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**Topic**

A workshop sewing vests and casual pants to prepare for the end of the year. Knowing how to sew a vest takes 2m of fabric and takes 20 hours; 1 casual pants takes 1.5 m of fabric and takes 5 hours. The factory is assigned to use no more than 900 m of fabric and the number of man-hours does not exceed 6 000 hours. According to the market survey, the number of pants sold is not less than the number of shirts and not more than twice the number of shirts. When exported to the market, 1 shirt makes a profit of 350,000 VND, 1 pair of pants makes a profit of 100,000 VND. How many vests and trousers does the workshop need to make to get the highest profit (knowing that the consumer market is always welcoming to the company’s products).

**Solution method – See details**

– Let the number of vests and casual pants to be sewn are x, y (pieces), respectively.

– Find inconsistencies in sewing time for pants and shirts

– Draw the solution domain of the system of inequalities

– Find the maximum value for the highest return

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**Detailed explanation**

Let the number of vests and casual pants to be sewn are x, y (pieces), respectively. \((x,y \in \mathbb{N})\)

The number of meters of fabric to make x shirts and y pants is: \(2x + 1.5y\left( m \right)\)

Since the factory is assigned to use no more than 900 m of fabric, we have: \(2x + 1.5y \le 900\left( 1 \right)\)

The number of hours to sew x shirts and y pants is: \(20x + 5y\) (hours)

Since the number of man-hours does not exceed 6 000 hours, we have: \(20x + 5y \le 6000\) or \(4x + y \le 1200\) (2)

According to market survey, we have:

The number of pants sold is not less than the number of shirts, so \(y \ge x\)(3)

The number of pants should not exceed 2 times the number of shirts, so \(y \le 2x\) (4)

From (1), (2), (3) and (4), we have a system of inequalities:

\(\left\{ {\begin{array}{*{20}{c}}{2x + 1.5y \le 900}\\{4x + y \le 1200}\\{y \ge x}\ \{y \le 2x}\\{x \ge 0}\\{y \ge 0}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{ 2x + 1.5y \le 900}\\{4x + y \le 1200}\\{x – y \le 0}\\{2x – y \ge 0}\\{x \ge 0}\\{ y \ge 0}\end{array}} \right.} \right.\)

Representing the solution domain of the system of inequalities on the coordinate system Oxy, we get:

The solution domain of the system of inequalities is quadrilateral OABC with \(O\left( {0;0} \right),A\left( {180;360} \right),B\left( {200;250} \ right),C\left( {240;240} \right)\)

Profit from selling x shirts and y pants is 350x + 100y (thousand dong).

Set T = 350x + 100y.

We have the expression T = 350x + 100y whose maximum value is at one of the vertices of the quadrilateral OABC.

Calculate the value of the expression T at the vertices of the quadrilateral:

At O(0; 0), with x = 0 and y = 0, then T = 350.0 + 100.0 = 0;

At A(180; 360), with x = 180 and y = 360, then T = 350,180 + 100,360 = 99 000;

At B(225; 300), with x = 225 and y = 300, then T = 350,225 + 100,300 = 108 750;

At C(240; 240), with x = 240 and y = 240, then T = 350,240 + 100,240 = 108 000;

We get T reaching the maximum value of 108 750 000 VND when x = 225, y = 300.

So to get the highest profit, the factory needs to sew 225 vests and 300 casual pants.

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