Answer :

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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