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A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table below. \[\] $\begin{matrix} \text{kg per bag:}& \text{Brand P} &\text{Brand Q} \\ \text {Nitrogen} &3 & 3.5\\ \text {Phosphoric Acid}& 1 & 2\\ \text {Potash}&3 &1.5 \\ \text {Chlorine}& 1.5 & 2 \end{matrix}$ \[\] If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

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  • First formulate the objective function and identify the constraints from the problem statement, To solve a Linear Programming problem graphically, first plot the constraints for the problem. This is done by plotting the boundary lines of the constraints and identifying the points that will satisfy all the constraints.
  • Let $R$ be the feasible region for a linear programming problem and let $Z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
  • If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Let $x$ bags of brand P and y bags of brand Q are needed to minimize nitrogen contents.
$\therefore$ the objective function is to minimize.
We need at least 240kg of phosphoric acid
$\therefore x+2y\geq 240$
We require at least of 270kg of potash
$\therefore 3x+1.5y\geq 270$
We require at most of 310kg of chlorine
$\therefore 1.5x+2y\leq 310$
Hence the objective function is $Z=3x+3.5y$
$x+2y\geq 240,3x+1.5y\geq 270,1.5x+2y\leq 310,x,y\geq 0$
Step 2:
Now let us draw the graph for the above equations,
(i) the line $x+2y=240$ passes through the point $A(240,0)$ and $(0,120)$
Put $x=0,y=0$ in $x+2y\geq 240$,we get $0\geq 240$ which is not true.
This implies the region $x+2y\geq 240$ lies on and above AB
(ii) Consider the line $3x+1.5y \geq 270$
Clearly this passes through $C(90,0),D(0,180)$
Put $x=0,y=0$,we get
$0\geq 270$ which is not true.
This implies the region $3x+1.5y\geq 270$ lies on CD and above it.
(iii)$1.5x+2y\leq 310$
Put $x=0,y=0$
$\Rightarrow 0\leq 310$ which is not true.
$\Rightarrow$ region $1.5x+2y \leq 310$ lies on and below EF
(iv) $x\geq 0$ lies on and to the right of $y$-axis
(v) $y\geq 0$ lies on and above the $x$-axis.
Step 3:
The shaded area $P$ and $R$ represents the feasible region.
Now $P$ is the point of intersection of the lines,
Let us solve the above equation to find the co-ordinate of P
Multiply equ(2) by 2
Therefore $x=20$
Hence the coordinate of P are (20,140)
Step 4:
$Q$ is the point of intersection of the line
Multiply equ(3) by 3
$\therefore x=40$
Hence the coordinates of $Q$ is $(40,100)$
Step 5:
$R$ is the point of intersection of lines
$\therefore y=50$
Hence the coordinates of $R$ is $(140,50)$
Hence the corner points of the feasible region is $P(20,140),Q(40,100),R(140,50)$
Step 6:
Now let us calculate the minimum value of $Z$ as follows :
At the points (x,y) the value of the objective function subject to $Z=3x+3.5y$
At point $P(20,140)$ the value of the objective function $Z=3\times 20+3.5\times 140=550$
At point $Q(40,100)$ the value of the objective function $Z=3\times 40+3.5\times 100=470$
At point $R(140,50)$ the value of the objective function $Z=3\times 140+3.5\times 50=595$
Hence $Z$ is maximum at $R(140,50)$
To maximize the amount of nitrogen,140 bags of brand P and 50 bags of brand Q are required.
Maximum amount of nitrogen required=595kgs.
answered Aug 29, 2013 by sreemathi.v

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