The solution of the new math book for 9th class exercise 5.2 step by step with the help of the graph is shown below.
Question 1: Maximize \( f(x, y) = 2x + 5y \)
Subject to constraints:
\[2y – x \leq 8, \quad x – y \leq 4, \quad x \geq 0, \quad y \geq 0\]
Rewrite the inequalities as equalities as \( 2y – x = 8 \)
Put \( x = 0 \) into the equation:
\[ 2y = 8 \]
\[ y = 4 \]
\[(0, 4) \]
Put \( y = 0 \) into the equation:
\[ -x = 8 \]
\[ x = -8 \]
\[(-8, 0) \]
Draw the line passing through \((0, 4)\) and \((-8, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[2(0)-0 \leq 8\]
\[0\leq 8\]
\[\text{true}\]
\[ x – y = 4 \]
Put \( x = 0 \) into the equation
\[ -y = 4 \]
\[ y = -4 \]
\[(0, -4) \]
Put \( y = 0 \) into the equation:
\[ x = 4 \]
\[ (4, 0) \]
Draw the line passing through \((0, -4)\) and \((4, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[0-0 \leq 4\]
\[0\leq 4\]
\[\text{True} \]
\( x \geq 0 \): The region is to the right of the \( y \)-axis.
\( y \geq 0 \): The region is above the \( x \)-axis.
Then the feasible region on the graph of the both inequality is.
Find the corner points of the feasible region
The intersection of the lines gives the corner points:
Intersection of \( 2y – x = 8 \) and \( x – y = 4 \)
Solve the system of equations:
\[x = 2y – 8\]
Substitute into \( x – y = 4 \):
\[(2y – 8) – y = 4 \]
\[ y = 12\]
\[x = 2(12) – 8 = 16\]
Point \((16, 12)\)
Other corner points are:
\[(0,0),(4,0),(0,4)\]
Evaluate \( f(x, y) = 2x + 5y \) at each corner point
At \((0, 4)\): \( f(0, 4) = 2(0) + 5(4) = 20 \)
At \((0, 0)\): \( f(0, 0) = 2(0) + 5(0) = 0 \)
At \((4, 0)\): \( f(4, 0) = 2(4) + 5(0) = 8 \)
At \((16, 12)\): \( f(16, 12) = 2(16) + 5(12) = 92 \)
Conclusion
The maximum value of \( f(x, y) = 92 \) at the point \((16, 12)\).
Question 2: Maximize \( f(x, y) = x + 3y \)
Subject to constraints:
\[2x + 5y \leq 30, \quad 5x + 4y \leq 20, \quad x \geq 0, \quad y \geq 0\]
Rewrite the inequalities as equalities as \( 2x + 5y = 30 \)
Put \( x = 0 \) into the equation:
\[ 5y = 30 \]
\[ y = 6 \]
\[ (0, 6) \]
Put \( y = 0 \) into the equation:
\[ 2x = 30 \]
\[x = 15 \]
\[ (15, 0) \]
Draw the line passing through \((0, 6)\) and \((15, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[2(0) + 5(0) \leq 30\]
\[0\leq 30\]
\[\text{True}\]
\[ 5x + 4y = 20 \]
Put \( x = 0 \) into the equation:
\[ 4y = 20 \]
\[ y = 5 \]
\[(0, 5) \]
Put \( y = 0 \) into the equation:
\[ 5x = 20 \]
\[ x = 4 \]
\[(4, 0) \]
Draw the line passing through \((0, 5)\) and \((4, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[5(0) + 4(0) \leq 20\]
\[0\leq 20\]
\[\text{True}\]
\( x \geq 0 \): The region is to the right of the \( y \)-axis.
\( y \geq 0 \): The region is above the \( x \)-axis.
Then the feasible region on the graph of the both inequality is.
Find the corner points of the feasible region
The intersection of the lines gives the corner points:
\[(0,0),(4,0),(0,5)\]
Evaluate \( f(x, y) = x + 3y \) at each corner point
At \((0, 5)\): \( f(0, 5) = 0 + 3(5) = 15 \)
At \((0, 0)\): \( f(0, 0) = 0 + 3(0) = 0 \)
At \((4, 0)\): \( f(4, 0) = 4 + 3(0) = 4 \)
Conclusion
The maximum value of \( f(x, y) = 15 \) at the point \((0, 5)\).
Question 3: Maximize \( z = 2x + 3y \)
Subject to constraints:
\[2x + y \leq 4, \quad 4x – y \leq 4, \quad x \geq 0, \quad y \geq 0\]
Rewrite the inequalities as equalities
\[ 2x + y = 4 \]
Put \( x = 0 \) into the equation:
\[ 2(0) + y = 4 \]
\[y = 4 \]
\[(0, 4) \]
Put \( y = 0 \) into the equation:
\[ 2x + 0 = 4 \]
\[ x = 2 \]
\[(2, 0) \]
Draw a solid line passing through \((0, 4)\) and \((2, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[2(0) + (0) \leq 4\]
\[0\leq 4\]
\[\text{True}\]
\[ 4x – y = 4 \]
Put \( x = 0 \) into the equation:
\[ 4(0) – y = 4 \]
\[y = -4 \]
\[(0, -4)\]
Put \( y = 0 \)
\[ 4x – 0 = 4 \]
\[x = 1 \]
\[(1, 0) \]
Draw a solid line passing through \((1, 0)\) and (0,-4).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[4(0)- 0 \leq 4\]
\[0\leq 4\]
\[\text{True}\]
\( x \geq 0 \): Region to the right of the \( y \)-axis.
