Answer:
a) Minimize [tex]Cost=90x_1+120x_2[/tex]
subject to
[tex]0.2x_1+0.3x_2\geq8[/tex]
[tex]0.2x_1+0.25x_2\geq6[/tex]
[tex]0.15x_1+0.1x_2\geq5[/tex]
[tex]x_1\geq0\\x_2\geq0[/tex]
b) Attached
c) The optimum value that minimizes cost is x1=28 and x2=8.
Step-by-step explanation:
The objective function is the cost of extraction and needs to be minimized.
The cost of extraction is the sum of the cost of extraction of ore type 1 and the cost of extraction of ore type 2:
[tex]Cost=90x_1+120x_2[/tex]
Being x1 the tons of ore type 1 extracted and x2 the tons of ore type 2.
The constraints are the amount of minerals that need to be in the final mix
Copper:
[tex]0.2x_1+0.3x_2\geq8[/tex]
Zinc
[tex]0.2x_1+0.25x_2\geq6[/tex]
Magnesium
[tex]0.15x_1+0.1x_2\geq5[/tex]
Of course, x1 and x2 has to be positive numbers.
[tex]x_1\geq0\\x_2\geq0[/tex]
The feasible region can be seen in the attached graph.
The orange line is the magnesium constraint. The red line is the copper constraint. The green line is the zinc constraint.
The optimal solution is found in one of the intersection points between two constraints that belong to the limits of the feasible region.
In this case, the cost can be calculated for the 3 points that satisfies the conditions.
The optimum value that minimizes cost is x1=28 and x2=8.
