A factory makes 3 different chairs Chair A requires 5 hours

A factory makes 3 different chairs. Chair A requires 5 hours to construct, at a cost of $20 per desk to the company. Chair B requires 5 hours to construct, at $30 per desk. Chair C requires 10 hours to construct, at $40 per desk. There is a total of 500 hours per week for construction, and weekly budget of $1500 to cover production cost. How many chairs of each type can the factory produce each week if it is to use all the weekly budget and labour hours available?

Solution

Let the company construct, per week, x number of chair A, y number of chair B and z number of chair C.Now, if the company is to use all the weekly budget and labour hours available, then 5x+5y + 10z = 500 or, x+y +2z = 100…(1). Also, 20x+30y+40z = 1500 or, 2x +3y+4z = 150…(2). On multiplying the 1^st equation by 2 and then subtracting the result from the 2^nd equation ( to eliminate z and x), we get (2x+3y+4z)-(2x+2y+4z) = 150-200 or, y = -50. Apparently , the given data is erroneous as y cannot be negative.

Note:

After correcting the data, we can obtain 2 equations in x and z by substituting the correct value of y. Then x and z can be determined.