24. A dietician wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg of food are given below:

Food	Vitamin A	Vitamin B	Vitamin C
X	1	2	3
Y	2	2	1

One kg of food X costs ?. 16, and one kg of food Y costs ?. 20. Find the least cost of the mixture which will produce the required diet.

Let the mixture contain x kg of food X and y kg of food Y, respectively.

The mathematical formulation of the given problem can be written as given below:

$Minimisez=16x+20y\dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+2y\ge 10\dots \dots \dots \dots .\left(ii\right)$

$x+y\ge 6\dots \dots \dots \dots \left(iii\right)$

$3x+y\ge 8\dots \dots \dots \dots .\left(iv\right)$

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

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

A (10, 0), B (2, 4), C (1, 5) and D (0, 8) are the corner points of the feasible region.

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

Corner Point	z = 16x + 20y
A (10, 0)	160
B (2, 4)	112	Minimum

C (1, 5)	116
D (0, 8)	160

Since the feasible region is unbounded, 112 may or may not be the minimum value of z.

For this purpose, we draw a graph of the inequality, $16x+20y<112or4x+5y<28$ , and check whether the resulting half-plane has points in common with the feasible region or not

...more

Let the mixture contain x kg of food X and y kg of food Y, respectively.

The mathematical formulation of the given problem can be written as given below:

$Minimisez=16x+20y\dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+2y\ge 10\dots \dots \dots \dots .\left(ii\right)$

$x+y\ge 6\dots \dots \dots \dots \left(iii\right)$

$3x+y\ge 8\dots \dots \dots \dots .\left(iv\right)$

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

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

A (10, 0), B (2, 4), C (1, 5) and D (0, 8) are the corner points of the feasible region.

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

Corner Point	z = 16x + 20y
A (10, 0)	160
B (2, 4)	112	Minimum

C (1, 5)	116
D (0, 8)	160

Since the feasible region is unbounded, 112 may or may not be the minimum value of z.

For this purpose, we draw a graph of the inequality, $16x+20y<112or4x+5y<28$ , 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 $4x+5y<28$ .

Hence, the minimum value of z is 112 at (2, 4).

Therefore, the mixture should contain 2 kg of food X and 4 kg of food Y.

The minimum cost of the mixture is ? 112.

less

<p>Let the mixture contain x kg of food X and y kg of food Y, respectively.</p><p>The mathematical formulation of the given problem can be written 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>1</mn><mn>6</mn><mi>x</mi><mo>+</mo><mn>2</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></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>≥</mo><mn>1</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><mi>y</mi><mo>≥</mo><mn>6</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>3</mn><mi>x</mi><mo>+</mo><mi>y</mi><mo>≥</mo><mn>8</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"><br><math><mrow><mrow><mrow><mo></mo></mrow></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/1732774085phpXJRkq2_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732774085phpXJRkq2.jpeg" alt="" width="269" height="218"></picture></div></div><p>A (10, 0), B (2, 4), C (1, 5) and D (0, 8) 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 = 16x + 20y</p></td><td>&nbsp;</td></tr><tr><td><p>A (10, 0)</p></td><td><p>160</p></td><td>&nbsp;</td></tr><tr><td><p>B (2, 4)</p></td><td><p>112</p></td><td><p>Minimum</p></td></tr></tbody></table><table><tbody><tr><td><p>C (1, 5)</p></td><td><p>116</p></td><td>&nbsp;</td></tr><tr><td><p>D (0, 8)</p></td><td><p>160</p></td><td>&nbsp;</td></tr></tbody></table><p>Since the feasible region is unbounded, 112 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>1</mn><mn>6</mn><mi>x</mi><mo>+</mo><mn>2</mn><mn>0</mn><mi>y</mi><mo>&lt;</mo><mn>1</mn><mn>1</mn><mn>2</mn><mi>o</mi><mi>r</mi><mn>4</mn><mi>x</mi><mo>+</mo><mn>5</mn><mi>y</mi><mo>&lt;</mo><mn>2</mn><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>Here, it can be seen that the feasible region has no common point with&nbsp;<span title="Click to copy mathml"><math><mrow><mn>4</mn><mi>x</mi><mo>+</mo><mn>5</mn><mi>y</mi><mo>&lt;</mo><mn>2</mn><mn>8</mn></mrow></math></span>&nbsp;.</p><p>Hence, the minimum value of z is 112 at (2, 4).</p><p>Therefore, the mixture should contain 2 kg of food X and 4 kg of food Y.</p><p>The minimum cost of the mixture is ? 112.</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment