Let the manufacturer produces $x$ pedestal lamps and $y$ wooden shades; then the time taken by $x$ pedestal lamps and $y$ wooden shades on grinding/ cutting machines $=(2 x+y)$ hours and time taken by $x$ pedestal lamps and $y$ shades on the sprayer $=(3 x+2 y)$ hours.
Since grinding/ cutting machine is available for at the most 12 hours, $2 x+y \leq 12$ and sprayer is available for at the most 20 hours.
We have: $3 x+2 y \leq 20$.
Profit from the sale of $x$ lamps and $y$ shades.
$\mathrm{Z}=5 \mathrm{x}+3 \mathrm{y}$
So, our problem is tomaximizeZ $=5 x+3$ subject toconstraints $3 \mathrm{x}+2 \mathrm{y} \leq 20,2 \mathrm{x}+\mathrm{y} \leq 12, \mathrm{x}, \mathrm{y} \geq 0$.
Consider $\quad 3 \mathrm{x}+2 \mathrm{y} \leq 20$
Let $\quad 3 \mathrm{x}+2 \mathrm{y}=20 \Rightarrow \mathrm{y}=\frac{20-3 x}{2}$


Now the area represented by $3 x+2 y \leq 20$ is the half-plane containing $(0,0)$ as $(0,0)$ stisfies the inequations.
Again consider
$$
2 \mathrm{x}+\mathrm{y} \leq 12
$$
Let
$$
2 x+y=12 \Rightarrow y=12-2 x
$$

The inequations consists of the half-plane containing $(0,0)$ as $(0,0)$ satisfies it.

The double shaded region OPCAO is our solution where $\mathrm{O}(0,0), \mathrm{P}(6,0), \mathrm{C}(4,4), \mathrm{A}(0,10)$.
Now $\quad Z=5 x+3 y$
At $\mathrm{O}(0,0), \quad Z=0$
At $\mathrm{P}(6,0), \quad Z=30$
At $\mathrm{C}(4,4), \quad \mathrm{Z}=5 \times 4+3 \times 4=32$
At $\mathrm{A}(0,10), \quad \mathrm{Z}=5 \times 0+3 \times 10=30$
Now maximum $Z=32$ at $x=4, y=4$.
Hence maximum $Z=₹ 32$ when he manufacture 4 pedestal lamps and 4 wooden shades.