Question

A diet of a sick person must contains at least $48$ units of vitamin $A$ and $64$ units of vitamin $B$. Two foods $F_1$ and $F_2$ are available. Food $F_1$ costs Rs. $6$ per unit and Food $F_2$ costs Rs. $10$ per unit. One unit of food $F_1$ contains $6$ units of vitamin $A$ and $7$ units of vitamin $B$. One unit of food $F_2$ 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

Given, Vitamin $A\ge 48$ units and Vitamin $B\ge 64$ units.
$\therefore F_1=6A+7B$ and $F_2=8A+12B$
Let the required number of units for $F_1$ be $x$ and for $F_2$ be $y$.
Thus $x{F}_1+y{F}_2\ge 48A+64B$
$\Rightarrow x(6A+7B)+y(8A+12B)\ge 48A+64B$
$\Rightarrow A(6x+8y)+B(7x+12y)\ge 48A+64B$
By comparing, we get
$6x+8y\ge 48$
$3x+4y\ge 24$ ......... (i)
and $7x+12y\ge 64$ ........ (ii)
To find the intersection, consider equality. Multiplying equation (i) by $3$ and subtracting equation (ii) from it,
$9x+12y=72$
$7x+12y=64$
Thus $2x=8$

$\Rightarrow x=4$
Thus $3\left(4\right)+4y=24$
$\Rightarrow 12+4y=24$
$\Rightarrow 4y=12$
$\Rightarrow y=3$
Also, $F_1=Rs.6$ and $F_2=Rs.10$
Hence, $4F_1+3F_2=Rs\left[4\right(6)+3(10\left)\right]$
$=Rs[24+30]=Rs.54$