A manufacture produces two models of bike model X and model Y. Model X takes a 6 man hours to make per unit, while model Y takes 10 man hours per unit. There is a total of 450 man hour availbale per week. Handling and marketing costs are 2000 and 1000 per unit of model X and Y, respectively. The tota funds available for these purposes are 8000 per week. Profits per unit for models X and Y are 1000 and manufacturer produce, so, as to yeild a maximum profit? Find the maximum profit.

Answer

N/a

Explanation

Let the manufacturer produces x number of models X and y number of model Y bikes. Model X takes a 6 man hours to make per unit and model Y takes 10 man hours to make per unit. There is a total of 450 man hour available per week
`therefore 6x+10y le 450`
`Rightarrow 3x+5y le 225....(i) `
For models X and Y, handling and marketing costs are 2000 and 1000 respectively, total funds available for these purposes are 80000 per week.
`therefore 2000x+1000y ge 80000`
`Rightarrow 2x+y ge 80`
Also, `x ge 0, y ge....(ii)`
Hence the profits per unit fo r models X and Y are 1000 and 500 respectively `therefore "Required LPP i"`
Maximise Z=1000x+500y
Subject to `3x+5y le 225, 2x+y le 80, x ge 0, y ge 0`
From the shaded feasible region, it is clear that coordinates of corner points are (0,) (40,0),(25,30) and (0,45)
On solving 3x+5y=225 and 2x+y=80, we get
x=25, y=30

So, the manufacturer should poroduce 25 bikes of model X and 30 bikes of model Y to get a maximum profit of 40000.
Since in question it is asked that each model bikes should be produced.