Q.A farmer wants his crop to receive at least 10 units of nitrogen and at least 15 units of phosphorus. Fertiliser $P$ costs ₹4 per bag and supplies 1 unit of nitrogen and 3 units of phosphorus; fertiliser $Q$ costs ₹6 per bag and supplies 2 units of nitrogen and 1 unit of phosphorus. Formulate this as an LPP and solve it graphically to find the minimum cost.

Formulate with $x,y$ = bags of $P,Q$; minimise cost $Z=4x+6y$ subject to the nitrogen and phosphorus requirements; evaluate the corner points of the feasible region.

Corner points: $(10,0)$, $(4,3)$, $(0,15)$. $Z$ values: $40$, $34$, $90$. Minimum is $Z=34$ at $(4,3)$.

Formulation. Let $x$ = bags of fertiliser $P$, $y$ = bags of fertiliser $Q$. Minimise $Z=4x+6y$. Nitrogen: $x+2y\ge10$. Phosphorus: $3x+y\ge15$. Non-negativity: $x\ge0,y\ge0$.

Corner points. On the x-axis ($y=0$): $x\ge10$; $3x\ge15\Rightarrow x\ge5$. The larger governs: $(10,0)$ (checking $3(10)+0=30\ge15$ ✓).

On the y-axis ($x=0$): $2y\ge10\Rightarrow y\ge5$; $y\ge15$. The larger governs: $(0,15)$ (checking $0+2(15)=30\ge10$ ✓).

Intersection of the two lines. Solving $x+2y=10$ and $3x+y=15$: from the second, $y=15-3x$; substituting, $x+2(15-3x)=10 \Rightarrow x+30-6x=10 \Rightarrow -5x=-20 \Rightarrow x=4,\ y=15-12=3$ — the third corner point, $(4,3)$.

Evaluating $Z=4x+6y$.

The smallest value is $Z=34$ at $(4,3)$. Since both constraints push the feasible region away from the origin in the direction that increases $Z=4x+6y$, no feasible point beyond $(4,3)$ can give a smaller cost — $34$ is genuinely the minimum.

Independent check. Substitute $(4,3)$ into both constraints: $4+2(3)=10\ge10$ ✓ (exactly meets nitrogen); $3(4)+3=15\ge15$ ✓ (exactly meets phosphorus) — both requirements exactly satisfied, confirming a genuine corner point.

Solving the two boundary equations by substituting an expression back into the same equation it came from (instead of into the other equation) produces a trivial identity rather than the actual intersection point — always substitute the expression for one variable into the other remaining equation.