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.
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’s take x an y to be the number of models of bike produced by the manufacturer.

From the question we have,

Model x takes

    \[6\]

man-hours to make per unit

Model y takes

    \[10\]

man-hours to make per unit

Total man-hours available =

    \[450\]

So,

    \[6x\text{ }+\text{ }10y\le 450\Rightarrow 3x\text{ }+\text{ }5y\le 225\]

….. (i)

The handling and marketing cost of model x and y are Rs

    \[2000\]

and Rs

    \[1000\]

respectively.

And, the total funds available is Rs

    \[80,000\]

per week

So,

    \[2000x\text{ }+\text{ }1000y\le 80000\Rightarrow 2x\text{ }+\text{ }y\le 80\]

… (ii)

And,

    \[x\ge 0,\text{ }y\ge 0\]

Now, the total profit (Z) per unit of models x and y are Rs

    \[1000\]

and Rs

    \[500\]

repectively

    \[Z\text{ }=\text{ }1000x\text{ }+\text{ }500y\]

Hence, the required LPP is

Maximize

    \[Z\text{ }=\text{ }1000x\text{ }+\text{ }500y\]

subject to the constraints

    \[3x\text{ }+\text{ }5y\le 225,\text{ }2x\text{ }+\text{ }y\le 80\text{ }and\text{ }x\ge 0,\text{ }y\ge 0\]

Now, let’s construct a constrain table for the above:

Next, on solving equation (i) and (ii) we get

    \[x\text{ }=\text{ }25\text{ }and\text{ }y\text{ }=\text{ }30\]

After plotting all the constraint equations, we observe that the feasible region is OABC, whose corner points are

    \[O\left( 0,\text{ }0 \right),\text{ }A\left( 40,\text{ }0 \right),\text{ }B\left( 25,\text{ }30 \right)\text{ }and\text{ }C\left( 0,\text{ }45 \right).\]

On evaluating the value of Z, we get

Therefore, from the above table it’s seen that the maximum profit is Rs

    \[40,000\]

.

The maximum profit can be achieved by producing

    \[25\]

bikes of model x and

    \[30\]

bikes of model Y or by producing

    \[40\]

bikes of model x.