Question

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 content of one kg of food is given below:

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

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

Answer 1

Hint: We solve this by the method of linear inequality. We know that ‘at least’ means that ‘greater than or equal to’, we use this to make an inequality. From the given table we obtain three inequality. We get one inequality from costs of the food, we minimize this inequality to get the least cost of the mixture.

Complete step-by-step answer:
Let, the mixture contains x units of food X.
The mixture contains y units of the food Y.
From the given data,
 \[ \bullet \] Vitamin A content in one kg of food:
Food X contains \[ \to 1 \times x\] units
Food Y contains \[ \to 2 \times y\] units
Total requirement is \[ \to \] at least 10
  \[\therefore x + 2y \geqslant 10\] . ----- (1)
 \[ \bullet \] Vitamin B content in one kg of food
Food X contains \[ \to 2 \times x\] units
Food Y contains \[ \to 2 \times y\] units
Total requirement is \[ \to \] at least 12
 \[\therefore 2x + 2y \geqslant 12\] (Divide by 2)
 \[\therefore x + y \geqslant 6\] ------- (2)
 \[ \bullet \] Vitamin C content in one kg of food
Food X contains \[ \to 3 \times x\] units
Food Y contains \[ \to 1 \times y\] units
Total requirement is \[ \to \] at least 8
 \[\therefore 3x + y \geqslant 8\] ----- (3).
Also we have, \[x \geqslant 0\] and \[y \geqslant 0\] .
As, we need to find the minimize cost of the mixture, we use the function: Minimize Z
Now food X cost \[ \to \] Rs. 16
Food Y cost \[ \to \] Rs. 20.
 \[\therefore Z = 16x + 20y\] . -------- (4)
Combining all constraints:
Min \[Z = 16x + 20y\] subject to the constraints,
 \[x + 2y \geqslant 10\] , \[x + y \geqslant 6\] , \[3x + y \geqslant 8\] , \[x \geqslant 0\] and \[y \geqslant 0\] .
Now, we have

 \[x + 2y \geqslant 10\]

x	0	10
y	5	0

\[x + y \geqslant 6\]

x	0	6
y	6	0

 \[3x + y \geqslant 8\]

x	0	2.66
y	8	0

We find the above values by putting \[x = 0\] , we find the value of \[y\] and putting \[y = 0\] we find the value of \[x\] .
Let’s see for the first inequality,
 \[x = 0\] in inequality (1).
 \[ \Rightarrow 0 + 2y = 10\]
 \[ \Rightarrow y = \dfrac{{10}}{2}\]
 \[ \Rightarrow y = 5\]
Now put \[y = 0\]
 \[ \Rightarrow x + 2 \times 0 = 10\]
 \[ \Rightarrow x = 10\] .
Similarly we can find for the other two inequality and listed in the above tabular column.
Now let’s draw the graph,
Scale: X-axis=1 unit =2 units
  Y-axis=1 unit = 2 units.
Since all the inequalities are ‘greater than equal to’ it is obvious that we shade the above the line.

Now from the graph we have,

Corner Points	Value of Z
(0, 8)	160
(1, 5)	116
(2, 4)	112
(10, 0)	120

We obtained the value of Z by substituting x and y value in the inequality (4).
Lets see for corner points (2, 4)
 \[Z = 16x + 20y\] , put \[x = 2\] and \[y = 4\]
 \[ \Rightarrow Z = (16 \times 2) + (20 \times 4)\]
 \[ \Rightarrow Z = 32 + 80\]
 \[ \Rightarrow Z = 112\] . Similarly we can find other corner points and are listed in the above tabular column.
We can see the feasible region is unbounded.
Hence 112 may or may not be the minimum value of Z.
For this we need to draw graph inequality:
 \[16x + 20y < 112\] (Divide by 4)
 \[4x + 5Y < 28\] .

x	7	0
y	0	5.6

Since, there is no point in common Between the Feasible region and inequality.
Hence, the minimum value of Z is 112.
We have a minimum value at the corner points (2, 4).
Hence the cost of mixture will be minimum if 2 packets of food X and 4 packets of Food Y are used. Minimum cost is Rs. 112
So, the correct answer is “112”.

Note: If we see in the above graph we did not shade on the negative X-axis and negative Y-axis. This is because \[x \geqslant 0\] and \[y \geqslant 0\] . Feasible means shaded region. We converted the given word problem into inequalities and then we solved it. Careful while drawing a graph, as we can see above the problem depends on the graph we drawn.