Exercise 5.2 — Class 9 Mathematics Solutions

Solution

1. Formulate associated equations and find intercepts.

   Given constraints:

   $$\begin{array}{cc}2y-x& \le 8\text{— (1)}\\ x-y& \le 4\text{— (2)}\\ x& \ge 0,y\ge 0\text{— (3)}\end{array}$$

   1. For inequality (1), the associated equation is:

   $$\begin{array}{cc}2y-x& =8\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}2(0)-x& =8\\ -x& =8\\ x& =-8\end{array}$$

   So, point $A(-8,0)$ lies on line (1).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}2y-0& =8\\ 2y& =8\\ y& =4\end{array}$$

   So, point $B(0,4)$ lies on line (1).

   2. For inequality (2), the associated equation is:

   $$\begin{array}{cc}x-y& =4\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}x-0& =4\\ x& =4\end{array}$$

   So, point $C(4,0)$ lies on line (2).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}0-y& =4\\ -y& =4\\ y& =-4\end{array}$$

   So, point $D(0,-4)$ lies on line (2).

   ---

2. Perform test point check at $(0,0)$.

   • For inequality (1), substitute $(0,0)$:

   $$\begin{array}{cc}2(0)-(0)& \le 8\\ 0& \le 8(\text{True})\end{array}$$

   Since the statement is true, the closed half-plane of $2y-x\le 8$ contains the origin $(0,0)$ (the region below the line $2y-x=8$).

   • For inequality (2), substitute $(0,0)$:

   $$\begin{array}{cc}(0)-(0)& \le 4\\ 0& \le 4(\text{True})\end{array}$$

   Since the statement is true, the closed half-plane of $x-y\le 4$ contains the origin $(0,0)$ (the region above the line $x-y=4$).

   ---

3. Determine the feasible region.

   The non-negative constraints $x\ge 0$ and $y\ge 0$ restrict the feasible region entirely to the first quadrant. The intersection of the shaded half-planes in the first quadrant forms a closed polygonal feasible region bounded by the coordinate axes and the boundary lines.

   ---

4. Find all corner points (vertices) of the feasible region.

   1. The origin:

   $$\begin{array}{c}O(0,0)\end{array}$$

   2. On the $x$-axis ($y=0$):

   $$\begin{array}{c}C(4,0)\end{array}$$

   3. On the $y$-axis ($x=0$):

   $$\begin{array}{c}B(0,4)\end{array}$$

   4. Intersection point $E$ of the lines $2y-x=8$ and $x-y=4$: Solve simultaneously:

   $$\begin{array}{cc}-x+2y& =8\text{— (i)}\\ x-y& =4\text{— (ii)}\end{array}$$

   Adding equations (i) and (ii):

   $$\begin{array}{cc}(-x+2y)+(x-y)& =8+4\\ y& =12\end{array}$$

   Substitute $y=12$ into equation (ii):

   $$\begin{array}{cc}x-12& =4\\ x& =4+12\\ x& =16\end{array}$$

   Thus, the point of intersection is $E(16,12)$.

   The corner points of the feasible region are:

   $$\begin{array}{c}O(0,0),C(4,0),E(16,12),B(0,4)\end{array}$$

   ---

5. Evaluate the objective function $f(x,y)=2x+5y$ at each corner point.

   Corner Point $(x,y)$	Objective Function $f(x,y)=2x+5y$	Value
   $O(0,0)$	$f(0,0)=2(0)+5(0)=0+0$	$0$
   $C(4,0)$	$f(4,0)=2(4)+5(0)=8+0$	$8$
   $B(0,4)$	$f(0,4)=2(0)+5(4)=0+20$	$20$
   $E(16,12)$	$f(16,12)=2(16)+5(12)=32+60$	$92$

   ---

6. Conclude the optimal solution.

   Comparing all evaluated values, the maximum value of the objective function is $92$, which occurs at the corner point $(16,12)$.

Solution

1. Formulate associated equations and find intercepts.

   Given constraints:

   $$\begin{array}{cc}2x+5y& \le 30\text{— (1)}\\ 5x+4y& \le 20\text{— (2)}\\ x& \ge 0,y\ge 0\text{— (3)}\end{array}$$

   1. For inequality (1), the associated equation is:

   $$\begin{array}{cc}2x+5y& =30\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}2x+5(0)& =30\\ 2x& =30\\ x& =15\end{array}$$

   So, point $A(15,0)$ lies on line (1).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}2(0)+5y& =30\\ 5y& =30\\ y& =6\end{array}$$

   So, point $B(0,6)$ lies on line (1).

