Question

A manufacturer produces two models of bikes -,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 available per week. Handling and marketing costs are Rs 2000 and Rs 1000 per units for models X and Y, respectively. The total funds available for these purposes are Rs 80000 per week. Profits per units 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.

Answer 1

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

$\begin{array}{cc}\text{Corner points}& \text{Value of Z =1000x+500y}\\ (0,0)& 0\\ (40,0)& 40000\leftarrow Maximum\\ (25,30)& 25000+15000=40000\leftarrow Maximum\\ (0,45)& 22500\end{array}$
So, the manufacturer should produce 25 bikes of model X and 30 bikes of model Y to get a maximum profit of Rs 40000,
Since, in question it is asked that each model bikes should be produced.