Q. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes $2 h$ on grinding/cutting machine and $3 h$ on the sprayer to manufacture a pedestal lamp. It takes $1 h$ on the grinding/cutting machine and $2 h$ on the sprayer to manufacture a shade.On any day, the sprayer is available for atmost $20 h$ and the grinding/cutting machine for atmost $12 h$. The profit from the sale of a lamp is $₹ 5$ and that from a shade is $₹ 3$. Assuming that the manufacturer can sell all the lamps and shades that he produces, the number of pedestal lamps and wooden shades to maximise the profit are respectively.
Linear Programming
Solution:
Let the manutacturer produces $x$ pedestal lamps and $y$ wouderi shades everyday. We consistruct lhe folluwing lable
Item
Number of packages
Time on grinding/cuttin g machine (inh)
Time onsprayer(in h)
Prófit (in ₹)
A
x
2x
3x
5x
B
y
y
2 y
3 y
Total
x+y
2x + y
3x + 2y
5 x + 3 y
Availability
12
20
The profits on a lamp is $₹ 5$ and on the shades $₹ 3$.
Our problem is to maximise $Z=5 x+3 y$...(i)
Subject to the constraints are $2 x+y \leq 12$...(ii)
$ 3 x+2 y \leq 20$...(iii)
$x \geq 0, y \geq 0$...(iv)
Firstly, draw the graph of the line $2 x+y=12$
x
0
6
y
12
0
Putting $(0,0)$ in the inequality $2 x+y \leq 12$, we have
$2 \times 0+0 \leq 12 \Rightarrow 0 \leq 12$(which is true)
So, the half plane is towards the origin.
Since, $ x, y \geq 0$
So, the feasible region lies in the first quadrant.
Secondary, draw the graph of the line $3 x+2 y=20$
x
0
$\frac{20}{3}=6.6$
y
10
0
Putting $(0,0)$ in the inequality $3 x+2 y \leq 20$, we have
$3 \times 0+2 \times 0 \leq 20$
$\Rightarrow 0 \leq 20 $ (which is true)
So, the half plane is towards the origin.
On solving equations $2 x+y=12$ and $3 x+2 y=20$, we get $B(4,4)$.
$\therefore$ Feasible region is OABCO.
The corner point of the feasible region are $O(0,0), A(6,0)$, $B(4,4)$ and $C(0,10)$. The values of $Z$ at these points are as follows
Corner point
$z=5 x+3y$
$O(0,0)$
0
$A(6,0)$
30
$B(4,4)$
$32 \rightarrow$ Maximum
$C(0,10)$
30
The maximum value of $Z$ is $₹ 32$ at $B(4,4)$.
Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profits.
| Item | Number of packages | Time on grinding/cuttin g machine (inh) | Time onsprayer(in h) | Prófit (in ₹) |
|---|---|---|---|---|
| A | x | 2x | 3x | 5x |
| B | y | y | 2 y | 3 y |
| Total | x+y | 2x + y | 3x + 2y | 5 x + 3 y |
| Availability | 12 | 20 |
| x | 0 | 6 |
| y | 12 | 0 |
| x | 0 | $\frac{20}{3}=6.6$ |
| y | 10 | 0 |
| Corner point | $z=5 x+3y$ |
|---|---|
| $O(0,0)$ | 0 |
| $A(6,0)$ | 30 |
| $B(4,4)$ | $32 \rightarrow$ Maximum |
| $C(0,10)$ | 30 |