Chancellor Manufacturing makes two pumps for use in household reef aquariums. The Standard pump (Model A) requires 4.5 lbs of stainless steel while the Deluxe lightweight pump (Model B) 3.0 lbs of stainless steel. There are 63 lbs of stainless steel available during this production period. The combined production of both pumps must be at least 12 pumps. There are orders for at least 6 of the lightweight Model B. Lastly, management has decided that no more than 15 Model B pumps be produced. Required:
a. Write the LP formulation which provides the optimal production schedule for the standard and deluxe pumps such that the cost is minimized.
b. Write the above LP formulation in Standard Form.
c. Using your grid page, graph the above LP formulation. Be sure to label axes, constraints, and the feasible region.

The answer is below

Step-by-step explanation:

Chancellor Manufacturing makes two pumps for use in household reef aquariums. The Standard pump (Model A) requires 4.5 lbs of stainless steel while the Deluxe lightweight pump (Model B) 3.0 lbs of stainless steel. There are 63 lbs of stainless steel available during this production period. The combined production of both pumps must be at least 12 pumps. There are orders for at least 6 of the lightweight Model B. Lastly, management has decided that no more than 15 Model B pumps be produced.

The cost to produce one Standard pump is $150 and to produce one Deluxe pump is $210.

Solution:

a) Let x represent model pump A and let y represent model pump B.

Since There are 63 lbs of stainless steel available, hence:

4.5x + 3y ≤ 63

Both pumps must be at least 12 pumps. Therefore:

x + y ≤ 12

There are orders for at least 6 of the lightweight Model B:

y ≥ 6

Also,  no more than 15 Model B pumps be produced:

y ≤ 15

The cost to produce one Standard pump is $150 and to produce one Deluxe pump is $210. The cost equation is:

Minimize Cost = 150x + 210y

b)

Hence the LP formation is:

Minimize Cost = 150x + 210y

4.5x + 3y ≤ 63

x + y ≤ 12

y ≥ 6

y ≤ 15

x ≥ 0.

c) The LP problem was solved using geogebra online graphing tool.

The points that satisfy the problem are:

(0, 6), (0,12), (6, 6)

At (0,6); cost = 150(0) + 210(6) = 1260

At (0,6); cost = 150(0) + 210(12) = 2520

At (6,6); cost = 150(6) + 210(6) = 2160

Hence the minimum cost is at (0, 6)