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Solve Lesson 17 Page 30 Math 10 SBT – Kite>

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Topic

a) Representation of the solution domain of the system of inequalities \(\left\{ {\begin{array}{*{20}{c}}{x + y \le 5}\\{3x + 2y \le 12}\\{x \ge 1}\\{ y \ge 0}\end{array}} \right.\left( {III} \right)\)

b) Find x, y are solutions of the system of inequalities (III) such that \(F = 3x + 7y\) maximum and minimum values.

Solution method – See details

a)

  • Step 1: Draw a line \(d:x – 2y = 4\).
  • Step 2: Take a point \(M\left( {{x_o};{y_o}} \right)\) that is not on d (we usually take the origin O if \(c \ne 0\). (a{x_o} + b{y_o}\) and compare with c
  • Step 3: Conclusion
    • If \(a{x_o} + b{y_o} < c\) then the half-plane (excluding line d) containing the point M is the solution domain of the inequality \(ax + by < c\)
    • If \(a{x_o} + b{y_o} > c\) then the half-plane (excluding d) containing no point M is the solution domain of the inequality \(ax + by > c\)

b)

– Representation of the solution domain of the system of inequalities on the coordinate system

\(F\left( {x;y} \right)\) reaches max or min at one of the vertices so we only need to compute the value of \(F\left( {x;y} \right)\) at those vertices

Detailed explanation

a) We draw four lines:

dfirst: x + y = 5 is a straight line passing through two points with coordinates (0; 5) and (5; 0);

d2: 3x + 2y = 12 is a line passing through two points with coordinates (4; 0) and (0; 6);

d3: x = 1 is a line parallel to the vertical axis and passing through the point (1; 0);

d4: y = 0 is the horizontal axis.

We define each solution domain of each inequality in the system, crossing out the parts that are not in the solution domain of each inequality.

The solution domain of the system of inequalities is the region in quadrilateral ABCD with A(1; 0), B(1; 4), C(2; 3) and D(4; 0) as shown in the following figure:

b) We have the expression F = 3x + 7y reaching the maximum value, the smallest value at one of the vertices of the quadrilateral ABCD.

At A(1; 0) with x = 1 and y = 0, then F = 3.1 + 7.0 = 3;

At B(1; 4) with x = 1 and y = 4, then F = 3.1 + 7.4 = 31;

At C(2; 3) with x = 2 and y = 3, then F = 3.2 + 7.3 = 27;

At D(4; 0) with x = 4 and y = 0, then F = 3.4 + 7.0 = 12.

So the maximum value of F is 31 at x = 1 and y = 4, the minimum value of F is 3 at x = 1 and y = 0

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