Let no. of tablet $X=x$

no. of tablet $Y=y$

Let $Z$ be the total cost

$Z=2 x+y \text { (Minimize) }$

Subject to

$6 x+2 y \geqslant 18 \text { ie. } 3 x+y \geqslant 9 \text { (iron constraints) }$

Linear constraints

$\begin{aligned} & 3 x+y \geqslant 9 --(i) \\ & x+y \geqslant 7 --(ii) \\ & x+2 y \geqslant 8 --(iii) \end{aligned}$

Associated equations for (i), (ii) (iii) are

$3 x+y=9 \Rightarrow \frac{x}{3}+\frac{y}{9}=1$

$x+y=7 \Rightarrow \frac{x}{7}+\frac{y}{7}=1$

$x+2 y=8 \Rightarrow \frac{x}{8}+\frac{y}{4}=1$

Origin test for (i) $3 \times 0+0=0 \geqslant 9 \text { (false) }$

$\therefore$ solution region for (i) does'nt include origin.

origin test for (ii) $0+0=0 \geqslant 7 \text { (false) }$

$\therefore$ Solution region of (ii) does'nt include origin.

orign test for (iv) $0+2 \times 0=0 \geqslant 8 \text { (false) }$

origin does not include in solution region.

Corner points are

A(8,0), B(6,1), C(1,6) & D(0,9)

$\therefore$ value of $Z$ at corner points

$Z_A=2 \times 8+0=16$

$Z_B=2 \times 6+1=13$

$Z_C=2 \times 1+6=8 \rightarrow$ { Minimum.}

$Z_D=2 \times 0+9=9$

$\begin{aligned} & z<8 \\ \Rightarrow & 2 x+y<8 \end{aligned}$

Since Half plane $2 x+y<8$ have no any common point with the open feasible region.

$\therefore Z_c=8$ will be the minimum value of $Z$ at $C(1,6)$

$\therefore$ No. of tablet $X=1$

No. of tablet $Y=6$

A dietician has to develop a special diet using two foods $P$ and $Q$. Each packet of food $P$ contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of food Q contains 3 unit of calcium ,20 units of iron, 4 units of cholestrol and 3 units of vitamin A.

The diet requires atleast 240 units 4 calcium, at least 460 units of iron and atmost 300 units of cholestrol.

How many packets of each food should be used to minimize the amount of Vit A?

What is the minimum amount of Vit A?

Let the no. of food packet P = x

and the no. of food packet Q = y

Objective function for Vit. A

$Z=6 x+3 y \text { (Minimize) }$

Subject to

$12 x+3 y \geqslant 240 \text { i.e. } 4 x+y \geqslant 80$ (calcium constraints)

$4 x+20 y \geqslant 460 \text { i.e. } x+5 y \geqslant 115$ (Iron constraints)

$6 x+4 y \leqslant 300$ i.e. $3 x+2 y \leqslant 150$ (cholestrol constraints)

$x \geqslant 0, y \geqslant 0$ (non-negative constraints)

Linear Constraints are

$\begin{aligned} & 4 x+y \geqslant 80 \text { - (i) } \\ & x+5 y \geqslant 115 \text { - (ii) } \\ & 3 x+2 y \leqslant 150 \text { - (iii) } \end{aligned}$

Associated eqn for (i), (ii) & (vii) are

$4 x+y=80 \Rightarrow \frac{x}{20}+\frac{y}{80}=1$

$x+5 y=115 \Rightarrow \frac{x}{115}+\frac{y}{23}=1$

$3 x+2 y=150 \Rightarrow \frac{x}{50}+\frac{y}{75}=1$

origin test for (i)

$4 \times 0+0=0 \geqslant 80 \text { (false) }$

origin does not include in the region.

origin test tor (ii)

$0+5 \times 0=0 \geqslant 115 \text { (false) }$

origin does not include in the region.

Origin test for(ii)

$3 \times 0+2 \times 0=0 \leqslant 150 \text { (true) }$

$\therefore$ origin iclude in the solution region.

$\therefore$ Feasible region $A B C$ is bounded and convex.

$\therefore$ Corner points are

$A(15,20), B(40,15), C(2,72)$

value of Z(= 6x + 3y) at corner points

$Z_A=6 \times 15+3 \times 20=90+60=150$

$Z_B=6 \times 40+3 \times 15=240+45=285$

$Z_C=6 \times 2+3 \times 72=12+216=228$

Minimum value of $Z$ exist at

$A(15,20)$

$\text { i.e. } Z_{\text {min }}=150$

Solution: Let fertilizer of type $A$ used $=x \mathrm{~kg}$.

and fertilizer of type $B$ used =$y$ kg.

Let fertilizer A used = x kg

Let fertilizer B used = y kg

Total cost,

$Z=10 x+8 y \text { (Minimize) }$

Linear Constraints are

$3 x+y \geqslant 300 ....(i)$

$x+y \geqslant 240 ....(ii)$

Associated eqn for (i) & (ii)

$3 x+y=300 \Rightarrow \frac{x}{100}+\frac{y}{300}=1$

$x+y=240 \Rightarrow \frac{x}{240}+\frac{y}{240}=1$

Origin test for (i) $3 \times 0+0=0 \geqslant 300 \text { (False) }$

Origin- does not include

origin test for (ii) $0+0=0 \geqslant 240 \text { (false) }$

$\therefore$ orign does not include

Feasible region is open region.

Corner points are

$A(240,0), B(30,210), C(0,300)$

$Z=10 k+8 y$

$Z_A=10 \times 40+8 \times 0=2400$

$Z_B=10 \times 30+8 \times 210=1980 \rightarrow { Min. }$

$Z_C=10 \times 0+8 \times 300=2400$

Since $Z_B=1980$

$\therefore 10 x+8 y<1980$

Half plane $10 x+8 y<1980$ does'nt include any point of open feasible region.

Min of Z exist.

$\therefore Z_{\text {min. }}=1980$

fertilizer $A$ used $=30 \mathrm{~kg}$.

fertilizer $B$ used $=210 kg$.