Question

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type $A$ require $5$ minutes each for cutting and $10$ minutes each for assembling. Souvenirs of type $B$ require $8$ minutes each for cutting and $8$ minutes each for assembling. There are $3$ hours $20$ minutes available for cutting and $4$ hours for assembling. The profit is $Rs.5$ each for type $A$ and $Rs.6$ each for type $B$ souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

Answer 1

Let number of souvenirs A to be made is $X$ and number of souvenirs B to be made is $Y$

Since, souvenirs A requires $5$ minutes and souvenirs B requires $8$ minutes for cutting. Also, there is maximum $3$ hours $20$ minutes (200 minutes) for cutting.

$\therefore 5X+8Y\le 200...\left(1\right)$

Since, souvenirs A requires $10$ minutes and souvenirs B requires $8$ minutes for assembling. Also, there is maximum $4$ hours (240 minutes) for assembling.

$\therefore 10X+8Y\le 240$

$\Rightarrow 5X+4Y\le 120...\left(2\right)$

Since, count of objects can't be negative.

$\therefore X\ge 0,Y\ge 0...\left(3\right)$

We have to maximize profit of the company.

Here, profit on souvenirs A is $5Rs$ and on souvenirs B is $6Rs$

So, objective function is $Z=5X+6Y$

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=5X+6Y$
A $(0,25)$	$150$
B $(8,20)$	$160$ (maximum)
C $(24,0)$	$120$

Hence, company should produce 8 souvenirs A and 20 souvenirs B in a day to maximize his profit. Also, maximum profit will be $160Rs$