Q.A furniture dealer deals in tables and chairs. He has Rs.1,50,000 to invest and a space to store at most 60 pieces. A table costs him Rs.1500 and a chair Rs.750. Construct the inequations and find the feasible solution.

Let $x$ = number of tables and $y$ = number of chairs. The budget gives $1500x+750y\le1{,}50{,}000$, which simplifies (divide by 750) to $2x+y\le200$. The storage space gives $x+y\le60$. Together with $x\ge0,y\ge0$, we graph both lines: $2x+y=200$ meets the axes at $(100,0)$ and $(0,200)$, while $x+y=60$ meets them at $(60,0)$ and $(0,60)$. Since even the extreme cases $x=60,y=0$ (cost Rs.90,000) and $x=0,y=60$ (cost Rs.45,000) both use far less than the Rs.1,50,000 budget, the line $2x+y=200$ never actually cuts into the triangle bounded by $x+y\le60$ and the axes — the budget constraint is 'slack' (non-binding) throughout. So the true feasible re …

Write the cost inequation and the piece-count inequation, simplify the cost inequation for easier plotting, then graph both t …

A frequent error is assuming every stated constraint must be 'active' at some vertex of the region — here the budget genuinely never binds, and the honest graphical answ …