  # Linear Equations – Assembly Manufacturing Process

Linear Equations Applied to Manufacturing Process

Pandora Vehicle Sdn. Bhd. produces two models of bicycles: model A and model B.

• Assembly Time.

- Model A requires 2 hours of assembly time.

- Model B requires 3 hours of assembly time.

• Cost of Parts.

- The parts for model A cost RM25 per bike

- the parts for model B cost RM30 per bike

• Time and Money Budgeted to produce these two models

- total of 34 hours of assembly time per day

- RM365 is available per day

How many of each model can be made in a day?

### Comments for Linear Equations – Assembly Manufacturing Process

 Sep 03, 2012 Linear Equations Applied to Manufacturing Process by: Staff Answer:Part IA = number of Model A bicycles manufacturedB = number of Model B bicycles manufacturedT = total number of bicycles (A + B) manufactured The objective function:      • maximize total units manufactured:        T = A + B Constraints     • Equation which models Assembly Time:          - maximum time available = 34 hours per day         - Model A requires 2 hours of assembly time.          - Model B requires 3 hours of assembly time.        (2 hours) * (number of Model A bicycles) + (3 hours) * (number of Model B bicycles) ≤ 34 hours       2A + 3B ≤ 34      • Equation which models the Cost of Parts:          - maximum cost cannot exceed RM365 per day          - Model A requires parts which cost RM25.          - Model B requires parts which cost RM30.       (RM25) * (number of Model A bicycles) + (RM30) * (number of Model B bicycles) ≤ RM365       25A + 30B ≤ 365 Mathematically, the question before us is:     • Solve for the values of A and B which maximize the following objective function :         T = A + B      • Subject to the following restrictions:         2A + 3B ≤ 34          25A + 30B ≤ 365          A ≥ 0          B ≥ 0 ---------------------------------------------

 Sep 03, 2012 Linear Equations Applied to Manufacturing Process by: Staff --------------------------------------------- Part II Plot the restrictions:          2A + 3B ≤ 34          (the red line and all the area below the red line)          25A + 30B ≤ 365          (the green line and all the area below the green line)          A ≥ 0          (all the area to the right of the vertical blue line, which is the “B-axis”)          B ≥ 0          (all the area above the horizontal pink line, which is the “A-axis”)      • When these boundaries are plotted, the area inside the boundaries forms the Bounded Feasible Region:          Bounded Feasible Region          (shaded yellow area) • The corner points of the bounded feasible region (shown as the yellow shaded area) define a maximum value and a minimum value for the objective function (T = A + B). ---------------------------------------------

 Sep 03, 2012 Linear Equations Applied to Manufacturing Process by: Staff --------------------------------------------- Part III      • The corner points are (in A,B format):          (0,0), (0,11⅓), (5,8), (14.6,0)      • Find the maximum value of T by computing the values of T at the corner points.          T = A + B        Corner point: (0,0)          T = 0 + 0 = 0, minimum value of T        Corner point: (0,11⅓)          T = 0 + 11⅓ = 11⅓        Corner point: (5,8)          T = 5 + 8 = 13        Corner point: (14.6,0)          T = 14.6 + 0 = 14.6, maximum value of T        Maximum value of T is: 14.6; when A = 14.6 and B = 0        Minimum value of T is: 0; when A = 0 and B = 0 Final Answer Maximum possible Number of A units = 14 Maximum possible Number of B units = 11 Maximum possible Number of TOTAL units = 14 Thanks for writing. Staff www.solving-math-problems.com