   2. For inequality (2), the associated equation is:

   $$\begin{array}{cc}5x+4y& =20\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}5x+4(0)& =20\\ 5x& =20\\ x& =4\end{array}$$

   So, point $C(4,0)$ lies on line (2).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}5(0)+4y& =20\\ 4y& =20\\ y& =5\end{array}$$

   So, point $D(0,5)$ lies on line (2).

   ---

2. Perform test point check at $(0,0)$.

   • For inequality (1), substitute $(0,0)$:

   $$\begin{array}{cc}2(0)+5(0)& \le 30\\ 0& \le 30(\text{True})\end{array}$$

   The closed half-plane of $2x+5y\le 30$ contains the origin $(0,0)$.

   • For inequality (2), substitute $(0,0)$:

   $$\begin{array}{cc}5(0)+4(0)& \le 20\\ 0& \le 20(\text{True})\end{array}$$

   The closed half-plane of $5x+4y\le 20$ contains the origin $(0,0)$.

   ---

3. Determine the feasible region.

   The non-negative constraints $x\ge 0,y\ge 0$ restrict the feasible region to the first quadrant. Notice that throughout the first quadrant, the line $5x+4y=20$ lies strictly inside (closer to the origin than) the line $2x+5y=30$. Specifically, at the intercepts:

   $$\begin{array}{cc}x\text{-intercepts: }4& <15\\ y\text{-intercepts: }5& <6\end{array}$$

   The intersection of the two lines occurs at $(-\frac{20}{17},\frac{110}{17})$, where $x<0$ (Quadrant II), so they do not intersect inside Quadrant I. Therefore, the constraint $5x+4y\le 20$ fully defines the outer boundary of the feasible region in the first quadrant.

   ---

4. Find all corner points (vertices) of the feasible region.

   The vertices of the triangular feasible region are:

   1. The origin:

   $$\begin{array}{c}O(0,0)\end{array}$$

   2. On the $x$-axis:

   $$\begin{array}{c}C(4,0)\end{array}$$

   3. On the $y$-axis:

   $$\begin{array}{c}D(0,5)\end{array}$$

   ---

5. Evaluate the objective function $f(x,y)=x+3y$ at each corner point.

   Corner Point $(x,y)$	Objective Function $f(x,y)=x+3y$	Value
   $O(0,0)$	$f(0,0)=0+3(0)=0+0$	$0$
   $C(4,0)$	$f(4,0)=4+3(0)=4+0$	$4$
   $D(0,5)$	$f(0,5)=0+3(5)=0+15$	$15$

   ---

6. Conclude the optimal solution.

   Comparing all evaluated values, the maximum value of the objective function is $15$, which occurs at the corner point $(0,5)$.

Solution

1. Formulate associated equations and find intercepts.

   Given constraints:

   $$\begin{array}{cc}2x+y& \le 4\text{— (1)}\\ 4x-y& \le 2\text{— (2)}\\ x& \ge 0,y\ge 0\text{— (3)}\end{array}$$

   1. For inequality (1), the associated equation is:

   $$\begin{array}{cc}2x+y& =4\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}2x+0& =4\\ 2x& =4\\ x& =2\end{array}$$

   So, point $A(2,0)$ lies on line (1).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}2(0)+y& =4\\ y& =4\end{array}$$

   So, point $B(0,4)$ lies on line (1).

   2. For inequality (2), the associated equation is:

   $$\begin{array}{cc}4x-y& =2\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}4x-0& =2\\ 4x& =2\\ x& =\frac{1}{2}\end{array}$$

   So, point $C(\frac{1}{2},0)$ lies on line (2).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}4(0)-y& =2\\ -y& =2\\ y& =-2\end{array}$$

   So, point $D(0,-2)$ lies on line (2).

   ---

2. Perform test point check at $(0,0)$.

   • For inequality (1), substitute $(0,0)$:

   $$\begin{array}{cc}2(0)+(0)& \le 4\\ 0& \le 4(\text{True})\end{array}$$

   The closed half-plane contains the origin $(0,0)$.

   • For inequality (2), substitute $(0,0)$:

   $$\begin{array}{cc}4(0)-(0)& \le 2\\ 0& \le 2(\text{True})\end{array}$$

   The closed half-plane contains the origin $(0,0)$.

