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Solve Lesson 27 Page 32 SBT Math 10 – Kite>

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Topic

a) Express the solutions of the following inequalities: \(\left\{ {\begin{array}{*{20}{c}}{3x – y \le 9}\\{3x + 6y \le 30}\\{x \ge 0}\\{0 \le y \le 4}\end{array}} \right.\left( I \right)\)

b) Find x, y is the solution of the system of inequalities (I) such that \(F = 3x + 4y\) reaches the maximum value

Solution method – See details

a) Determine the solution domain of each bpt. The solution domain of the bpt system is the intersection of those solution domains.

Representation of the root domain of bpt \(ax + by < c\)

Step 1: Draw a line \(d:ax + by = c\)

Step 2: Take a point \(M\left( {{x_o};{y_o}} \right)\) that is not on d (we usually take the origin O if \(c \ne 0\)). Calculate \(a{x_o} + b{y_o}\) 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) Calculate the value of \(F\left( {x;y} \right)\) at the vertices of the polygon root domain.

Detailed explanation

Draw straight lines:

d1: 3x – y = 9 passes through two points with coordinates (3; 0) and (0; 9).

d2: 3x + 6y = 30 passes through two points (10; 0) and (0; 5).

d3: x = 0 is the vertical axis.

d4: y = 0 is the horizontal axis

d5: y = 4 passes through the point (0; 4) and is parallel to the horizontal axis.

Cross out the parts that are not in the solution domain of each inequality.

The solution domain of the system of inequalities is the OABCD pentagonal domain with O(0; 0), A(0; 4), B(2; 4), C(4; 3), D(3; 0):

b) Replace x, y are the coordinates of points O, A, B, C, D respectively into the expression F:

\(O(0,0)\)

\(A(0;4)\)

\(B(2;4)\)

\(C(4;3)\)

\(D(3;0)\)

\(F = 3x + 4y\)

\(0\)

\(16\)

\(22\)

\(24\)

\(9\)

F reaches its maximum value of 24 at \(x = 4,y = 3\)

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