\( y \geq 0 \): Region above the \( x \)-axis.
Then the feasible region on the graph of the both inequality is.
Find the corner points of the feasible region
Intersection of \( 2x + y = 4 \) and \( 4x – y = 4 \):
Solve the system of equations:
\[2x + y = 4\]
\[4x – y = 4\]
Add the equations:
\[6x = 8 \]
\[ x = \frac{4}{3}\]
Substitute \( x = \frac{4}{3} \) into \( 2x + y = 4 \):
\[2\left(\frac{4}{3}\right) + y = 4 \]
\[ \frac{8}{3} + y = 4 \]
\[ y = \frac{4}{3}\]
\[\left(\frac{4}{3}, \frac{4}{3}\right)\]
Other corner points:
\[(0, 4), (0, 0), \text{and}~ (1, 0)\]
Evaluate \( z = 2x + 3y \) at each corner point
At \((0, 4)\): \( z = 2(0) + 3(4) = 12 \)
At \((0, 0)\): \( z = 2(0) + 3(0) = 0 \)
At \((1, 0)\): \( z = 2(1) + 3(0) = 2 \)
At \(\left(\frac{4}{3}, \frac{4}{3}\right)\)
\[z = 2\left(\frac{4}{3}\right) + 3\left(\frac{4}{3}\right) = \frac{8}{3} + \frac{12}{3} = \frac{20}{3}\]
\[\approx 6.67\]
Conclusion
The maximum value of \( z = 12 \) at the point \((0, 4)\).
Question 4: Minimize \( z = 2x + y \)
Subject to constraints:
\[x + y \geq 3, \quad 7x + 5y \leq 35, \quad x \geq 0, \quad y \geq 0\]
Rewrite the inequalities as equalities
\[ x + y = 3 \]
Put \( x = 0 \) into the equation:
\[ 0 + y = 3 \]
\[ y = 3 \]
\[(0, 3) \]
Put \( y = 0 \) into the equation:
\[ x + 0 = 3 \]
\[x = 3 \]
\[(3, 0) \]
Draw a solid line passing through \((0, 3)\) and \((3, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[(0)+ 0 \geq 3\]
\[0\geq 3\]
\[\text{False}\]
\[ 7x + 5y = 35 \]
Put \( x = 0 \) into the equation:
\[ 7(0) + 5y = 35 \]
\[ y = 7 \]
\[(0, 7) \]
Put \( y = 0 \) into the equation
\[ 7x + 0 = 35 \]
\[ x = 5 \]
\[(5, 0) \]
Draw a solid line passing through \((0, 7)\) and \((5, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[7(0)+5( 0) \leq 35\]
\[0\leq 35\]
\[\text{True}\]
\( x \geq 0 \): Region to the right of the \( y \)-axis.
\( y \geq 0 \): Region above the \( x \)-axis.
Then the feasible region on the graph of the both inequality is.
Find the corner points of the feasible region
\[(0, 3), (3,0),(5,0),(0,7)\]
Evaluate \( z = 2x + y \) at each corner point
At \((0, 3)\): \( z = 2(0) + 3 = 3 \)
At \((3, 0)\): \( z = 2(3) + 0 = 6 \)
At \((5, 0)\): \( z = 2(5) + 0 = 10 \)
At \((0,7)\): \( z = 2(0) + 7 = 7 \)
Conclusion
The minimum value of \( z = 3 \) at the point \((0, 3)\).
Question 5: Maximize \( f(x, y) = 2x + 3y \)
Subject to constraints:
\[2x + y \leq 8, \quad x + 2y \leq 14, \quad x \geq 0, \quad y \geq 0\]
Rewrite the inequalities as equalities
\[ 2x + y = 8 \]
Put \( x = 0 \):
\[ 2(0) + y = 8 \]
\[ y = 8 \]
\[(0, 8) \]
Put \( y = 0 \) into the equation:
\[ 2x + 0 = 8 \]
\[x = 4 \]
\[(4, 0) \]
Draw a solid line passing through \((0, 8)\) and \((4, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[2(0)+ 0 \leq 8\]
\[0\leq 8\]
\[\text{True}\]
\[ x + 2y = 14 \]
Put \( x = 0 \) into the equation:
\[ 0 + 2y = 14 \]
\[y = 7 \]
\[(0, 7) \]
Put \( y = 0 \) into the equation:
\[ x + 0 = 14 \]
\[ x = 14 \]
\[(14, 0) \]
Draw a solid line passing through \((0, 7)\) and \((14, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[0+ 2(0) \leq 14\]
\[0\leq 14\]
\[\text{True}\]
\( x \geq 0 \): Region to the right of the \( y \)-axis.
