Answer: To find the number of each model that should be manufactured weekly for maximum profit, and the maximum profit itself, we can set up a linear programming problem.
Let's use the following variables:
x = number of standard models manufactured weekly
y = number of deluxe models manufactured weekly
Objective function (to maximize profit):
Z = 1000x + 2500y
Subject to the following constraints:
Manufacturing time constraint: 4x + 8y ≤ 72 (hours available per week)
Minimum production constraint: x + y ≥ 10 (at least 10 dishwashers must be manufactured weekly)
Non-negativity constraint: x ≥ 0, y ≥ 0 (cannot manufacture a negative number of dishwashers)
Now, we can graph these constraints and find the feasible region:
Manufacturing time constraint: 4x + 8y ≤ 72
Divide both sides by 4: x + 2y ≤ 18
Plot the line x + 2y = 18 and shade the region below it.
Minimum production constraint: x + y ≥ 10
Plot the line x + y = 10 and shade the region above it.
Non-negativity constraint: x ≥ 0, y ≥ 0
The region where x and y are both non-negative.
Now, we find the points where the two lines intersect in the feasible region. The coordinates of these points will be the corner points of the feasible region. Evaluate the objective function (Z) at these corner points to find the maximum profit.
Let's find the corner points:
Intersection of x + 2y = 18 and x + y = 10
Subtract the equations to get: y = 8
Plug y = 8 into x + y = 10: x + 8 = 10
x = 2
Intersection of x + y = 10 and y = 0
y = 0
x = 10
Intersection of x + 2y = 18 and y = 0
y = 0
x = 18
Now, evaluate the objective function (Z) at each corner point:
Z = 1000x + 2500y = 1000(2) + 2500(8) = 2000 + 20000 = 22000
Z = 1000x + 2500y = 1000(10) + 2500(0) = 10000 + 0 = 10000
Z = 1000x + 2500y = 1000(18) + 2500(0) = 18000 + 0 = 18000
The maximum profit (Z) is N$22,000, and it occurs when 2 standard models and 8 deluxe models are manufactured weekly.