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**Hint:**We assume the number of bags of each brand as different variables. Form three equations using the given data indicating the weight of phosphoric acid, potash and chlorine. Write the equation of weight of nitrogen. Convert all the obtained data in the form of linear programming problems with constraints. Use graph methods to find the vertices of the shaded region. Use the vertices of the shaded region to check the minimum value of nitrogen by substituting the coordinates of the vertices in the minimization equation of nitrogen.

* Linear programming problem (LPP): A LPP contains a set of inequalities as constraints which on solving give us an optimized solution for the problem.

**Complete step-by-step solution:**

Let us assume the number of bags of brand P as ‘x’ and number of bags of brand Q as ‘y’.

We form three equations for each of the chemicals.

Phosphoric Acid:

Weight of brand P phosphoric acid\[ = 1\]kg

Weight of brand Q phosphoric acid\[ = 2\]kg

Since we are given the garden needs at least 240 kg of phosphoric acid.

So the equation will be formed indicating the sum of weight of two brands of phosphoric acid should be greater than 240kg.

\[ \Rightarrow x + 2y \geqslant 240\]...................… (1)

Potash:

Weight of brand P potash\[ = 3\]kg

Weight of brand Q potash\[ = 1.5\]kg

Since we are given the garden needs at least 270 kg of potash.

So the equation will be formed indicating the sum of weight of two brands of potash should be greater than 270kg.

\[ \Rightarrow 3x + 1.5y \geqslant 270\]................… (2)

Chlorine:

Weight of brand P chlorine\[ = 1.5\]kg

Weight of brand Q chlorine\[ = 2\]kg

Since we are given the garden needs at most 310 kg of chlorine.

So the equation will be formed indicating the sum of weight of two brands of chlorine should be less than 310kg.

\[ \Rightarrow 1.5x + 2y \leqslant 310\]...............… (3)

Nitrogen:

Weight of brand P nitrogen\[ = 3\]kg

Weight of brand Q nitrogen\[ = 3.5\]kg

We have to minimize the quantity of nitrogen used.

Let z be the minimum amount of nitrogen used.

\[ \Rightarrow z = 3x + 3.5y\]..............… (4)

Also, we know that \[x \geqslant 0,y \geqslant 0\] as the number of bags of each brand is greater than or equal to 0.

Then from equation (1), (2), (3) and (4)

We can write linear programming problem as

Minimize:\[z = 3x + 3.5y\]with respect to the constraints:

\[x + 2y \geqslant 240\]

\[3x + 1.5y \geqslant 270\]

\[1.5x + 2y \leqslant 310\]

\[x \geqslant 0,y \geqslant 0\]

Now we will solve this linear programming problem using graphical methods.

We will plot equations (1), (2) and (3) on the graph

From equation (1):

\[ \Rightarrow x + 2y = 240\]

Put \[x = 0\]

\[ \Rightarrow 2y = 240\]

\[ \Rightarrow y = 120\]

Point becomes\[(0,120)\]

Now put \[y = 0\]

\[ \Rightarrow x = 240\]

Point becomes \[(240,0)\]

\[ \Rightarrow \]Points joining equation (1) are \[(0,120)\] and \[(240,0)\]................… (4)

From equation (2):

\[ \Rightarrow 3x + 1.5y = 270\]

Put \[x = 0\]

\[ \Rightarrow 1.5y = 270\]

\[ \Rightarrow y = 180\]

Point becomes\[(0,180)\]

Now put \[y = 0\]

\[ \Rightarrow 3x = 270\]

\[ \Rightarrow x = 90\]

Point becomes\[(90,0)\]

\[ \Rightarrow \]Points joining equation (2) are \[(0,180)\] and \[(90,0)\]..............… (5)

From equation (3):

\[ \Rightarrow 1.5x + 2y = 310\]

Put \[x = 0\]

\[ \Rightarrow 2y = 310\]

\[ \Rightarrow y = 155\]

Point becomes\[(0,155)\]

Now put \[y = 0\]

\[ \Rightarrow 1.5x = 310\]

\[ \Rightarrow x = 206.66\]

Point becomes\[(206.66,0)\]

\[ \Rightarrow \]Points joining equation (3) are \[(0,155)\] and \[(206.66,0)\]..............… (6)

Plot the equations with the help of (4), (5) and (6)

Red line: Equation (1); Green line: Equation (2); Yellow line: Equation (3)

Now we shade the region from each equation by checking if the point of origin lies in the shaded region or not.

\[\because x + 2y \geqslant 240\]

Put \[x = 0,y = 0\]

\[ \Rightarrow 0 \geqslant 240\]

This is false. So shaded region is away from origin

\[\because 3x + 1.5y \geqslant 270\]

Put \[x = 0,y = 0\]

\[ \Rightarrow 0 \geqslant 270\]

This is false. So shaded region is away from origin

\[\because 1.5x + 2y \leqslant 310\]

Put \[x = 0,y = 0\]

\[ \Rightarrow 0 \leqslant 310\]

This is true. So shaded region is towards the origin

Now we find the endpoints of the shaded area.

