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Reshma wishes to mix two types of food $P$ and $Q$ in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin $A$ and 11 units of vitamin B. Food $P$ costs $₹ 60 / \mathrm{kg}$ and Food $Q$ costs $₹ 80 / \mathrm{kg}$. Food $P$ contains 3 units/kg of Vitamin A and 5 units/ kg of Vitamin B while food $Q$ contains 4 units/kg of Vitamin $A$ and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
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Let $\mathrm{x} \mathrm{kg}$ of food $\mathrm{P}$ and $\mathrm{y} \mathrm{kg}$ of food $\mathrm{Q}$ are mixed, we have
Quantity of vitamin A in $\mathrm{x} \mathrm{kg}$ food $\mathrm{P}$ and $\mathrm{y} \mathrm{kg}$ food $\mathrm{Q}=3 \mathrm{x}+4 \mathrm{y}$
Quantity of vitamin required at least $=8$ units $\Rightarrow \mathrm{Q}=3 \mathrm{x}+4 \mathrm{y} \geq 8$
Quantity of vitamin B in $x \mathrm{~kg}$ food $P$ and $\mathrm{y} \mathrm{kg}$ of food $\mathrm{Q}=5 \mathrm{x}+2 \mathrm{y}$
Quantity of vitamin required at least $=11$ units $\Rightarrow 5 \mathrm{x}+2 \mathrm{y} \geq 11$
Cost of $x \mathrm{~kg}$ of food $P$ and $y \mathrm{~kg}$ of food
$\mathrm{Q}=60 \mathrm{x}+80 \mathrm{y}$
$\Rightarrow$ Objective function $Z=60 x+80 y$, subject to Constraints are $3 x+4 y \geq 8 \quad 5 x+2 y \geq 11$ and $x, y \geq 0$
(i) The line $3 \mathrm{x}+4 \mathrm{y}=8$ passes through $\mathrm{A}\left(\frac{8}{3}, 0\right)$ and $B(0,2)$ putting $x=0, y=0$ in $3 \mathrm{x}+4 \mathrm{y} \geq 8$ we get $0 \geq 8$ which is not true
$\Rightarrow$ The region represented by $3 x+4 y \geq 8$ lies on and above AB.
(ii) The line $5 \mathrm{x}+2 \mathrm{y}=11$ passes through $\mathrm{C}\left(\frac{11}{5}, 0\right)$ and $\mathrm{D}\left(0, \frac{11}{2}\right)$. Putting $x=0, y=0$ in $5 x+2 y \geq 11,0 \geq 11$ which is not true
$\Rightarrow$ Region $5 \mathrm{x}+2 \mathrm{y} \geq 11$ lies on and above CD.
(iii) $x \geq 0$ is the region on and to the right of $y$-axis.
(iv) $y \geq 0$ is the region on and above $x$-axis. The shaded area represents the feasible region YDPAX, where $P$ is the intersection of $\mathrm{AB}$ and $\mathrm{CD}$. These lines are
Now objective function $\mathrm{Z}=60 \mathrm{x}+80 \mathrm{y}$,
At $\mathrm{D}\left(0, \frac{11}{2}\right), \mathrm{Z}=80 \times \frac{11}{2}=440$
At $P\left(2, \frac{1}{2}\right), \quad Z=120 \times 40=160 \min ^{\mathrm{m}}$ cost,
$\operatorname{At~} \mathrm{A}\left(\frac{8}{3}, 0\right), \quad \mathrm{Z}=60 \times \frac{8}{3}+0=160 \rightarrow \min ^{\mathrm{m}}$ cost
Minimum value of $Z=160$. But feasible region is unbounded
$\therefore \quad$ Consider the inequality $60 \mathrm{x}+80 \mathrm{y} < 160$
or $3 x+4 y < 8$
This shows that this region is below the line
AD : $3 x+4 y=8$
$\Rightarrow$ There is no point in common between feasible region and $3 x+4 y < 8$. Hence the minimum cost $z$ is $₹ 160$
at all points lying on the segment joining $\mathrm{A}\left(\frac{8}{3}, 0\right)$ and $\mathrm{P}\left(2, \frac{1}{2}\right)$.
Quantity of vitamin A in $\mathrm{x} \mathrm{kg}$ food $\mathrm{P}$ and $\mathrm{y} \mathrm{kg}$ food $\mathrm{Q}=3 \mathrm{x}+4 \mathrm{y}$
Quantity of vitamin required at least $=8$ units $\Rightarrow \mathrm{Q}=3 \mathrm{x}+4 \mathrm{y} \geq 8$
Quantity of vitamin B in $x \mathrm{~kg}$ food $P$ and $\mathrm{y} \mathrm{kg}$ of food $\mathrm{Q}=5 \mathrm{x}+2 \mathrm{y}$
Quantity of vitamin required at least $=11$ units $\Rightarrow 5 \mathrm{x}+2 \mathrm{y} \geq 11$
Cost of $x \mathrm{~kg}$ of food $P$ and $y \mathrm{~kg}$ of food
$\mathrm{Q}=60 \mathrm{x}+80 \mathrm{y}$
$\Rightarrow$ Objective function $Z=60 x+80 y$, subject to Constraints are $3 x+4 y \geq 8 \quad 5 x+2 y \geq 11$ and $x, y \geq 0$
(i) The line $3 \mathrm{x}+4 \mathrm{y}=8$ passes through $\mathrm{A}\left(\frac{8}{3}, 0\right)$ and $B(0,2)$ putting $x=0, y=0$ in $3 \mathrm{x}+4 \mathrm{y} \geq 8$ we get $0 \geq 8$ which is not true
$\Rightarrow$ The region represented by $3 x+4 y \geq 8$ lies on and above AB.
(ii) The line $5 \mathrm{x}+2 \mathrm{y}=11$ passes through $\mathrm{C}\left(\frac{11}{5}, 0\right)$ and $\mathrm{D}\left(0, \frac{11}{2}\right)$. Putting $x=0, y=0$ in $5 x+2 y \geq 11,0 \geq 11$ which is not true
$\Rightarrow$ Region $5 \mathrm{x}+2 \mathrm{y} \geq 11$ lies on and above CD.
(iii) $x \geq 0$ is the region on and to the right of $y$-axis.
(iv) $y \geq 0$ is the region on and above $x$-axis. The shaded area represents the feasible region YDPAX, where $P$ is the intersection of $\mathrm{AB}$ and $\mathrm{CD}$. These lines are
Now objective function $\mathrm{Z}=60 \mathrm{x}+80 \mathrm{y}$,
At $\mathrm{D}\left(0, \frac{11}{2}\right), \mathrm{Z}=80 \times \frac{11}{2}=440$
At $P\left(2, \frac{1}{2}\right), \quad Z=120 \times 40=160 \min ^{\mathrm{m}}$ cost,
$\operatorname{At~} \mathrm{A}\left(\frac{8}{3}, 0\right), \quad \mathrm{Z}=60 \times \frac{8}{3}+0=160 \rightarrow \min ^{\mathrm{m}}$ cost
Minimum value of $Z=160$. But feasible region is unbounded
$\therefore \quad$ Consider the inequality $60 \mathrm{x}+80 \mathrm{y} < 160$
or $3 x+4 y < 8$
This shows that this region is below the line
AD : $3 x+4 y=8$
$\Rightarrow$ There is no point in common between feasible region and $3 x+4 y < 8$. Hence the minimum cost $z$ is $₹ 160$
at all points lying on the segment joining $\mathrm{A}\left(\frac{8}{3}, 0\right)$ and $\mathrm{P}\left(2, \frac{1}{2}\right)$.
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