A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg. food are given below:

Food	Vitamin A	Vitamin B	Vitamin C
X	1	2	3
Y	2	2	1

One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?

Let the dietician mix x kg of food X and y kg of food Y.
Let Z be the cost.
Table

We are to minimise
Z = 16x + 20y
subject to the constraints
x + 2y ≥ 10
2x + 2y ≥ 12 or x + y ≥ 6
3x + y ≥ 8
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + 2y = 10
For x = 0, 2 y = 10 or y = 5
For y = 0, x = 10
∴ line meets OX in A(10, 0) and OY in L(0, 5).
Also we draw the graph of x + y = 6
For x = 0, y = 6
For y = 0, x = 6
∴ line meets OX in B(6, 0) and OY in M(0, 6).
Again we draw the graph of 3x + y = 8
For x = 0,  y = 8
For y = 0,  3x = 8  or x = 

Since feasible region satisfies all the constraints.
 

∴ shaded region is the feasible region which is unbounded and corner points are A(10, 0), D(2, 4), E(1, 5), N(0, 8).
At A(10, 0), Z = 16 × 10 + 20 × 0 = 160 + 0 = 160
At D(2, 4), Z = 16 × 2 + 20 × 4 = 32 + 80 = 112
At E(1, 5), Z = 16 × 1 + 20 × 5 = 16 + 100 = 116
At N(0, 8), Z = 16 × 0 + 20 × 8 = 0 + 160 = 160
∴ smallest value = 112 at (2, 4)
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph of
16 x + 20 y < 112 or 4 x + 5 y < 28    ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = 112 at (2, 4)
∴ minimum cost is Rs. 112 when 2 kg of food X and 4 kg. of food Y are mixed.