For point A:

At point A, lines of equation (2) and equation (3) intersect

Equation (2): \[3x + 1.5y = 270\]

Shift all values except with variable y to one side of the equation

\[ \Rightarrow 1.5y = 270 - 3x\]

Divide both sides by 1.5

\[ \Rightarrow \dfrac{{1.5y}}{{1.5}} = \dfrac{{270 - 3x}}{{1.5}}\]

Cancel same factors from numerator and denominator on both sides of the equation

\[ \Rightarrow y = 180 - 2x\]

Equation (3): \[1.5x + 2y = 310\]

Substitute the value of \[y = 180 - 2x\]in equation (3)

\[ \Rightarrow 1.5x + 2(180 - 2x) = 310\]

\[ \Rightarrow 1.5x + 360 - 4x = 310\]

Shift similar terms on one side of the equation

\[ \Rightarrow 1.5x - 4x = 310 - 360\]

\[ \Rightarrow - 2.5x = - 50\]

Cancel same sign from both sides

\[ \Rightarrow 2.5x = 50\]

Divide both sides by 2.5

\[ \Rightarrow \dfrac{{2.5x}}{{2.5}} = \dfrac{{50}}{{2.5}}\]

\[ \Rightarrow x = 20\]

\[\therefore y = 180 - 2(20)\]

\[ \Rightarrow y = 180 - 40\]

\[ \Rightarrow y = 140\]

The point of intersection becomes \[(20,140)\]

For point B:

At point A, lines of equation (1) and equation (2) intersect

Equation (2): \[3x + 1.5y = 270\]

Shift all values except with variable y to one side of the equation

\[ \Rightarrow 1.5y = 270 - 3x\]

Divide both sides by 1.5

\[ \Rightarrow \dfrac{{1.5y}}{{1.5}} = \dfrac{{270 - 3x}}{{1.5}}\]

Cancel same factors from numerator and denominator on both sides of the equation

\[ \Rightarrow y = 180 - 2x\]

Equation (1): \[x + 2y = 240\]

Substitute the value of \[y = 180 - 2x\] in equation (1)

\[ \Rightarrow x + 2(180 - 2x) = 240\]

\[ \Rightarrow x + 360 - 4x = 240\]

Shift similar terms on one side of the equation

\[ \Rightarrow x - 4x = 240 - 360\]

\[ \Rightarrow - 3x = - 120\]

Cancel same sign from both sides

\[ \Rightarrow 3x = 120\]

Divide both sides by 3

\[ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{120}}{3}\]

\[ \Rightarrow x = 40\]

\[\therefore y = 180 - 2(40)\]

\[ \Rightarrow y = 180 - 80\]

\[ \Rightarrow y = 100\]

The point of intersection becomes \[(40,100)\]

For point C:

At point A, lines of equation (1) and equation (3) intersect

Equation (1): \[x + 2y = 240\]

Shift all values except with variable x to one side of the equation

\[ \Rightarrow x = 240 - 2y\]

Substitute the value of \[x = 240 - 2y\]in equation (3)

\[ \Rightarrow 1.5(240 - 2y) + 2y = 310\]

\[ \Rightarrow 360 - 3y + 2y = 310\]

Shift similar terms on one side of the equation

\[ \Rightarrow 2y - 3y = 310 - 360\]

\[ \Rightarrow - y = - 50\]

Cancel same sign from both sides

\[ \Rightarrow y = 50\]

\[\therefore \Rightarrow x = 240 - 2(50)\]

\[ \Rightarrow x = 240 - 100\]

\[ \Rightarrow x = 140\]

The point of intersection becomes \[(140,50)\]

We have vertices of shaded region as \[A(20,140);B(40,100);C(140,50)\]

We form a table which tells us the value of z on these points

_{ Points } | _{ }\[z = 3x + 3.5y\] |

_{ }\[A(20,140)\] | \[z = 3 \times 20 + 3.5 \times 140\]\[ \Rightarrow z = 550\] |

_{ }\[B(40,100)\] | \[z = 3 \times 40 + 3.5 \times 100\]\[ \Rightarrow z = 470\] |

_{ }\[C(140,50)\] | \[z = 3 \times 140 + 3.5 \times 50\]\[ \Rightarrow z = 595\] |

So from the table, minimum value of z is at point \[B(40,100)\] i.e. 470

So, the number of bags of brand P is 40 and the number of brands of bag Q is 100.

**\[\therefore \]Nitrogen will be minimum when there are 40 bags from P and 100 bags from Q.**

Also, the minimum amount of nitrogen required in the garden is 470kg.

Also, the minimum amount of nitrogen required in the garden is 470kg.

**Note:**Students many times make mistakes in shading the region by just looking at the sign of ‘less than equal to’ or ‘greater than equal to’. Keep in mind we use a single point like origin to check if we have to shade in that direction or opposite direction.

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