How many sets of each type of television should the company produce to make a max prof what is the max profit?

Question

A company makes color televisions. It produces a bargin set that sells for $100 profit and a deluxe set that sells for $150 profit. On the assembly line the bargain set requires 3 hours, the deluxe set takes 5 hours. The Cabinet shop spends 1 hour on the cabinet for the bargain set and 3 hours for the deluxe set. Both sets require 2 hours of time for testing and packing. The company has available 3900 hours for assembly line, 2100 hours in the cabinet shop and 2200 hours in testing and packaging. How many sets of each type should the company produce to make a maximum profit? what is the maximum profit?

Answer 1

Rather than work it out for you, I will present the method to find the answer:
http://en.wikipedia.org/wiki/Simplex_algorithm_method

Your Z is profit, in dollars. x is bargain sets, y is deluxe sets. So your maximization equation is: Z = 100x + 150y

Your constraints are:
3x + 5y <= 3900 (assembly line hours)
x + 3y <= 2100 (cabinet shop hours)
2x + 2y <= 2200 (you can reduce this to x + y <= 1100 if you like - for the testing and packaging).

Now simply follow the procedure detailed for the simplex algorithm and you will know the max profit.

Answer 2

Goodness, this is linear programming. Anyways, if you already know Simplex Method (which would be the most convenient method to solve this one), the answer to that would be

bargain: 300
deluxe: 600
maximum profit: 120000

Also, if you are not used with Simplex, you could use graphing and find the intersection of the "condition" lines. Just plug the intersection with the objective function, and you got it.

Objective function: Profit = 100x + 150y (x is the number of bargain and y is the number of deluxe)

3x + 5y <= 3900 (assembly condition)
x + 3y <= 2100 (cabinet)
2x + 2y <= 2200 (testing and packaging)
x >= 0 and y >= 0