19. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs ?. 4 per unit food and F2 costs ?. 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

Let the diet contain x units of food F₁ and y units of food F₂. Therefore,

x ≥ 0 and y ≥ 0

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

	Vitamin A (units)	Mineral (units)	Cost per unit (Rs)
Food F1 (x)	3	4	4
Food F2 (y)	6	3	6
Requirement	80	100

The cost of food F₁ is Rs 4 per unit and of Food F₂  is ? 6 per unit. Therefore, the constraints are

$\begin{array}{}\end{array}3x +6y \ge 80$

$4x +3y \ge 100$

$x, y \ge 0$

$Totalcostofthediet,Z=4x +6y$

The mathematical formulation of the given problem is

Minimise $Z=4x +6y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}3x + 6y \ge 80 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}4x + 3y \ge 100 \dots \left(3\right)\hfill \end{array}$

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

The feasible region determined by the constraints is as follows.

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are  $A\left(\frac{8}{3},0\right),B\left(2,\frac{1}{2}\right),C\left(0,\frac{11}{2}\right)$ .

The corner points are $A\left(\frac{80}{3},0\right),B\left(24,\frac{4}{3}\right),C\left(0,\frac{100}{3}\right)$ .

The values of Z at these corner points ar

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Let the diet contain x units of food F₁ and y units of food F₂. Therefore,

x ≥ 0 and y ≥ 0

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

	Vitamin A (units)	Mineral (units)	Cost per unit (Rs)
Food F1 (x)	3	4	4
Food F2 (y)	6	3	6
Requirement	80	100

The cost of food F₁ is Rs 4 per unit and of Food F₂  is ? 6 per unit. Therefore, the constraints are

$\begin{array}{}\end{array}3x +6y \ge 80$

$4x +3y \ge 100$

$x, y \ge 0$

$Totalcostofthediet,Z=4x +6y$

The mathematical formulation of the given problem is

Minimise $Z=4x +6y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}3x + 6y \ge 80 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}4x + 3y \ge 100 \dots \left(3\right)\hfill \end{array}$

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

The feasible region determined by the constraints is as follows.

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are  $A\left(\frac{8}{3},0\right),B\left(2,\frac{1}{2}\right),C\left(0,\frac{11}{2}\right)$ .

The corner points are $A\left(\frac{80}{3},0\right),B\left(24,\frac{4}{3}\right),C\left(0,\frac{100}{3}\right)$ .

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

As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality, $4x +6y <104or2x +3y <52$ , 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 $2x +3y <52$

Therefore, the minimum cost of the mixture will be ? 104.

less

<p>Let the diet contain x units of food F₁&nbsp;and&nbsp;<em>y</em>&nbsp;units of food 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>Vitamin A (units)</p></td><td><p>Mineral (units)</p></td><td><p>Cost per unit</p><p>(Rs)</p></td></tr><tr><td><p>Food F₁&nbsp;(<em>x</em>)</p></td><td><p>3</p></td><td><p>4</p></td><td><p>4</p></td></tr><tr><td><p>Food F₂&nbsp;(<em>y</em>)</p></td><td><p>6</p></td><td><p>3</p></td><td><p>6</p></td></tr><tr><td><p>Requirement</p></td><td><p>80</p></td><td><p>100</p></td><td><p>&nbsp;</p></td></tr></tbody></table><p>The cost of food F₁&nbsp;is Rs 4 per unit and of Food F₂&nbsp; is ? 6 per unit. Therefore, the constraints are</p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"></mtable><mrow><mn>3</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>6</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo><mn>8</mn><mn>0</mn></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mn>4</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>3</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo><mn>1</mn><mn>0</mn><mn>0</mn></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>x</mi><mo>,</mo><mo>&nbsp;</mo><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo><mn>0</mn></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi>T</mi><mi>o</mi><mi>t</mi><mi>a</mi><mi>l</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>t</mi><mi>o</mi><mi>f</mi><mi>t</mi><mi>h</mi><mi>e</mi><mi>d</mi><mi>i</mi><mi>e</mi><mi>t</mi><mo>,</mo><mi>Z</mi><mo>=</mo><mn>4</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>6</mn><mi>y</mi></mrow></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>4</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>6</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>6</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<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>4</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>3</mn><mi>y</mi><mo>&nbsp;</mo><mo>≥</mo>&nbsp;<mn>1</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 constraints is as follows.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1732773907phpPRmisv_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732773907phpPRmisv.jpeg" alt="" width="292" height="266"></picture></div></div><p>It can be seen that the feasible region is unbounded.</p><p>The corner points of the feasible region are&nbsp;&nbsp;<span title="Click to copy mathml"><math><mrow><mi>A</mi><mrow><mo>(</mo><mrow><mfrac><mrow><mn>8</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>2</mn><mo>,</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mo>,</mo><mi>C</mi><mrow><mo>(</mo><mrow><mn>0</mn><mo>,</mo><mfrac><mrow><mn>1</mn><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow></math></span>&nbsp;.</p><p>The corner points are&nbsp;<span title="Click to copy mathml"><math><mrow><mi>A</mi><mrow><mo>(</mo><mrow><mfrac><mrow><mn>8</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>2</mn><mn>4</mn><mo>,</mo><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mo>,</mo><mi>C</mi><mrow><mo>(</mo><mrow><mn>0</mn><mo>,</mo><mfrac><mrow><mn>1</mn><mn>0</mn><mn>0</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow></math></span>&nbsp;.</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/1732773925phpPn15qX_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732773925phpPn15qX.jpeg" alt="" width="242" height="209"></picture></div></div><p>As the feasible region is unbounded, therefore, 104 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>4</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>6</mn><mi>y</mi><mo>&nbsp;</mo><mo>&lt;</mo><mn>1</mn><mn>0</mn><mn>4</mn><mi>o</mi><mi>r</mi><mn>2</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>3</mn><mi>y</mi><mo>&nbsp;</mo><mo>&lt;</mo><mn>5</mn><mn>2</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>2</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo><mn>3</mn><mi>y</mi><mo>&nbsp;</mo><mo>&lt;</mo><mn>5</mn><mn>2</mn></mrow></math></span></p><p>Therefore, the minimum cost of the mixture will be ? 104.</p>

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