Q.A dietician wishes to mix two types of food, $F_1$ and $F_2$, to prepare a diet containing at least 8 units of protein and at least 8 units of iron. Food $F_1$ contains 1 unit of protein and 2 units of iron per kg and costs Rs 3 per kg; Food $F_2$ contains 2 units of protein and 1 unit of iron per kg and costs Rs 4 per kg. How many kg of each food should be mixed to minimize the cost while meeting both requirements?

Let $x=$ kg of $F_1$ and $y=$ kg of $F_2$.

At $y=0$: protein needs $x\ge8$, iron needs $x\ge4$; binding $x\ge8$, vertex $(8,0)$ — check iron: $16\ge8$, slack.

At $x=0$: protein needs $y\ge4$, iron needs $y\ge8$; binding $y\ge8$, vertex $(0,8)$ — check protein: $16\ge8$, slack.

Intersection: $x+2y=8,\ 2x+y=8$. From the second, $y=8-2x$; substituting, $x+2(8-2x)=8\Rightarrow x+16-4x=8\Rightarrow-3x=-8\Rightarrow x=\dfrac83,\ y=8-\dfrac{16}{3}=\dfrac{8}{3}$, giving $\left(\dfrac83,\dfrac83\right)$.

Minimum $=\dfrac{56}{3}\approx$ Rs $18.67$, at $x=y=\dfrac83\approx2.67$ kg.

Set up the two nutrient requirements as ≥ constraints, find all three corner points of the unbounded region, and evaluate the cost function at each, keeping the smallest.

Rounding $\frac83$ to a whole number too early and evaluating Z at the rounded (and therefore infeasible) point instead of the exact fraction.