\( y \geq 0 \): Region above the \( x \)-axis.
Then the feasible region on the graph of the both inequality is.
Find the corner points of the feasible region
Intersection of \( 2x + y = 8 \) and \( x + 2y = 14 \):
Solve the system of equations:
\[y = 8 – 2x\]
Substitute \( y = 8 – 2x \) into \( x + 2y = 14 \):
\[x + 2(8 – 2x) = 14 \]
\[x + 16 – 4x = 14 \]
\[ -3x = -2 \]
\[ x = \frac{2}{3}\]
Substitute \( x = \frac{2}{3} \) into \( y = 8 – 2x \):
\[y = 8 – 2\left(\frac{2}{3}\right) = 8 – \frac{4}{3} = \frac{20}{3}\]
Point
\[\left(\frac{2}{3}, \frac{20}{3}\right)\]
Other corner points:
\[(0, 0), (4, 0), \text{and} (0, 7)\]
Evaluate \( f(x, y) = 2x + 3y \) at each corner point
At \((0, 0)\): \( f(0, 0) = 2(0) + 3(0) = 0 \)
At \((4, 0)\): \( f(4, 0) = 2(4) + 3(0) = 8 \)
At \((0, 7)\): \( f(0, 7) = 2(0) + 3(7) = 21 \)
At \(\left(\frac{2}{3}, \frac{20}{3}\right)\):
\[f\left(\frac{2}{3}, \frac{20}{3}\right) = 2\left(\frac{2}{3}\right) + 3\left(\frac{20}{3}\right) \]
\[= \frac{4}{3} + \frac{60}{3} = \frac{64}{3} \]
\[\approx 21.33\]
Conclusion
The maximum value of \( f(x, y) = \frac{64}{3} \approx 21.33 \) at the point \(\left(\frac{2}{3}, \frac{20}{3}\right)\).
Question 6: Find minimum and maximum values of \( z = 3x + y \)
Subject to constraints:
\[3x + 5y \geq 15, \quad x + 6y \geq 9, \quad x \geq 0, \quad y \geq 0\]
Rewrite the inequalities as equalities
\[ 3x + 5y = 15 \]
Put \( x = 0 \) into the equation:
\[ 3(0) + 5y = 15 \]
\[ y = 3 \]
\[(0, 3) \]
Put \( y = 0 \) into the equation
\[ 3x + 0 = 15 \]
\[ x = 5 \]
\[(5, 0) \]
Draw a solid line passing through \((0, 3)\) and \((5, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[3(0)+ 5(0) \geq 15\]
\[0\geq 15\]
\[\text{False}\]
\[ x + 6y = 9 \]
Put \( x = 0 \) into the equation:
\[ 0 + 6y = 9 \]
\[y = \frac{3}{2} \]
\[(0, \frac{3}{2}) \]
Put \( y = 0 \) into the equation:
\[ x + 0 = 9 \]
\[ x = 9 \]
\[(9, 0) \]
Draw a solid line passing through \((0, \frac{3}{2})\) and \((9, 0)\).
Take a test points
Substitute \((x,y)=(0,0)\) into the inequality
\[0+ 6(0) \geq 9\]
\[0\geq 9\]
\[\text{False}\]
\( x \geq 0 \): Region to the right of the \( y \)-axis.
\( y \geq 0 \): Region above the \( x \)-axis.
Then the feasible region on the graph of the both inequality is.
Find the corner points of the feasible region
Intersection of \( 3x + 5y = 15 \) and \( x + 6y = 9 \):
Solve the system of equations:
\[y = \frac{15 – 3x}{5}\]
Substitute \( y = \frac{15 – 3x}{5} \) into \( x + 6y = 9 \):
\[x + 6\left(\frac{15 – 3x}{5}\right) = 9 \]
\[ x + \frac{90 – 18x}{5} = 9\]
\[5x + 90 – 18x = 45 \]
\[ -13x = -45 \]
\[x = \frac{45}{13}\]
Substitute \( x = \frac{45}{13} \) into \( y = \frac{15 – 3x}{5} \):
\[y = \frac{15 – 3\left(\frac{45}{13}\right)}{5} = \frac{15 – \frac{135}{13}}{5} = \frac{\frac{195}{13} – \frac{135}{13}}{5} = \frac{\frac{60}{13}}{5} = \frac{12}{13}\]
Point:
\[\left(\frac{45}{13}, \frac{12}{13}\right)\]
Other corner points:
\[(0, 3), ~\text{and}~ (9, 0)\]
Step 3: Evaluate \( z = 3x + y \) at each corner point
At \((0, 3)\): \( z = 3(0) + 3 = 3 \)
At \((9, 0)\): \( z = 3(9) + 0 = 27 \)
At \(\left(\frac{45}{13}, \frac{12}{13}\right)\):
\[z = 3\left(\frac{45}{13}\right) + \frac{12}{13} = \frac{135}{13} + \frac{12}{13} = \frac{147}{13} \approx 11.31\]
Conclusion
The minimum value of \( z = 3 \) at the point \((0, 3)\).
The maximum value of \( z = 27 \) at the point \((9, 0)\).
