SOLUTION: A company manufactures two types of​ washers, top load and front load. The company can manufacture a maximum of 18 washers per day. It makes a profit of​ $20 on top load machin

Algebra ->  Graphs -> SOLUTION: A company manufactures two types of​ washers, top load and front load. The company can manufacture a maximum of 18 washers per day. It makes a profit of​ $20 on top load machin      Log On


   

Question 1147010: A company manufactures two types of​ washers, top load and front load. The company can manufacture a maximum of 18 washers per day. It makes a profit of​ $20 on top load machines and​ $25 on front load machines. No more than 5 front load machines can be manufactured due to production restrictions. To meet consumer​ demand, the company must manufacture at least 2 front load machines and 2 top load machines per day. What is the maximum​ profit?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x = number of top loaders.
y = number of front loaders.
constraint equations are:
x + y <= 18
y <= 5
x >= 2
y >= 2
objective function is:
profit = 20x + 25y

using the desmos.com calculator, graph the opposite of these inequalities.
the area on the graph that is not shaded is your region of feasibility.
the corner points of this region are where the maximum profit will lie.

here's the graph.

$$$

the highest pofit is when x = 13 and y = 5.
that's the point (13,5) on the graph.
the profit is 13 * 20 + 5 * 25 = 385
all the constraint are met, as shown below.
x + y <= 18
y <= 5
x >= 2
y >= 2

solution is that maximum profit is 385 dollars.