11. Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ?. 60/kg and Food Q costs ?. 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while Food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.

Let the mixture contain x kg of food P and y kg of food Q. Therefore, x ≥ 0 and y ≥ 0

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

	Vitamin A (units/kg)	Vitamin B (units/kg)	Cost (Rs/kg)
Food P	3	5	60
Food Q	4	2	80
Requirement (units/kg)	8	11

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Therefore, the constraints are

$3x + 4y \ge 8$

$5x + 2y \ge 11$

Total cost, Z, of purchasing food is, $Z=60x +80y$

The mathematical formulation of the given problem is

Minimise $Z=60x +80y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}3x + 4y \ge 8 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}5x + 2y \ge 11 \dots \left(3\right)\hfill \end{array}$

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

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 of the feasible region are A(8/3,0) ,B(2,1/2) and C(0,11/2)

The values of Z at these

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Let the mixture contain x kg of food P and y kg of food Q. Therefore, x ≥ 0 and y ≥ 0

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

	Vitamin A (units/kg)	Vitamin B (units/kg)	Cost (Rs/kg)
Food P	3	5	60
Food Q	4	2	80
Requirement (units/kg)	8	11

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Therefore, the constraints are

$3x + 4y \ge 8$

$5x + 2y \ge 11$

Total cost, Z, of purchasing food is, $Z=60x +80y$

The mathematical formulation of the given problem is

Minimise $Z=60x +80y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}3x + 4y \ge 8 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}5x + 2y \ge 11 \dots \left(3\right)\hfill \end{array}$

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

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 of the feasible region are A(8/3,0) ,B(2,1/2) and C(0,11/2)

The values of Z at these corner points are as follows.

As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.

For this, we graph the inequality, $60x +80y <160or3x +4y <8$ , 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 $3x +4y <8$

Therefore, the minimum cost of the mixture will be ? 160 at the line segment joining the points (8/3,0) and (2,1/2).

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<p>Let the mixture contain x kg of food P and y kg of food Q. Therefore, x&nbsp;≥ 0 and&nbsp;y&nbsp;≥ 0</p><p>The given information can be compiled in a table as follows.</p><table><tbody><tr><td><p>&nbsp;</p></td><td><p>Vitamin A (units/kg)</p></td><td><p>Vitamin B (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>5</p></td><td><p>60</p></td></tr><tr><td><p>Food Q</p></td><td><p>4</p></td><td><p>2</p></td><td><p>80</p></td></tr><tr><td><p>Requirement (units/kg)</p></td><td><p>8</p></td><td><p>11</p></td><td><p>&nbsp;</p></td></tr></tbody></table><p>The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Therefore, the constraints are</p><p><span title="Click to copy mathml"><math><mn>3</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>4</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<mn>8</mn>&nbsp;</math><br></span></p><p><span title="Click to copy mathml"><math><mn>5</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>2</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<mn>1</mn><mn>1</mn>&nbsp;</math><br><br></span></p><p>Total cost, Z, of purchasing food is,&nbsp;<span title="Click to copy mathml"><math><mrow><mi>Z</mi><mo>=</mo><mn>6</mn><mn>0</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>8</mn><mn>0</mn><mi>y</mi></mrow></math></span></p><p>The mathematical formulation of the given problem is</p><p>Minimise&nbsp;<span title="Click to copy mathml"><math><mrow><mi>Z</mi><mo>=</mo><mn>6</mn><mn>0</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>8</mn><mn>0</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>3</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>4</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<mn>8</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>5</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>2</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<mn>1</mn><mn>1</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/1732773556php4AvE1d_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732773556php4AvE1d.jpeg" alt="" width="343" height="336"></picture></div></div><p>It can be seen that the feasible region is unbounded.</p><p>The corner points of the feasible region are A(8/3,0) ,B(2,1/2) and C(0,11/2)</p><p>The values of Z at these corner points are as follows.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1732773574php94JCUr_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732773574php94JCUr.jpeg" alt="" width="301" height="261"></picture></div></div><p>As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.</p><p>For this, we graph the inequality,&nbsp;<span title="Click to copy mathml"><math><mrow><mn>6</mn><mn>0</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>8</mn><mn>0</mn><mi>y</mi><mo>&nbsp;</mo><mo>&lt;</mo><mn>1</mn><mn>6</mn><mn>0</mn><mi>o</mi><mi>r</mi><mn>3</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>4</mn><mi>y</mi><mo>&nbsp;</mo><mo>&lt;</mo><mn>8</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&nbsp;<span title="Click to copy mathml"><math><mrow><mn>3</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>4</mn><mi>y</mi><mo>&nbsp;</mo><mo>&lt;</mo><mn>8</mn></mrow></math></span></p><p>Therefore, the minimum cost of the mixture will be ? 160 at the line segment joining the points (8/3,0) and (2,1/2).</p>

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