   ---

3. Determine the feasible region.

   The non-negative constraints $x\ge 0,y\ge 0$ restrict the feasible region to the first quadrant. The feasible region is bounded by:

   • The $y$-axis ($x=0$) from $y=0$ to $y=4$
   • The line $2x+y=4$ from $(0,4)$ to the intersection point $E(1,2)$
   • The line $4x-y=2$ from $E(1,2)$ to $C(\frac{1}{2},0)$
   • The $x$-axis ($y=0$) from $C(\frac{1}{2},0)$ to $O(0,0)$

   ---

4. Find all corner points (vertices) of the feasible region.

   1. The origin:

   $$\begin{array}{c}O(0,0)\end{array}$$

   2. Point on the $x$-axis: Since $4x-y\le 2$ requires $x\le \frac{1}{2}$ when $y=0$:

   $$\begin{array}{c}C(\frac{1}{2},0)\end{array}$$

   3. Point on the $y$-axis:

   $$\begin{array}{c}B(0,4)\end{array}$$

   4. Intersection point $E$ of the lines $2x+y=4$ and $4x-y=2$: Solve simultaneously:

   $$\begin{array}{cc}2x+y& =4\text{— (i)}\\ 4x-y& =2\text{— (ii)}\end{array}$$

   Adding equations (i) and (ii):

   $$\begin{array}{cc}(2x+y)+(4x-y)& =4+2\\ 6x& =6\\ x& =1\end{array}$$

   Substitute $x=1$ into equation (i):

   $$\begin{array}{cc}2(1)+y& =4\\ 2+y& =4\\ y& =2\end{array}$$

   Thus, the intersection point is $E(1,2)$.

   The corner points are:

   $$\begin{array}{c}O(0,0),C(\frac{1}{2},0),E(1,2),B(0,4)\end{array}$$

   ---

5. Evaluate the objective function $z=2x+3y$ at each corner point.

   Corner Point $(x,y)$	Objective Function $z=2x+3y$	Value
   $O(0,0)$	$z=2(0)+3(0)=0+0$	$0$
   $C(\frac{1}{2},0)$	$z=2\left(\frac{1}{2}\right)+3(0)=1+0$	$1$
   $E(1,2)$	$z=2(1)+3(2)=2+6$	$8$
   $B(0,4)$	$z=2(0)+3(4)=0+12$	$12$

   ---

6. Conclude the optimal solution.

   Comparing all evaluated values, the maximum value of the objective function is $12$, which occurs at the corner point $(0,4)$.

Solution

1. Formulate associated equations and find intercepts.

   Given constraints:

   $$\begin{array}{cc}x+y& \ge 3\text{— (1)}\\ 7x+5y& \le 35\text{— (2)}\\ x& \ge 0,y\ge 0\text{— (3)}\end{array}$$

   1. For inequality (1), the associated equation is:

   $$\begin{array}{cc}x+y& =3\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}x+0& =3\\ x& =3\end{array}$$

   So, point $A(3,0)$ lies on line (1).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}0+y& =3\\ y& =3\end{array}$$

   So, point $B(0,3)$ lies on line (1).

   2. For inequality (2), the associated equation is:

   $$\begin{array}{cc}7x+5y& =35\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}7x+5(0)& =35\\ 7x& =35\\ x& =5\end{array}$$

   So, point $C(5,0)$ lies on line (2).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}7(0)+5y& =35\\ 5y& =35\\ y& =7\end{array}$$

   So, point $D(0,7)$ lies on line (2).

   ---

2. Perform test point check at $(0,0)$.

   • For inequality (1), substitute $(0,0)$:

   $$\begin{array}{cc}(0)+(0)& \ge 3\\ 0& \ge 3(\text{False})\end{array}$$

   The closed half-plane does not include the origin $(0,0)$ (the region lies on and above the line $x+y=3$).

   • For inequality (2), substitute $(0,0)$:

   $$\begin{array}{cc}7(0)+5(0)& \le 35\\ 0& \le 35(\text{True})\end{array}$$

   The closed half-plane contains the origin $(0,0)$ (the region lies on and below the line $7x+5y=35$).

