20. There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs ?. 6/kg and F2 costs ?. 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Let the farmer buy x kg of fertilizer F₁ and y kg of fertilizer F₂. Therefore,

x ≥ 0 and y ≥ 0

The given information can be complied in a table as follows.

	Nitrogen (%)	Phosphoric Acid (%)	Cost (Rs/kg)
F1 (x)	10	6	6

F2 (y)	5	10	5
Requirement (kg)	14	14

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.

$\therefore 10\mathrm{\%}of x +5\mathrm{\%}of y \ge 14$

$\begin{array}{l}\frac{x}{10}+\frac{y}{20}\ge 14\\ 2x+y\ge 280\end{array}$

F₁ consists of 6% phosphoric acid and F₂  consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.

$\therefore 6\mathrm{\%}of x +10\mathrm{\%}of y \ge 14$

$\begin{array}{l}\frac{6x}{100}+\frac{10y}{100}\ge 14\\ 3x+56y\ge 700\end{array}$

Total cost of fertilizers, $Z=6x +5y$

The mathematical formulation of the given problem is

Minimize $Z=6x +5y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}2x + y \ge 280 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + 5y \ge 700 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

The feasible region determined by the system of constra

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Let the farmer buy x kg of fertilizer F₁ and y kg of fertilizer F₂. Therefore,

x ≥ 0 and y ≥ 0

The given information can be complied in a table as follows.

	Nitrogen (%)	Phosphoric Acid (%)	Cost (Rs/kg)
F1 (x)	10	6	6

F2 (y)	5	10	5
Requirement (kg)	14	14

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.

$\therefore 10\mathrm{\%}of x +5\mathrm{\%}of y \ge 14$

$\begin{array}{l}\frac{x}{10}+\frac{y}{20}\ge 14\\ 2x+y\ge 280\end{array}$

F₁ consists of 6% phosphoric acid and F₂  consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.

$\therefore 6\mathrm{\%}of x +10\mathrm{\%}of y \ge 14$

$\begin{array}{l}\frac{6x}{100}+\frac{10y}{100}\ge 14\\ 3x+56y\ge 700\end{array}$

Total cost of fertilizers, $Z=6x +5y$

The mathematical formulation of the given problem is

Minimize $Z=6x +5y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}2x + y \ge 280 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + 5y \ge 700 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

The feasible region determined by the system of constraints is as follows.

It can be seen that the feasible region is unbounded.

The corner points are  $A\left(\frac{700}{3},0\right),B\left(100,80\right),C\left(0,280\right)$ .

The values of Z at these points are as follows.

 

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality, $6x +5y <1000$ , and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with

$6x +5y <1000$

Therefore, 100 kg of fertiliser F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is ? 1000.

