A manufacturer produces two Models of bikes - Model X and Model Y. Model X takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

Let number of bikes per week of model X and Y be x and y respectively.


Given that model X takes 6 man-hours.


So, time taken by x bikes of model X = 6x hours.


Given that model Y takes 10 man-hours.


So, time taken by y bikes of model X = 10y hours.


Total man-hour available per week = 450


So, 6x + 10y ≤ 450


⇒ 3x + 5y ≤ 225


Handling and marketing cost of model X and Y is Rs 2000 and Rs 1000 per unit respectively.


So, total handling and marketing cost of x units of model X and y units of model y is 2000x + 1000y


Maximum amount available for handling and marketing per week is Rs 80000.


So, 2000x + 1000y ≤ 80000


⇒ 2x + y ≤ 80


Profits per unit for Models X and Y are Rs 1000 and Rs 500, Respectively.


Let total profit = Z


So, Z = 1000x + 500y


Also, as units will be positive numbers so x, y ≥ 0


So, we have,


Z = 1000x + 500y


With constraints,


3x + 5y ≤ 225


2x + y ≤ 80


x, y ≥ 0


We need to maximize Z, subject to the given constraints.


Now let us convert the given inequalities into equation.


We obtain the following equation


3x + 5y ≤ 225


⇒ 3x + 5y = 225


2x + y ≤ 80


⇒ 2x + y = 80


x ≥ 0


⇒ x=0


y ≥ 0


⇒ y=0


The region represented by 3x + 5y ≤ 225:


The line 3x + 5y = 225 meets the coordinate axes (75,0) and (0,45) respectively. We will join these points to obtain the line 3x + 5y = 225. It is clear that (0,0) satisfies the inequation 3x + 5y ≤ 225. So, the region containing the origin represents the solution set of the inequation 3x + 5y ≤ 225.


The region represented by 2x + y ≤ 80:


The line 2x + y = 80 meets the coordinate axes (40,0) and (0,80) respectively. We will join these points to obtain the line


2x + y = 80.


It is clear that (0,0) satisfies the inequation 2x + y ≤ 80. So, the region containing the origin represents the solution set of the inequation 2x + y ≤ 80


Region represented by x≥0 and y≥0 is first quadrant, since every point in the first quadrant satisfies these inequations.


Plotting these equations graphically, we get



Feasible region is ABCD


Value of Z at corner points A, B, C and D –



So, value of Z is maximum on-line BC, the maximum value is 40000.So manufacturer must produce 25 number of models X and 30 number of model Y.


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