Q.A farmer wants his crop to receive at least 10 units of nitrogen and at least 15 units of phosphorus. Fertiliser $P$ costs Rs 4 per bag and supplies 1 unit of nitrogen and 3 units of phosphorus; fertiliser $Q$ costs Rs 6 per bag and supplies 2 units of nitrogen and 1 unit of phosphorus. Formulate this as a linear programming problem 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, then evaluate the corner points of the feasible region.

Formulation. Let $x$ = bags of fertiliser $P$, $y$ = bags of fertiliser $Q$. Minimise

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: $x=10$, giving $(10,0)$ (checking $3(10)+0=30\ge15$, true).

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

Intersection of the two lines. Solving $x+2y=10$ and $3x+y=15$: from the second, $y=15-3x$; substituting,

so the lines meet at $(4,3)$, the third corner point.

A frequent error here is solving the two boundary equations by substituting the wrong variable's expression back into the same equation it came from instead of into the other equatio …