Respuesta :
Answer:
100 soccer balls and 50 cork balls must be produced
Step-by-step explanation:
Let the number of soccer ball = p and number of cork ball = q
Production hours of p = 3 hours
Production hours of q = 4 hours
Then the production time equation would be:
3p + 4q ≤500
The number of the balls must be at least 150
p + q ≥150
x ≥ 0 y ≥ 0
Minimum production r = 9p + 7q
Therefore 100 soccer balls and 50 cork balls must be produced

The linear programming model is:
- Minimize [tex]\mathbf{Z = 9x + 7y}[/tex]
- Subject to [tex]\mathbf{3x + 4y \le500}[/tex], [tex]\mathbf{x + y \ge 150}[/tex] , [tex]\mathbf{x,y \ge 0}[/tex]
- Sporting Equipment, Inc should produce at least 100 soccer balls and at most 50 co rk balls.
Represent soccer balls with x and co rk balls with y.
The given parameters are:
Production hours:
[tex]\mathbf{x=3\ hours}[/tex]
[tex]\mathbf{y=4\ hours}[/tex]
[tex]\mathbf{Available = 500\ hours}[/tex]
So, we have:
[tex]\mathbf{3x + 4y \le500}[/tex]
Number of balls
[tex]\mathbf{Minimum = 150}[/tex]
So, we have:
[tex]\mathbf{x + y \ge 150}[/tex]
The cost is given as: $9 per soccer ball, and $7 per co rk ball.
So, the objective function is:
[tex]\mathbf{Z = 9x + 7y}[/tex]
Hence, the linear programming model is:
Minimize [tex]\mathbf{Z = 90x + 75y}[/tex]
Subject to
[tex]\mathbf{3x + 4y \le500}[/tex]
[tex]\mathbf{x + y \ge 150}[/tex]
[tex]\mathbf{x,y \ge 0}[/tex]
Express [tex]\mathbf{x + y \ge 150}[/tex] as an equation
[tex]\mathbf{x = 150 - y}[/tex]
Substitute [tex]\mathbf{x = 150 - y}[/tex] in [tex]\mathbf{3x + 4y \le500}[/tex]
[tex]\mathbf{3(150 - y) + 4y \le500}[/tex]
[tex]\mathbf{450 - 3y + 4y \le500}[/tex]
Subtract 450 from both sides
[tex]\mathbf{y \le50}[/tex]
Substitute 50 for y in [tex]\mathbf{x + y \ge 150}[/tex]
[tex]\mathbf{x + 50 \ge 150}[/tex]
Subtract 50 from both sides
[tex]\mathbf{x \ge 100}[/tex]
Hence, Sporting Equipment, Inc should produce at least 100 soccer balls and at most 50 co rk balls.
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