Chemical Engineering

Brief background In Malaysia, the production of dairy products is overseen by a plant manager. It is well known that the seller has purchased too much sugar, which is one of the raw materials. For the production of its primary products, milk and cheese, 75kg of sugars were purchased. The factory manager will have to pay RM 120.00 per kg of sugar for the milk and RM 210.00 per kg of cheese in order to produce these. The factory manager has chosen to allot RM 15,000.00 towards the expense. After the production process, these products are kept in the storage room before being delivered to end-users. This room has a small capacity of 4000dm3. For every kg of sugar used, an average of 110dm3 of milk and 30dm3 of butter are produced. It is known that the net profit for milk per dm3 is RM 1.30 and for butter per dm3 is RM 2.00 after a thorough computation of the net profit that can be generated. The factory manager then uses an optimization method to maximize the profit considering the previous constraints. Objective function, decision variable and constraints The objective function (OF) to be optimized in this example is to maximize the profit by the manufacturing of milk and cheese products, which can be shown as follows: Where: x represents the weight in kg of milk : y represents the weight in kg of butter Profit,P(x,y)=(110)(1.3)x+(30)(2)y=143x+60y The available cost allotted, storing space, and sugar weight are all susceptible to various constraints in order to maximize the profit. These are the equations: Constraint 1: 120x+210y15000 (Available cost allocated) Constraint 2: 110x+30y4000 (Available storage) Constraint 3:x+y75 (Limit of weight of sugar) Solve above problem using matlab fminsearch. Provide the coding for the matlab r2016a