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<p>Let the farmer buy x kg of fertilizer F₁&nbsp;and&nbsp;<em>y</em>&nbsp;kg of fertilizer F₂. Therefore,</p><p>x&nbsp;≥ 0 and&nbsp;y&nbsp;≥ 0</p><p>The given information can be complied in a table as follows.</p><table><tbody><tr><td><p>&nbsp;</p></td><td><p>Nitrogen (%)</p></td><td><p>Phosphoric Acid (%)</p></td><td><p>Cost (Rs/kg)</p></td></tr><tr><td><p>F₁&nbsp;(<em>x</em>)</p></td><td><p>10</p></td><td><p>6</p></td><td><p>6</p></td></tr></tbody></table><table><tbody><tr><td><p>F₂&nbsp;(<em>y</em>)</p></td><td><p>5</p></td><td><p>10</p></td><td><p>5</p></td></tr><tr><td><p>Requirement (kg)</p></td><td><p>14</p></td><td><p>14</p></td><td>&nbsp;</td></tr></tbody></table><p>F₁&nbsp;consists of 10% nitrogen and F₂&nbsp;consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.</p><p><span title="Click to copy mathml"><math><mrow><mo>∴</mo><mn>1</mn><mn>0</mn><mn>%</mn><mi>o</mi><mi>f</mi><mo>&nbsp;</mo><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>5</mn><mn>%</mn><mi>o</mi><mi>f</mi><mo>&nbsp;</mo><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo><mn>1</mn><mn>4</mn></mrow></math></span></p><p><span title="Click to copy mathml"><math><mtable columnalign="left"><mtr><mtd><mrow><mfrac><mrow><mi>x</mi></mrow><mrow><mn>1</mn><mn>0</mn></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>y</mi></mrow><mrow><mn>2</mn><mn>0</mn></mrow></mfrac><mo>≥</mo><mn>1</mn><mn>4</mn></mrow></mtd></mtr><mtr><mtd><mrow><mn>2</mn><mi>x</mi><mo>+</mo><mi>y</mi><mo>≥</mo><mn>2</mn><mn>8</mn><mn>0</mn></mrow></mtd></mtr></mtable></math></span></p><p>F₁&nbsp;consists of 6% phosphoric acid and F₂&nbsp; consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.</p><p><span title="Click to copy mathml"><math><mrow><mo>∴</mo><mn>6</mn><mn>%</mn><mi>o</mi><mi>f</mi><mo>&nbsp;</mo><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>1</mn><mn>0</mn><mn>%</mn><mi>o</mi><mi>f</mi><mo>&nbsp;</mo><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo><mn>1</mn><mn>4</mn></mrow></math></span></p><p><span title="Click to copy mathml"><math><mtable columnalign="left"><mtr><mtd><mrow><mfrac><mrow><mn>6</mn><mi>x</mi></mrow><mrow><mn>1</mn><mn>0</mn><mn>0</mn></mrow></mfrac><mo>+</mo><mfrac><mrow><mn>1</mn><mn>0</mn><mi>y</mi></mrow><mrow><mn>1</mn><mn>0</mn><mn>0</mn></mrow></mfrac><mo>≥</mo><mn>1</mn><mn>4</mn></mrow></mtd></mtr><mtr><mtd><mrow><mn>3</mn><mi>x</mi><mo>+</mo><mn>5</mn><mn>6</mn><mi>y</mi><mo>≥</mo><mn>7</mn><mn>0</mn><mn>0</mn></mrow></mtd></mtr></mtable></math></span></p><p>Total cost of fertilizers,&nbsp;<span title="Click to copy mathml"><math><mrow><mi>Z</mi><mo>=</mo><mn>6</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>5</mn><mi>y</mi></mrow></math></span></p><p>The mathematical formulation of the given problem is</p><p>Minimize&nbsp;<span title="Click to copy mathml"><math><mrow><mi>Z</mi><mo>=</mo><mn>6</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>5</mn><mi>y</mi><mo>&nbsp;</mo><mo>…</mo><mrow><mo>(</mo><mrow><mn>1</mn></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"><mtr columnalign="left"><mtd columnalign="left"><mn>2</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mo>&nbsp;</mo><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<mn>2</mn><mn>8</mn><mn>0</mn>&nbsp;<mo>…</mo>&nbsp;<mrow><mo>(</mo><mrow><mn>2</mn></mrow><mo>)</mo></mrow></mtd></mtr></mtable></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mn>3</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>5</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<mn>7</mn><mn>0</mn><mn>0</mn>&nbsp;<mo>…</mo>&nbsp;<mrow><mo>(</mo><mrow><mn>3</mn></mrow><mo>)</mo></mrow></mtd></mtr></mtable></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mi>x</mi><mo>,</mo><mo>&nbsp;</mo><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<mn>0</mn>&nbsp;<mo>…</mo>&nbsp;<mrow><mo>(</mo><mrow><mn>4</mn></mrow><mo>)</mo></mrow></mtd></mtr></mtable></mrow></math></span></p><p>The feasible region determined by the system of constraints is as follows.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1732773947phpecxUe5_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732773947phpecxUe5.jpeg" alt="" width="277" height="276"></picture></div></div><p>It can be seen that the feasible region is unbounded.</p><p>The corner points are&nbsp;&nbsp;<span title="Click to copy mathml"><math><mrow><mi>A</mi><mrow><mo>(</mo><mrow><mfrac><mrow><mn>7</mn><mn>0</mn><mn>0</mn></mrow><mrow><mn>3</mn></mrow></mfrac><mo>,</mo><mn>0</mn></mrow><mo>)</mo></mrow><mo>,</mo><mi>B</mi><mrow><mo>(</mo><mrow><mn>1</mn><mn>0</mn><mn>0</mn><mo>,</mo><mn>8</mn><mn>0</mn></mrow><mo>)</mo></mrow><mo>,</mo><mi>C</mi><mrow><mo>(</mo><mrow><mn>0</mn><mo>,</mo><mn>2</mn><mn>8</mn><mn>0</mn></mrow><mo>)</mo></mrow></mrow></math></span>&nbsp;.</p><p>The values of Z at these points are as follows.</p><p>&nbsp;</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1732773965phpUs06c9_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732773965phpUs06c9.jpeg" alt="" width="263" height="215"></picture></div></div><p>As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.</p><p>For this, we draw a graph of the inequality,&nbsp;<span title="Click to copy mathml"><math><mrow><mn>6</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>5</mn><mi>y</mi><mo>&nbsp;</mo><mo>&lt;</mo><mn>1</mn><mn>0</mn><mn>0</mn><mn>0</mn></mrow></math></span>&nbsp;, and check whether the resulting half plane has points in common with the feasible region or not.</p><p>It can be seen that the feasible region has no common point with</p><p><span title="Click to copy mathml"><math><mrow><mn>6</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>5</mn><mi>y</mi><mo>&nbsp;</mo><mo>&lt;</mo><mn>1</mn><mn>0</mn><mn>0</mn><mn>0</mn></mrow></math></span></p><p>Therefore, 100 kg of fertiliser F1&nbsp;and 80 kg of fertilizer F2&nbsp;should be used to minimize the cost. The minimum cost is ? 1000.</p>

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