This is a typical diet problem. Let the mixture contain x kg of Food ‘P’ and y kg of Food ‘Q’. Clearly, x, y ≥ 0. Let us construct the following table from the given data:

Resources 
P 
Q 
Requirements 
Vitamin A 
3 
4 
8 
Vitamin B 
5 
2 
11 
Cost (Rs./Kg) 
60 
80 

Since the mixture must contain at least 8 units of Vitamin A and 10 units of Vitamin B, we have the following constraints: 3x + 4y ≥ 8. and 5x + 2y ≥ 11.
The total cost of purchasing x kg of food P and y kg of food Q is Z = 60x + 80y.
Hence we need to Minimize Z = 50x + 60y, subject to the constraints
$(1) \quad 3x+4y \geq 8$
$(2) \quad 5x+2y \geq 11$
$\textbf{Plotting the constraints}$
Plot the straight lines 3x+4y = 8 and 4x+2y = 11.
First draw the graph of the line 3x+4y = 8.
If x = 0, y = 2 and if y = 0, x = 8/3.
At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 8. So the area associated with this inequality is unbounded and away from the origin.
Similarly, draw the graph of the line 5x + 2y = 11.
If x = 0, y = 11/2 and if y =0, x = 11/5.
At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 11. So the area associated with this inequality is unbounded and away from the origin.
$\textbf{Finding the feasible region}$
We can see that the feasible region is unbounded.
On solving the equations 3x + 4y = 8 and 5x + 2y = 11, we get,
3x + (2210x) = 8 $\rightarrow$ 7x = 14 $\rightarrow$ x = 2.
If x = 2, then y = (11 – 10) /2 = 1/2 .
x = 2 and y = $\frac{1}{2}$
Therefore the feasible region has the corner points $(\frac{8}{3}, 0), (2,\frac{1}{2}), (0, \frac{11}{2}).$
$\textbf{Solving using Corner Point Method}$
The values of Z at the corner points are summarized as follows:
Corner Point 
Z = 60x+80y 
(8/3, 0) 
160 (Min Value) 
(2, 1/2) 
160 (Min Value) 
(0, 11/2) 
440 
In the table we find that the smallest value of Z is 160. Can we say that this is the minimum value? Remember that the feasible region is unbounded.
Therefore we have to draw the graph of the inequality 60x + 80y < 160, i.e, 3x + 4y < 8 to check if the resulting half plane has any common point with the feasible region.
From the figure and table we can see that it has no points in common.
$\textbf{Thus, the minimum value of Z is 160 attained at two of the points}$
$\textbf{The optimum strategy would be to mix 2 kgs of P and ½ kg of Q or just use 8/3 kgs of P, at a cost of Rs. 160.}$