11}\right) = \frac{270}{11} + \frac{250}{11} = \frac{520}{11} \approx 47.27$

3. At $(12, 6)$: $C = 3(12) + 5(6) = 36 + 30 = 66$

Step 4: Determine the Minimum Value

The minimum value of $C$ occurs at $\left(\frac{90}{11}, \frac{50}{11}\right)$ with $C \approx 47.27$.

Solution

The minimum value is: $C = \frac{520}{11} \approx 47.27$ at $(x, y) = \left(\frac{90}{11}, \frac{50}{11}\right)$.

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Relative Questions

1. What are the main properties of the method of corners?
2. How would this solution change if we were maximizing $C$ instead?
3. What happens if additional constraints were added to the problem?
4. Can we use the graphical method for more than two variables?
5. Why is it essential to evaluate only the vertices in linear programming?

Tip

When using the method of corners, always ensure that you check every feasible corner point, as the minimum or maximum value for linear programming problems will occur at one of these vertices.