Q. A manufacturing company makes two types of television sets; one is black and white and the other is colour. The company has resources to make atmost 300 sets per week. It takes ₹ 1800 to make a black and white set and ₹ 2700 to make a coloured set. The company can spend not more than ₹ 648000 per week to make television sets. If it makes a profit of $₹ 510$ per black and white set and ₹ 675 per coloured set, how many sets of each type should be produced, so that the company has maximum profit?
Linear Programming
Solution:
Let $x$ and $y$ denote, respectively, the number of black and white sets and coloured sets made each week. Thus,
$x \geq 0, y \geq 0$
Since, the company can make atmost 300 sets a week, therefore,
$x+y \leq 300$
Weekly cost (in ₹) of manufacturing the set is
$1800 x+2700 y$
and the company can spend upto $₹ 648000$. Therefore,
$1800 x+2700 y \leq 648000 \text {, i.e., or } 2 x+3 y \leq 720$
The total profit on $x$ black and white sets and $y$ colour sets is $₹(510 x+675 y)$. Let $Z=510 x+675 y$. This is the objective function.
Thus, the mathematical formulation of the problem is
Maximise $Z=510 x+675 y$
The feasible region $O A B C$ is shown in the figure.
Since, the feasible region is bounded, therefore maximum of $Z$ must occur at the corner point of $O A B C$.
Corner point
Value of Z
$O(0,0)$
$510(0)+675(0)=0$
$A(300,0)$
$510(300)+675(0)=153000$
$B(180,120)$
$510(180)+675(120)=172800 \leftarrow$ Maximum
$C(0,240)$
$510(0)+675(240)=162000$
Thus, maximum $Z$ is 172800 at the point $(180,120)$, i.e., the company should produce 180 black and white television sets and 120 coloured television sets to get maximum profit.
| Corner point | Value of Z |
|---|---|
| $O(0,0)$ | $510(0)+675(0)=0$ |
| $A(300,0)$ | $510(300)+675(0)=153000$ |
| $B(180,120)$ | $510(180)+675(120)=172800 \leftarrow$ Maximum |
| $C(0,240)$ | $510(0)+675(240)=162000$ |