29. A fruit grower can use two types of fertilisers in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table, Tests indicate that the garden needs at least 240 kg of phosphoric acid, 270 kg of potash and at most 310 kg of chlorine.

If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added to the garden?

Kg per Bag
	Brand P	Brand Q
Nitrogen	3	3.5
Phosphoric acid	1	2
Potash	3	1.5
Chlorine	1.5	2

 

Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.

The problem can be formulated as given below:

$Minimisez=3x+3.5y\dots \dots \dots \dots .\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+2y \ge 240\dots \dots \dots ..\left(ii\right)$

$x+0.5y\ge 90\dots \dots ..\left(iii\right)$

$1.5x+2y\le 310\dots \dots \dots ..\left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots .\left(v\right)$

The feasible region determined by the system of constraints is given below:

A (240, 0), B (140, 50) and C (20, 140) are the corner points of the feasible region.

The values of z at these corner points are given below:

Corner Point	z = 3x + 3.5y
A (140, 50)	595
B (20, 140)	550
C (40, 100)	470	Minimum

The maximum value of z is 470 at (40, 100).

Therefore, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.

Hence, the minimum amount of nitrogen added to the garden is 470 kg.

<p>Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.</p><p>The problem can be formulated as given below:</p><p><span title="Click to copy mathml"><math><mrow><mi>M</mi><mi>i</mi><mi>n</mi><mi>i</mi><mi>m</mi><mi>i</mi><mi>s</mi><mi>e</mi><mi>z</mi><mo>=</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>3</mn><mo>.</mo><mn>5</mn><mi>y</mi><mo>…</mo><mo>…</mo><mo>…</mo><mo>…</mo><mo>.</mo><mrow><mo> (</mo><mrow><mi>i</mi></mrow><mo>)</mo></mrow></mrow></math></span></p><p>Subject to the constraints, </p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"></mtable><mrow><mi>x</mi><mo>+</mo><mn>2</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo><mn>2</mn><mn>4</mn><mn>0</mn><mo>…</mo><mo>…</mo><mo>…</mo><mo>.</mo><mo>.</mo><mrow><mo> (</mo><mrow><mi>i</mi><mi>i</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>x</mi><mo>+</mo><mn>0</mn><mo>.</mo><mn>5</mn><mi>y</mi><mo>≥</mo><mn>9</mn><mn>0</mn><mo>…</mo><mo>…</mo><mo>.</mo><mo>.</mo><mrow><mo> (</mo><mrow><mi>i</mi><mi>i</mi><mi>i</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mn>1</mn><mo>.</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>2</mn><mi>y</mi><mo>≤</mo><mn>3</mn><mn>1</mn><mn>0</mn><mo>…</mo><mo>…</mo><mo>…</mo><mo>.</mo><mo>.</mo><mrow><mo> (</mo><mrow><mi>i</mi><mi>v</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>x</mi><mo>, </mo><mi>y</mi><mo>≥</mo><mn>0</mn><mo>…</mo><mo>…</mo><mo>…</mo><mo>…</mo><mo>.</mo><mrow><mo> (</mo><mrow><mi>v</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p>The feasible region determined by the system of constraints is given below:</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1732774230phpVOmomY_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732774230phpVOmomY.jpeg" alt="" width="280" height="269"></picture></div></div><p>A (240, 0), B (140, 50) and C (20, 140) are the corner points of the feasible region.</p><p>The values of z at these corner points are given below:</p><table><tbody><tr><td><p>Corner Point</p></td><td><p>z = 3x + 3.5y</p></td><td>&nbsp;</td></tr><tr><td><p>A (140, 50)</p></td><td><p>595</p></td><td>&nbsp;</td></tr><tr><td><p>B (20, 140)</p></td><td><p>550</p></td><td>&nbsp;</td></tr><tr><td><p>C (40, 100)</p></td><td><p>470</p></td><td><p>Minimum</p></td></tr></tbody></table><p>The maximum value of z is 470 at (40, 100).</p><p>Therefore, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.</p><p>Hence, the minimum amount of nitrogen added to the garden is 470 kg.</p>

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