Question

A farmer has a supply of chemical fertilizer of type $A$ which contains $10\%$ nitrogen and $6\%$ phosphoric acid and of type $B$ which contains $5\%$ nitrogen and $10\%$ phosphoric acid. After soil test, it is found that at least $7$ kg of nitrogen and same quantity of phosphoric acid is required for a good crop. The fertilizer of type $A$ costs Rs. $5.00$ per kg and the type $B$ costs Rs. $8.00$ per kg. Using Linear programming, find how many kilograms of each type of the fertilizer should be bought to meet the requirement and for the cost to be minimum. Find the feasible region in the graph.

Answer 1

let the farmer bought $x$ kg of chemical $A$ and $y$ kg of $B$ , so

$x\ge \mathrm{0.....}\left(i\right)y\ge \mathrm{0.....}\left(ii\right)$

Minimun nitrogen required $=7$

$\Rightarrow \frac{10}{100}x+\frac{5}{100}y\ge 7\Rightarrow 2x+y\ge \mathrm{140......}\left(iii\right)$

Minimun phoshoric acid required $=7$ kg

$\Rightarrow \frac{6}{100}x+\frac{10}{100}y\ge 7\Rightarrow 3x+5y\ge \mathrm{350.....}\left(iv\right)$

we have to minimise the cost

$\Rightarrow z=5x+8y$

from the graph critical points are $A(50,40),B(0,140)$ and $C(116.62,0)$

At $A$

$z=5\left(50\right)+8\left(40\right)=570$

At $B$

$z=0+8\left(140\right)=1120$

At $C$

$z=5\left(116.62\right)+0=583$

Cost is minimum at $A$

So the farmer must bought $50$ kg of chemical $A$ and $40$ kg of chemical $B$