A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. (ii) What number of rackets and bats must be made if the factory is to work at full capacity? (ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
Consider the number of rackets be x and number of bats be y . As the quantities are always positive, so,
x ≥ 0 y ≥ 0
Tabulate the given data as,
Tennis Racket Bat Availability Machine Time 1.5 3 42 Craftsman’s time 3 1 24
The required constraints are,
1.5 x + 3 y ≤ 42 3 x + y ≤ 24 x ≥ 0 y ≥ 0
The objective function (profit) which needs to maximize is,
Z = 20 x + 10 y
(i)
Since the factory is to work at full capacity, so,
1.5 x + 3 y = 42 3 x + y = 24
Solving the above equations, value of x comes out to be 4 and value of y is 12.
Therefore, 4 rackets and 12 bats must be made.
(ii)
The line 1.5 x + 3 y ≤ 42 gives the intersection point as,
x 0 28 y 14 0
Also, when x = 0 , y = 0 for the line 1.5 x + 3 y ≤ 42 , then,
0 + 0 ≤ 42 0 ≤ 42
This is true, so the graph have the shaded region towards the origin.
The line 3 x + y ≤ 24 gives the intersection point as,
x 0 8 y 24 0
Also, when x = 0 , y = 0 for the line 3 x + y ≤ 24 , then,
0 + 0 ≤ 24 0 ≤ 24
This is true, so the graph have the shaded region towards the origin.
By the substitution method, the intersection points of the lines 1.5 x + 3 y ≤ 42 and 3 x + y ≤ 24 is ( 4 , 12 ) .
Plot the points of all the constraint lines,
It can be observed that the corner points are O ( 0 , 0 ) , A ( 8 , 0 ) , B ( 4 , 12 ) , C ( 0 , 14 ) .
Substitute these points in the given objective function to find the maximum value of Z.
Corner Points Z = 20 x + 10 y O ( 0 , 0 ) 0 A ( 8 , 0 ) 160 B ( 4 , 12 ) 200 (maximum) C ( 0 , 14 ) 140
The maximum value of Z is 200 at the point ( 4 , 12 ) .
Therefore, Rs200 is the maximum profit of the factory when it works to its full capacity.
Consider the number of rackets be
Tabulate the given data as,
| Tennis Racket | Bat | Availability | |
| Machine Time | 1.5 | 3 | 42 |
| Craftsman’s time | 3 | 1 | 24 |
The required constraints are,
The objective function (profit) which needs to maximize is,
(i)
Since the factory is to work at full capacity, so,
Solving the above equations, value of x comes out to be 4 and value of y is 12.
Therefore, 4 rackets and 12 bats must be made.
(ii)
The line
| x | 0 | 28 |
| y | 14 | 0 |
Also, when
This is true, so the graph have the shaded region towards the origin.
The line
| x | 0 | 8 |
| y | 24 | 0 |
Also, when
This is true, so the graph have the shaded region towards the origin.
By the substitution method, the intersection points of the lines
Plot the points of all the constraint lines,
It can be observed that the corner points are
Substitute these points in the given objective function to find the maximum value of Z.
| Corner Points | |
| | 0 |
| | 160 |
| | 200 (maximum) |
| | 140 |
The maximum value of Z is 200 at the point
Therefore, Rs200 is the maximum profit of the factory when it works to its full capacity.
