For the equation x+y=20, the shaded area and origin both lies on the same side of the live, therefore the compounding inequality is \(x+y\le 20\).
Similarly, \(3x+2y\le 48\)
Also,the shaded portion lines in Ist quadrant.
\(\therefore \quad x\ge 0\quad and\quad y\ge 0\)
Ans. \(x+y\le 20,3x+2y\le 48,x\ge 0,y\ge 0\)

Let the length of shortest side be x cm
Then, according to question, we have 
Length of third side = (x+2) cm
Since the perimeter of the triangle is more than 166 cm.
2x+x+(x+2)>166
\(\Rightarrow 4x+2>166\)
\( \Rightarrow 4x>164\quad [subtracting\quad 2\quad from\quad both\quad sides]\)
\(\Rightarrow x>41\quad [dividing\quad both\quad sides\quad by\quad 4]\)
Hence,the length of the shortest side should be greater than 41cm.

We know that, Profit=Revenue- Cost
In order to realise some profit, revenue should be greater than the cost
Thus, we should have R(x)>C(x)
\(\Rightarrow 60x+2000>20x+4000\)
\( \Rightarrow 60x+2000-20x+4000-20x\quad [subtracting\quad 20x\quad from\quad both\quad sides]\)
\(\Rightarrow 40x+2000>4000\)
\(\Rightarrow 40x+2000-2000>4000-2000[subtracting\quad 2000\quad from\quad both\quad sides]\)
\(\Rightarrow 40x>2000\Rightarrow \frac { 40x }{ 40 } >\frac { 2000 }{ 40 } [dividing\quad both\quad sides\quad by\quad 40]\)
\( \Rightarrow x>50\)
Hence,the manufacturer must sell more than 50 times to realise some profit.
 

\(8.2<\frac { 8.48+8.35+x }{ 3 } <8.5\)
Between 7.77 and 8.77

We have, \(\frac { 4 }{ x+1 } \le 3\le \frac { 6 }{ x+1 } \)
\(\Longrightarrow \)\(4\le 3\left( x+1 \right) \le 6\quad \left[ \because x+1\neq 0\Longrightarrow x\neq -1 \right] \)
\(\Longrightarrow \)\(\frac { 4 }{ 3 } \le x+1\le 2\)
\(\Longrightarrow \)\(\frac { 4 }{ 3 } -1\le x\le 2-1\Longrightarrow \frac { 1 }{ 3 } \le x\le 1\)
Ans. \(\left[ \frac { 1 }{ 3 } ,1 \right] \)