   ---

3. Determine the feasible region.

   The non-negative constraints $x\ge 0,y\ge 0$ restrict the feasible region to the first quadrant. In the first quadrant, the feasible region is the quadrilateral region lying between the two boundary lines $x+y=3$ and $7x+5y=35$, bounded by:

   • The $x$-axis segment between $x=3$ and $x=5$
   • The line $7x+5y=35$
   • The $y$-axis segment between $y=3$ and $y=7$
   • The line $x+y=3$

   ---

4. Find all corner points (vertices) of the feasible region.

   The vertices of the feasible region are:

   1. $A(3,0)$ (on the $x$-axis)
   2. $C(5,0)$ (on the $x$-axis)
   3. $D(0,7)$ (on the $y$-axis)
   4. $B(0,3)$ (on the $y$-axis)

   ---

5. Evaluate the objective function $z=2x+y$ at each corner point.

   Corner Point $(x,y)$	Objective Function $z=2x+y$	Value
   $A(3,0)$	$z=2(3)+0=6+0$	$6$
   $C(5,0)$	$z=2(5)+0=10+0$	$10$
   $D(0,7)$	$z=2(0)+7=0+7$	$7$
   $B(0,3)$	$z=2(0)+3=0+3$	$3$

   ---

6. Conclude the optimal solution.

   Comparing all evaluated values, the minimum value of the objective function is $3$, which occurs at the corner point $(0,3)$.

Solution

1. Formulate associated equations and find intercepts.

   Given constraints:

   $$\begin{array}{cc}2x+y& \le 10\text{— (1)}\\ x+2y& \le 14\text{— (2)}\\ x& \ge 0,y\ge 0\text{— (3)}\end{array}$$

   1. For inequality (1), the associated equation is:

   $$\begin{array}{cc}2x+y& =10\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}2x+0& =10\\ 2x& =10\\ x& =5\end{array}$$

   So, point $A(5,0)$ lies on line (1).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}2(0)+y& =10\\ y& =10\end{array}$$

   So, point $B(0,10)$ lies on line (1).

   2. For inequality (2), the associated equation is:

   $$\begin{array}{cc}x+2y& =14\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}x+2(0)& =14\\ x& =14\end{array}$$

   So, point $C(14,0)$ lies on line (2).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}0+2y& =14\\ 2y& =14\\ y& =7\end{array}$$

   So, point $D(0,7)$ lies on line (2).

   ---

2. Perform test point check at $(0,0)$.

   • For inequality (1), substitute $(0,0)$:

   $$\begin{array}{cc}2(0)+(0)& \le 10\\ 0& \le 10(\text{True})\end{array}$$

   The closed half-plane contains the origin $(0,0)$.

   • For inequality (2), substitute $(0,0)$:

   $$\begin{array}{cc}(0)+2(0)& \le 14\\ 0& \le 14(\text{True})\end{array}$$

   The closed half-plane contains the origin $(0,0)$.

   ---

3. Determine the feasible region.

   The non-negative constraints $x\ge 0,y\ge 0$ restrict the feasible region to the first quadrant. The feasible region is bounded by the axes and both boundary lines, forming a 4-sided polygon with vertices at the origin, the $x$-axis, the intersection of both lines, and the $y$-axis.

   ---

4. Find all corner points (vertices) of the feasible region.

   1. The origin:

   $$\begin{array}{c}O(0,0)\end{array}$$

   2. Point on the $x$-axis: Comparing $x$-intercepts ($5$ vs $14$), the region is restricted by $x\le 5$:

   $$\begin{array}{c}A(5,0)\end{array}$$

   3. Point on the $y$-axis: Comparing $y$-intercepts ($10$ vs $7$), the region is restricted by $y\le 7$:

   $$\begin{array}{c}D(0,7)\end{array}$$

   4. Intersection point $E$ of the lines $2x+y=10$ and $x+2y=14$: Solve simultaneously:

   $$\begin{array}{cc}2x+y& =10\text{— (i)}\\ x+2y& =14\text{— (ii)}\end{array}$$

   Multiply equation (i) by $2$:

   $$\begin{array}{cc}4x+2y& =20\text{— (iii)}\end{array}$$

   Subtract equation (ii) from equation (iii):

   $$\begin{array}{cc}(4x+2y)-(x+2y)& =20-14\\ 3x& =6\\ x& =2\end{array}$$

   Substitute $x=2$ into equation (i):

   $$\begin{array}{cc}2(2)+y& =10\\ 4+y& =10\\ y& =6\end{array}$$

   Thus, the intersection point is $E(2,6)$.

