Q.A dietician wishes to design a diet using two food items, Food I and Food II, that contains at least 8 units of a certain protein and at least 10 units of a certain mineral. One unit of Food I costs Rs 4 and one unit of Food II costs Rs 3. Every unit of both Food I and Food II supplies exactly 1 unit of the protein for each unit of food, while 1 unit of Food I supplies 2 units of the mineral and 1 unit of Food II supplies 1 unit of the mineral. (In symbols, with $x$ units of Food I and $y$ units of Food II: protein constraint $x + y \ge 8$, mineral constraint $2x + y \ge 10$.) Find the minimum cost diet that meets both requirements.

Let $x=$ units of Food I and $y=$ units of Food II.

Corner points. At $y=0$: protein needs $x\ge8$, mineral needs $x\ge5$; binding is $x\ge8$, giving vertex $(8,0)$ — check mineral: $16\ge10$, slack.

At $x=0$: protein needs $y\ge8$, mineral needs $y\ge10$; binding is $y\ge10$, giving vertex $(0,10)$ — check protein: $10\ge8$, slack.

Intersection: $x+y=8$ and $2x+y=10$; subtracting gives $x=2,\ y=6$, so $(2,6)$ — check both: $2+6=8$ ✓, $4+6=10$ ✓.

Evaluating $Z=4x+3y$:

$(8,0)\to32,\qquad (2,6)\to8+18=26,\qquad (0,10)\to30.$ …

Translate the protein and mineral requirements into ≥ constraints, find the three corner points of the resulting unbounded region, evaluate the linear cost function at each, and take the smallest v …

Forgetting to check that the candidate minimum corner also satisfies the OTHER constraint with slack, not just the one it lies on; mislabelling which intercep …