Question

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Explanation:

Part (a)

Step 1: Find the vertices of the feasible region

The constraints are \( x + 2y \leq 10 \), \( 3x + y \leq 15 \), \( x \geq 0 \), \( y \geq 0 \).

• Intersection of \( x = 0 \) and \( y = 0 \): \( (0,0) \)
• Intersection of \( x = 0 \) and \( x + 2y = 10 \): \( y = 5 \), so \( (0,5) \)
• Intersection of \( y = 0 \) and \( 3x + y = 15 \): \( x = 5 \), so \( (5,0) \)
• Intersection of \( x + 2y = 10 \) and \( 3x + y = 15 \):

Solve the system:
From \( 3x + y = 15 \), we get \( y = 15 - 3x \).
Substitute into \( x + 2y = 10 \):
\( x + 2(15 - 3x) = 10 \)
\( x + 30 - 6x = 10 \)
\( -5x = -20 \)
\( x = 4 \)
Then \( y = 15 - 3(4) = 3 \), so \( (4,3) \)

Step 2: Evaluate \( C = 6x + 4y \) at each vertex

• At \( (0,0) \): \( C = 6(0) + 4(0) = 0 \)
• At \( (0,5) \): \( C = 6(0) + 4(5) = 20 \)
• At \( (5,0) \): \( C = 6(5) + 4(0) = 30 \)
• At \( (4,3) \): \( C = 6(4) + 4(3) = 24 + 12 = 36 \)

Step 1: Find the vertices of the feasible region

The constraints are \( 4x + 3y \geq 24 \), \( 4x + y \leq 16 \), \( x \geq 0 \), \( y \geq 0 \).

• Intersection of \( 4x + 3y = 24 \) and \( 4x + y = 16 \):

Subtract the second equation from the first: \( 2y = 8 \), so \( y = 4 \)
Substitute \( y = 4 \) into \( 4x + y = 16 \): \( 4x + 4 = 16 \), \( 4x = 12 \), \( x = 3 \), so \( (3,4) \)

• Intersection of \( 4x + 3y = 24 \) and \( x = 0 \): \( y = 8 \), so \( (0,8) \)
• Intersection of \( 4x + y = 16 \) and \( y = 0 \): \( x = 4 \), so \( (4,0) \)

Step 2: Evaluate \( C = 6x + 3y \) at each vertex

• At \( (3,4) \): \( C = 6(3) + 3(4) = 18 + 12 = 30 \)
• At \( (0,8) \): \( C = 6(0) + 3(8) = 24 \)
• At \( (4,0) \): \( C = 6(4) + 3(0) = 24 \)

Answer:

The maximum value of \( C \) is \( 36 \) when \( x = 4 \) and \( y = 3 \).

Part (b)