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A diet is to contain at least 80 units of vitamin $A$ and 100 units of minerals. Two foods $F_1$ and $F_2$ are available. Food $F_1$ costs $₹ 4$ per unit food and $F_2$ costs $₹ 6$ per unit.One unit of food $F_1$ contains 3 units of vitamin Aand 4 units of minerals. One unit of food $F_2$ contains 6 units of vitamin $A$ and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
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Let there be $x$ units of food $F_1$ and $y$ units of food $F_2$.
vitamin $\mathrm{A}$ contained in food $\mathrm{F}_1$ and $\mathrm{F}_2=3 \mathrm{x}+6 \mathrm{y}$ least units of vitamin $A=80$ units quantity mineral required $=100$ units
$$
\therefore \quad 4 \mathrm{x}+3 \mathrm{y} \geq 100
$$
L. P. P is to minimize $Z=4 x+6 y$
Subject to Constraints are $3 x+6 y \geq 80,4 x+3 y \geq 100$ and $x, y \geq 0$
(i) The line $3 \mathrm{x}+6 \mathrm{y}=80$ passes through
A $\left(\frac{80}{3}, 0\right), \mathrm{B}\left(0, \frac{40}{3}\right)$ putting $x=0, y=0$, $0 \geq 80$ which is not true.
$\Rightarrow 3 \mathrm{x}+6 \mathrm{y} \geq 80$ lies on and above the line $\mathrm{AB}$.
(ii) The line $4 x+3 y=100$ passes through $\mathrm{C}(25,0), \mathrm{D}\left(0, \frac{100}{3}\right)$ putting $x=0, y=0$ in $4 x+3 y \geq 100,0 \geq 100$ which is not true. $\Rightarrow 4 x+3 y \geq 100$ lies on and above $C D$.
(iii) $x \geq 0$ lies on and to the right of $y$-axis.
(iv) $y \geq 0$ lies on and above $x$-axis.
Shaded area YDPAX is the feasible region which is unbounded where $P$ is the point of intersection of $A B$ and $\mathrm{CD}$ i.e.,
$3 x+6 y=80$
$4 x+3 y=100$
Mulitplying equ. (ii) by 2 and subtracting (i) from it
$5 x=200-80=120 \Rightarrow x=24$
form (i) $72+6 y=80,6 y=8$ or $y=\frac{4}{3}$
Point $\mathrm{P}$ is $\left(24, \frac{4}{3}\right)$, values of $\mathrm{z}=4 \mathrm{x}+6 \mathrm{y}$, At $\mathrm{D}\left(0, \frac{100}{3}\right), \mathrm{z}=0+6 \times \frac{100}{3}=200$ At $\mathrm{P}\left(24, \frac{4}{3}\right), \mathrm{z}=4 \times 24+6 \times \frac{4}{3}$ $=96+8=104 \mathrm{~min}^{\mathrm{m}}$
At $\mathrm{A}\left(\frac{80}{3}, 0\right), \mathrm{z}=4 \times \frac{80}{3}+0=\frac{320}{3}=106 \frac{2}{3}$
Minimum value of $\mathrm{Z}$ in 104 . Feasible region is unbounded.
Consider the inequality
$4 x+6 y < 104$ or $2 x+3 y < 52$.
The line $2 x+3 y=52$ passes through $(26,0),\left(0,17 \frac{1}{3}\right)$. No point of feasible region and
$4 x+6 y < 104$ is common to them.
$\Rightarrow$ Minimum cost of diet is $₹ 104$.
vitamin $\mathrm{A}$ contained in food $\mathrm{F}_1$ and $\mathrm{F}_2=3 \mathrm{x}+6 \mathrm{y}$ least units of vitamin $A=80$ units quantity mineral required $=100$ units
$$
\therefore \quad 4 \mathrm{x}+3 \mathrm{y} \geq 100
$$
L. P. P is to minimize $Z=4 x+6 y$
Subject to Constraints are $3 x+6 y \geq 80,4 x+3 y \geq 100$ and $x, y \geq 0$
(i) The line $3 \mathrm{x}+6 \mathrm{y}=80$ passes through
A $\left(\frac{80}{3}, 0\right), \mathrm{B}\left(0, \frac{40}{3}\right)$ putting $x=0, y=0$, $0 \geq 80$ which is not true.
$\Rightarrow 3 \mathrm{x}+6 \mathrm{y} \geq 80$ lies on and above the line $\mathrm{AB}$.
(ii) The line $4 x+3 y=100$ passes through $\mathrm{C}(25,0), \mathrm{D}\left(0, \frac{100}{3}\right)$ putting $x=0, y=0$ in $4 x+3 y \geq 100,0 \geq 100$ which is not true. $\Rightarrow 4 x+3 y \geq 100$ lies on and above $C D$.
(iii) $x \geq 0$ lies on and to the right of $y$-axis.
(iv) $y \geq 0$ lies on and above $x$-axis.
Shaded area YDPAX is the feasible region which is unbounded where $P$ is the point of intersection of $A B$ and $\mathrm{CD}$ i.e.,
$3 x+6 y=80$
$4 x+3 y=100$
Mulitplying equ. (ii) by 2 and subtracting (i) from it
$5 x=200-80=120 \Rightarrow x=24$
form (i) $72+6 y=80,6 y=8$ or $y=\frac{4}{3}$
Point $\mathrm{P}$ is $\left(24, \frac{4}{3}\right)$, values of $\mathrm{z}=4 \mathrm{x}+6 \mathrm{y}$, At $\mathrm{D}\left(0, \frac{100}{3}\right), \mathrm{z}=0+6 \times \frac{100}{3}=200$ At $\mathrm{P}\left(24, \frac{4}{3}\right), \mathrm{z}=4 \times 24+6 \times \frac{4}{3}$ $=96+8=104 \mathrm{~min}^{\mathrm{m}}$
At $\mathrm{A}\left(\frac{80}{3}, 0\right), \mathrm{z}=4 \times \frac{80}{3}+0=\frac{320}{3}=106 \frac{2}{3}$
Minimum value of $\mathrm{Z}$ in 104 . Feasible region is unbounded.
Consider the inequality
$4 x+6 y < 104$ or $2 x+3 y < 52$.
The line $2 x+3 y=52$ passes through $(26,0),\left(0,17 \frac{1}{3}\right)$. No point of feasible region and
$4 x+6 y < 104$ is common to them.
$\Rightarrow$ Minimum cost of diet is $₹ 104$.
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