23. A farmer mixes two brands, P and Q of cattle feed. Brand P, costing ?. 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ?. 200 per bag, contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements for nutrients A, B and C are 18 units, 45 units and 24 units, respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag. What is the minimum cost of the mixture per bag?

Let the farmer mix x bags of brand P and y bags of brand Q, respectively

The given information can be compiled in a table as given below:

	Vitamin A (units/kg)	Vitamin B (units/kg)	Vitamin C (units/kg)	Cost (Rs/kg)
Food P	3	2.5	2	250
Food Q	1.5	11.25	3	200
Requirement (units/kg)	18	45	24

The given problem can be formulated as given below:

$\begin{array}{}\end{array}Minimisez=250x+200y\dots \dots \dots \dots \left(i\right)$

$3x+1.5y\ge 18\dots \dots \dots \dots ..\left(ii\right)$

$2.5x+11.25y\ge 45\dots \dots \dots ..\left(iii\right)$

$2x+3y\ge 24\dots \dots \dots \dots ..\left(iv\right)$

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

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

A (18, 0), B (9, 2), C (3, 6) and D (0, 12) are the corner points of the feasible region.

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

Corner Point	z = 250x + 200y
A (18, 0)	4500
B (9, 2)	2650
C (3, 6)	1950	Minimum
D (0, 12)	2400

Here, the feasible region is unbounded; hence, 1950 may or may not be the minimum value of z.

For this purpose, we draw a graph of the inequality, $250x+200y<1950or5x+4y<39$ , and check whether the resulting half-plane has points in common with the feasi

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Let the farmer mix x bags of brand P and y bags of brand Q, respectively

The given information can be compiled in a table as given below:

	Vitamin A (units/kg)	Vitamin B (units/kg)	Vitamin C (units/kg)	Cost (Rs/kg)
Food P	3	2.5	2	250
Food Q	1.5	11.25	3	200
Requirement (units/kg)	18	45	24

The given problem can be formulated as given below:

$\begin{array}{}\end{array}Minimisez=250x+200y\dots \dots \dots \dots \left(i\right)$

$3x+1.5y\ge 18\dots \dots \dots \dots ..\left(ii\right)$

$2.5x+11.25y\ge 45\dots \dots \dots ..\left(iii\right)$

$2x+3y\ge 24\dots \dots \dots \dots ..\left(iv\right)$

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

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

A (18, 0), B (9, 2), C (3, 6) and D (0, 12) are the corner points of the feasible region.

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

Corner Point	z = 250x + 200y
A (18, 0)	4500
B (9, 2)	2650
C (3, 6)	1950	Minimum
D (0, 12)	2400

Here, the feasible region is unbounded; hence, 1950 may or may not be the minimum value of z.

For this purpose, we draw a graph of the inequality, $250x+200y<1950or5x+4y<39$ , and check whether the resulting half-plane has points in common with the feasible region or not.

Here, it can be seen that the feasible region has no common point with $5x+4y<39$ .

Hence, at point (3, 6), the minimum value of z is 1950.

Therefore, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost to ?. 1950.

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<p>Let the farmer mix x bags of brand P and y bags of brand Q, respectively</p><p>The given information can be compiled in a table as given below:</p><table><tbody><tr><td>&nbsp;</td><td><p>Vitamin A (units/kg)</p></td><td><p>Vitamin B (units/kg)</p></td><td><p>Vitamin C (units/kg)</p></td><td><p>Cost (Rs/kg)</p></td></tr><tr><td><p>Food P</p></td><td><p>3</p></td><td><p>2.5</p></td><td><p>2</p></td><td><p>250</p></td></tr><tr><td><p>Food Q</p></td><td><p>1.5</p></td><td><p>11.25</p></td><td><p>3</p></td><td><p>200</p></td></tr><tr><td><p>Requirement (units/kg)</p></td><td><p>18</p></td><td><p>45</p></td><td><p>24</p></td><td>&nbsp;</td></tr></tbody></table><p>The given problem can be formulated as given below:</p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"></mtable><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>2</mn><mn>5</mn><mn>0</mn><mi>x</mi><mo>+</mo><mn>2</mn><mn>0</mn><mn>0</mn><mi>y</mi><mo>…</mo><mo>…</mo><mo>…</mo><mo>…</mo><mrow><mo>(</mo><mrow><mi>i</mi></mrow><mo>)</mo></mrow></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mn>3</mn><mi>x</mi><mo>+</mo><mn>1</mn><mo>.</mo><mn>5</mn><mi>y</mi><mo>≥</mo><mn>1</mn><mn>8</mn><mo>…</mo><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><mn>2</mn><mo>.</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>1</mn><mn>1</mn><mo>.</mo><mn>2</mn><mn>5</mn><mi>y</mi><mo>≥</mo><mn>4</mn><mn>5</mn><mo>…</mo><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>2</mn><mi>x</mi><mo>+</mo><mn>3</mn><mi>y</mi><mo>≥</mo><mn>2</mn><mn>4</mn><mo>…</mo><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><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/1732774062phprRm6ew_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732774062phprRm6ew.jpeg" alt="" width="259" height="198"></picture></div></div><p>A (18, 0), B (9, 2), C (3, 6) and D (0, 12) 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 = 250x + 200y</p></td><td>&nbsp;</td></tr><tr><td><p>A (18, 0)</p></td><td><p>4500</p></td><td>&nbsp;</td></tr><tr><td><p>B (9, 2)</p></td><td><p>2650</p></td><td>&nbsp;</td></tr><tr><td><p>C (3, 6)</p></td><td><p>1950</p></td><td><p>Minimum</p></td></tr><tr><td><p>D (0, 12)</p></td><td><p>2400</p></td><td>&nbsp;</td></tr></tbody></table><p>Here, the feasible region is unbounded; hence, 1950 may or may not be the minimum value of z.</p><p>For this purpose, we draw a graph of the inequality,&nbsp;<span title="Click to copy mathml"><math><mrow><mn>2</mn><mn>5</mn><mn>0</mn><mi>x</mi><mo>+</mo><mn>2</mn><mn>0</mn><mn>0</mn><mi>y</mi><mo>&lt;</mo><mn>1</mn><mn>9</mn><mn>5</mn><mn>0</mn><mi>o</mi><mi>r</mi><mn>5</mn><mi>x</mi><mo>+</mo><mn>4</mn><mi>y</mi><mo>&lt;</mo><mn>3</mn><mn>9</mn></mrow></math></span>&nbsp;, and check whether the resulting half-plane has points in common with the feasible region or not.</p><p>Here, it can be seen that the feasible region has no common point with&nbsp;<span title="Click to copy mathml"><math><mrow><mn>5</mn><mi>x</mi><mo>+</mo><mn>4</mn><mi>y</mi><mo>&lt;</mo><mn>3</mn><mn>9</mn></mrow></math></span>&nbsp;.</p><p>Hence, at point (3, 6), the minimum value of z is 1950.</p><p>Therefore, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost to ?. 1950.</p>

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