13. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman’s time.

(i) What number of rackets and bats must be made if the factory is to work at full capacity?

(ii) If the profit on a racket and on a bat is Rs. 20 and Rs. 10 respectively, find maximum profit of the factory when it works at full capacity.

(i) Let the number of rackets and the number of bats to be made be x and y respectively.

The machine time is not available for more than 42 hours.

$\therefore 1.5x+3y\le 42....\left(1\right)$

The craftsman’s time is not available for more than 24 hours.

$\therefore 3x+y\le 24 ......\left(2\right)$

The factory is to work at full capacity. Therefore,

$\begin{array}{l}1.5x + 3y = 42\hfill \end{array}$

$\begin{array}{l}3x + y = 24\hfill \end{array}$

On solving these equations, we obtain

x = 4 and y = 12

Thus, 4 rackets and 12 bats must be made.

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

	Tennis Racket	Cricket Bat	Availability
Machine Time (h)	1.5	3	42
Craftsman’s Time (h)	3	1	24

$\begin{array}{l}\therefore 1.5x + 3y \le 42\hfill \end{array}$

$\begin{array}{l}3x + y \le 24\hfill \end{array}$

$\begin{array}{l}x, y \ge 0\hfill \end{array}$

The profit on a racket is Rs 20 and on a bat is Rs 10.

$\therefore Z=20x+10y$

The mathematical formulation of the given problem is

Maximize $Z=20x+10y\dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}1.5x + 3y \le 42 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + y \le 24 \dots (\mathrm{3)}\hfill \end{array}$

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

The feasible region determined by the sy

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(i) Let the number of rackets and the number of bats to be made be x and y respectively.

The machine time is not available for more than 42 hours.

$\therefore 1.5x+3y\le 42....\left(1\right)$

The craftsman’s time is not available for more than 24 hours.

$\therefore 3x+y\le 24 ......\left(2\right)$

The factory is to work at full capacity. Therefore,

$\begin{array}{l}1.5x + 3y = 42\hfill \end{array}$

$\begin{array}{l}3x + y = 24\hfill \end{array}$

On solving these equations, we obtain

x = 4 and y = 12

Thus, 4 rackets and 12 bats must be made.

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

	Tennis Racket	Cricket Bat	Availability
Machine Time (h)	1.5	3	42
Craftsman’s Time (h)	3	1	24

$\begin{array}{l}\therefore 1.5x + 3y \le 42\hfill \end{array}$

$\begin{array}{l}3x + y \le 24\hfill \end{array}$

$\begin{array}{l}x, y \ge 0\hfill \end{array}$

The profit on a racket is Rs 20 and on a bat is Rs 10.

$\therefore Z=20x+10y$

The mathematical formulation of the given problem is

Maximize $Z=20x+10y\dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}1.5x + 3y \le 42 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + y \le 24 \dots (\mathrm{3)}\hfill \end{array}$

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

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

The corner points are A (8, 0), B (4, 12), C (0, 14), and O (0, 0).

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

Thus, the maximum profit of the factory when it works to its full capacity is Rs 200.

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<p><strong>(i) </strong>Let the number of rackets and the number of bats to be made be&nbsp;x&nbsp;and&nbsp;y&nbsp;respectively.</p><p>The machine time is not available for more than 42 hours.</p><p><span title="Click to copy mathml"><math><mrow><mo>∴</mo><mn>1</mn><mo>.</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>3</mn><mi>y</mi><mo>≤</mo><mn>4</mn><mn>2</mn><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mrow><mo>(</mo><mrow><mn>1</mn></mrow><mo>)</mo></mrow></mrow></math></span></p><p>The craftsman’s time is not available for more than 24 hours.</p><p><span title="Click to copy mathml"><math><mrow><mo>∴</mo><mn>3</mn><mi>x</mi><mo>+</mo><mi>y</mi><mo>≤</mo><mn>2</mn><mn>4</mn><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>&nbsp;</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mrow><mo>(</mo><mrow><mn>2</mn></mrow><mo>)</mo></mrow></mrow></math></span></p><p>The factory is to work at full capacity. Therefore,</p><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mn>1</mn><mo>.</mo><mn>5</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>3</mn><mi>y</mi><mo>&nbsp;</mo><mo>=</mo>&nbsp;<mn>4</mn><mn>2</mn></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><mo>&nbsp;</mo><mi>y</mi><mo>&nbsp;</mo><mo>=</mo>&nbsp;<mn>2</mn><mn>4</mn></mtd></mtr></mtable></mrow></math></span></p><p>On solving these equations, we obtain</p><p>x&nbsp;= 4 and&nbsp;y&nbsp;= 12</p><p>Thus, 4 rackets and 12 bats must be made.</p><p><strong>(i) </strong>The given information can be complied in a table as follows.</p><table><tbody><tr><td><p>&nbsp;</p></td><td><p>Tennis Racket</p></td><td><p>Cricket Bat</p></td><td><p>Availability</p></td></tr><tr><td><p>Machine Time (h)</p></td><td><p>1.5</p></td><td><p>3</p></td><td><p>42</p></td></tr><tr><td><p>Craftsman’s Time (h)</p></td><td><p>3</p></td><td><p>1</p></td><td><p>24</p></td></tr></tbody></table><p><span title="Click to copy mathml"><math><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mo>∴</mo><mn>1</mn><mo>.</mo><mn>5</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>3</mn><mi>y</mi><mo>&nbsp;</mo><mo>≤</mo>&nbsp;<mn>4</mn><mn>2</mn></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><mo>&nbsp;</mo><mi>y</mi><mo>&nbsp;</mo><mo>≤</mo>&nbsp;<mn>2</mn><mn>4</mn></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></mtd></mtr></mtable></mrow></math></span></p><p>The profit on a racket is Rs 20 and on a bat is Rs 10.</p><p><span title="Click to copy mathml"><math><mrow><mo>∴</mo><mi>Z</mi><mo>=</mo><mn>2</mn><mn>0</mn><mi>x</mi><mo>+</mo><mn>1</mn><mn>0</mn><mi>y</mi></mrow></math></span></p><p>The mathematical formulation of the given problem is</p><p>Maximize&nbsp;<span title="Click to copy mathml"><math><mrow><mi>Z</mi><mo>=</mo><mn>2</mn><mn>0</mn><mi>x</mi><mo>+</mo><mn>1</mn><mn>0</mn><mi>y</mi><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>1</mn><mo>.</mo><mn>5</mn><mi>x</mi><mo>&nbsp;</mo><mo>+</mo>&nbsp;<mn>3</mn><mi>y</mi><mo>&nbsp;</mo><mo>≤</mo>&nbsp;<mn>4</mn><mn>2</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><mo>&nbsp;</mo><mi>y</mi><mo>&nbsp;</mo><mo>≤</mo>&nbsp;<mn>2</mn><mn>4</mn>&nbsp;<mo>…</mo>&nbsp;<mrow><mo>(</mo><mrow><mn>3)</mn></mrow></mrow></mtd></mtr></mtable></mrow></math></span></p><p><span title="Click to copy mathml"><br><math><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mo></mo></mrow></mtd><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/1732773651phpmtjPal_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732773651phpmtjPal.jpeg" alt="" width="257" height="232"></picture></div></div><p>The corner points are A (8, 0), B (4, 12), C (0, 14), and O (0, 0).</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/1732773686phpdru71a_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732773686phpdru71a.jpeg" alt="" width="235" height="194"></picture></div></div><p>Thus, the maximum profit of the factory when it works to its full capacity is Rs 200.</p>

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