Question

A diet of a sick person must contain at least 48 units of vitamin A and 64units of vitamin B. Two foods F1 and F2 are available. Food F1 costs Rs 6 per unit and food F2 costs Rs 10 per unit. One unit of food F1 contains 6 units of vitamin A and 7 units of vitamin B. One unit of food F2 contains 8 units of vitamin A and 12 units of vitamin B. Find the minimum cost for the diet that consists of mixture of these two foods and also meeting the minimum nutritional requirements

Answer 1

Hint: First assume let x units of food F1 is required and y units of food F2 is required. And also assume units of vitamin A as A and units of vitamin B as B then make two equations one for food F1 and other for food F2. Now, write an equation which consists of a mixture of foods F1 and F2 and apply the minimum vitamins requirement condition.

Complete step-by-step answer:
Let us assume x units of food F1 is required and y units of F2 is required.
Also assume “A” units of vitamin A and “B” units of vitamin B are required.

It is given that food F1 contains 6 units of vitamin A and 7 units of vitamin B so writing this in the form of equation as follows:
F1 = 6A + 7B
Food F2 contains 8 units of vitamin A and 12 units of vitamin B so writing this in the form of equation as follows:
F2 = 8A + 12B
Now, it is also given that the minimum vitamin requirement for a sick person is 48 units of vitamin A and 64 units of vitamin B so applying this condition to the mixture of foods F1 and F2 we get in the form of equation as follows:
$x{{F}_{1}}+y{{F}_{2}}\ge 48A+64B$
Substituting the values of F1 and F2 in the above equation we get,
$\begin{align}
  & x\left( 6A+7B \right)+y\left( 8A+12B \right)\ge 48A+64B \\
 & \Rightarrow \left( 6x+8y \right)A+\left( 7x+12y \right)B\ge 48A+64B \\
\end{align}$
We are applying the minimum requirement criteria so ignore the inequality and then equating the coefficients of A and B in the above equation we get,
$\begin{align}
  & 6x+8y=48 \\
 & 7x+12y=64 \\
\end{align}$
We can write the equation 6x + 8y = 48 as 3x + 4y = 24. So, now we have two simultaneous equations to solve:
$\begin{align}
  & 3x+4y=24............Eq.(1) \\
 & 7x+12y=64...........Eq.(2) \\
\end{align}$
Multiply eq. (1) by 3 and then subtract eq. (2) from this multiplied eq. (1).
$\begin{align}
  & \text{ }9x+12y=72 \\
 & \dfrac{-7x-12y=-64}{2x+0=8} \\
\end{align}$
From the above calculation, x = 4. Substituting this value of x in eq. (1) we get,
$\begin{align}
  & 3\left( 4 \right)+4y=24 \\
 & \Rightarrow 4y=12 \\
 & \Rightarrow y=3 \\
\end{align}$
From the above calculation, we get x = 4 and y = 3.
Cost of per unit of food F1 is Rs 6 so cost of 4 units of food F1 is Rs 24.
Cost of per unit of food F2 is Rs 10 so cost of 3 units of food F2 is Rs 30.
Hence, the minimum cost of diet fulfilling minimum requirements is Rs 54 (24 + 30).

Note: In the above question we have used the inequality ≥ because the diet contains at least 48 units of vitamin A and 64 units of vitamin B so the equation having inequality ≥ means equal to or greater than.
The meaning of “at least x” is minimum x or greater than x.