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Problem A company makes two types of BMX bicycles, a Pro Bike and a Trick Bike. The relevant manufacturing data are given in the following table: Department Labour hrs per bike Maximum labour hrs available per dayTrick bike Pro bikeFabricating 6 4 108 Finishing 1 1 24 (a) If the profit on a Trick Bike is €40 and the profit on a Pro Bike is €30, how many of each type of bike should be manufactured per day to maximize the profit? What is the maximum daily profit? Let, x = number of Pro Bike manufactured per dayy = number of Trick Bike manufactured per dayThe linear programming model is defined as below:Maximize profit: P = 30x + 40ySubject to constraint: (Fabricating Department) (Finishing Department)x, y ≥ 0 (Non-negativity)The solution exists at corner points.
Solving equations and , the point of intersection is (18, 6). The other corner points are (0, 18) and (24, 0). Figure 1 shows the feasible region.Below table summarizes the profits for each corner pointsxyP = 30x + 40y01830(0) + 40(18) = €72024030(24) + 40(0) = €72018630(18) + 40(6) = €780 (Max)Thus, 18 Pro Bike and 6 Trick Bike should be manufactured per day to maximize the profit. The maximum daily profit is €780.The labour hours for both Fabricating and Finishing departments will be fully utilizedFigure 1: Feasible region and corner points(b) Investigate the effect on the production schedule and the maximum profit if the profit on a Pro Bike decreases to €25.
Now only the profit equation will change and all other things will remain same. Therefore,Maximize profit: P = 25x + 40yBelow table summarizes the profits for each corner pointsxyP = 25x + 40y01825(0) + 40(18) = €720 (Max)24025(24) + 40(0) = €60018625(18) + 40(6) = €690In this case, 18 Trick Bike and no Pro Bike should be manufactured per day to maximize the profit. The maximum daily profit is €720. Now, the labour hours for the Fabricating department will be fully utilized, however, 6 labour hours for the Finishing department will remain unutilized.
Problem 2 A piece of equipment was bought for €50,000. Two years later, the value had dropped to €35,000. Construct an exponential decay model of the form , where V(t) is the value after t years. Use this model to answer the following questions. (a) What was the value after one year? (b) When will the value go below €20,000?When t = 0 years, the equipment value is €50,000 and when t = 2, the equipment value is €35,000. Therefore, (Taking natural logarithm of both side) Therefore, the exponential decay model of the form is given by(a) The value after one year was around €41,833.(b) The value will go below €20,000 after 5.
14 years (in 6th year) from date of purchase. (Taking natural logarithm of both side) = 5 years 1 month 20.4 days
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