Question

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?

• A. 120 black and white and 180 colour television sets
• B. 300 black and white television sets
• C. 240 colour television sets
• D. 180 black and white and 120 colour television sets

Answer 1

Answer: (D)

Let $x$ and $y$ denote, respectively, the number of black and white sets and coloured sets made each week. Thus,
$x\ge 0,y\ge 0$
Since, the company can make atmost 300 sets a week, therefore,
$x+y\le 300$
Weekly cost (in ₹) of manufacturing the set is
$1800x+2700y$
and the company can spend upto $₹648000$. Therefore,
$1800x+2700y\le 648000\text{, i.e., or }2x+3y\le 720$
The total profit on $x$ black and white sets and $y$ colour sets is $₹(510x+675y)$. Let $Z=510x+675y$. This is the objective function.
Thus, the mathematical formulation of the problem is
Maximise $Z=510x+675y$

The feasible region $OABC$ is shown in the figure.
Since, the feasible region is bounded, therefore maximum of $Z$ must occur at the corner point of $OABC$.

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.