   The corner points are:

   $$\begin{array}{c}O(0,0),A(5,0),E(2,6),D(0,7)\end{array}$$

   ---

5. Evaluate the objective function $f(x,y)=2x+3y$ at each corner point.

   Corner Point $(x,y)$	Objective Function $f(x,y)=2x+3y$	Value
   $O(0,0)$	$f(0,0)=2(0)+3(0)=0+0$	$0$
   $A(5,0)$	$f(5,0)=2(5)+3(0)=10+0$	$10$
   $E(2,6)$	$f(2,6)=2(2)+3(6)=4+18$	$22$
   $D(0,7)$	$f(0,7)=2(0)+3(7)=0+21$	$21$

   ---

6. Conclude the optimal solution.

   Comparing all evaluated values, the maximum value of the objective function is $22$, which occurs at the corner point $(2,6)$.

Solution

1. Formulate associated equations and find intercepts.

   Given constraints:

   $$\begin{array}{cc}3x+5y& \ge 15\text{— (1)}\\ x+3y& \le 9\text{— (2)}\\ x& \ge 0,y\ge 0\text{— (3)}\end{array}$$

   1. For inequality (1), the associated equation is:

   $$\begin{array}{cc}3x+5y& =15\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}3x+5(0)& =15\\ 3x& =15\\ x& =5\end{array}$$

   So, point $A(5,0)$ lies on line (1).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}3(0)+5y& =15\\ 5y& =15\\ y& =3\end{array}$$

   So, point $B(0,3)$ lies on line (1).

   2. For inequality (2), the associated equation is:

   $$\begin{array}{cc}x+3y& =9\end{array}$$

   • To find the $x$-intercept, set $y=0$:

   $$\begin{array}{cc}x+3(0)& =9\\ x& =9\end{array}$$

   So, point $C(9,0)$ lies on line (2).

   • To find the $y$-intercept, set $x=0$:

   $$\begin{array}{cc}0+3y& =9\\ 3y& =9\\ y& =3\end{array}$$

   So, point $B(0,3)$ lies on line (2).

   (Notice that both boundary lines share the same $y$-intercept $B(0,3)$, which is their point of intersection).

   ---

2. Perform test point check at $(0,0)$.

   • For inequality (1), substitute $(0,0)$:

   $$\begin{array}{cc}3(0)+5(0)& \ge 15\\ 0& \ge 15(\text{False})\end{array}$$

   The closed half-plane does not include the origin $(0,0)$ (the region lies on and above line 1).

   • For inequality (2), substitute $(0,0)$:

   $$\begin{array}{cc}0+3(0)& \le 9\\ 0& \le 9(\text{True})\end{array}$$

   The closed half-plane contains the origin $(0,0)$ (the region lies on and below line 2).

   ---

3. Determine the feasible region.

   The non-negative constraints $x\ge 0,y\ge 0$ restrict the feasible region to the first quadrant. In the first quadrant:

   • The region lies above $3x+5y=15$ and below $x+3y=9$.
   • On the $x$-axis ($y=0$), the region spans between $x=5$ and $x=9$, defined by the segment connecting $A(5,0)$ and $C(9,0)$.
   • On the $y$-axis ($x=0$), both boundaries meet at the single vertex $B(0,3)$.
   • Thus, the feasible region is a closed triangle bounded by the points $B(0,3)$, $A(5,0)$, and $C(9,0)$.

   ---

4. Find all corner points (vertices) of the feasible region.

   The three corner points of the feasible region are:

   1. $B(0,3)$ (common $y$-intercept and line intersection)
   2. $A(5,0)$ ($x$-intercept of $3x+5y=15$)
   3. $C(9,0)$ ($x$-intercept of $x+3y=9$)

   ---

5. Evaluate the objective function $z=3x+y$ at each corner point.

   Corner Point $(x,y)$	Objective Function $z=3x+y$	Value
   $B(0,3)$	$z=3(0)+3=0+3$	$3$
   $A(5,0)$	$z=3(5)+0=15+0$	$15$
   $C(9,0)$	$z=3(9)+0=27+0$	$27$

   ---

6. Conclude the optimal solution.

   Comparing all evaluated values:

   • The minimum value of $z$ is $3$, which occurs at the corner point $(0,3)$.
   • The maximum value of $z$ is $27$, which occurs at the corner point $(9,0)$.