Does anyone know how to solve this word problem involving a system of equations?

Question

A business manufactures two parts, X and Y. Machines A and B are needed to make each part. To make part X, machine A is needed 3 hours and machine B is needed 1 hour. to make part Y, machine A is needed 1 hour and machine B is needed 2 hours. Machine A is available 60 hours per week and machine B is available 50 hours per week. The profit from part X is $300 and the profit from Part Y is $250. How many parts of each type should be made to maximize weekly profit?

Answer 1

At most can make 20 of Part X. However, this leaves Machine B unused for 30 hours. By reducing the number of Part X by 1 it is possible to make 3 Part Ys for an additional profit. So the maximum profit will be when both machines are fully utilised.

The quick way to work this out is to set up a spreadsheet and work out the machine utilisation and profit for combinations of Parts X & Y.

Let the number of Part X made be x and the number of Part Y made be y.

Then for Machine A:

3x + y = 60
(3 hours for each Part X and 1 hour for each Part Y)

For Machine B:

x + 2y = 50
(1 hour for each Part X and 2 hours for each Part Y)

For Machine B this is the same as:

3x + 6y = 150 (multiply everything by 3)

Subtract Machine B from Machine A:

3x + y = 60
3x + 6y = 150
-----------------------------
-5y = -90
y = 18

Substitute back into Machine B:

x + 2y = 50
x + 36 = 50
x = 14

Therefore maximum profit ($8,700) is when 14 Part X and 18 Part Y are made.

Answer 2

X = 3A +B
Y = A + 2B

To maximise efficiency
produce nX + mY

nX = 3nA + nB
mY = mA + 2mB

But
m + 3n = 60
2m + n = 50

Solving
m = 18 and n = 14

So maximum efficiency when 14 X and 18Y are produced.