Review Exercise 5 — Class 9 Mathematics Solutions

Solution

Solution

1. Understand the definition of a linear equation.

   A linear equation in one variable $x$ is an algebraic statement of the form:

   $$\begin{array}{cc}ax+b& =0\end{array}$$

   where $a,b\in ℝ$ with $a\ne 0$. It satisfies two conditions:

   1. It contains an equality relation $(=)$.
   2. The highest exponent (degree) of the variable $x$ is $1$.

2. Evaluate each given option.

   • Option (a): $5x>7$ contains a strict inequality sign $(>)$, so it is a linear inequality, not an equation.
   • Option (b): $4x-2<1$ contains a strict inequality sign $(<)$, making it a linear inequality.
   • Option (c): $2x+1=1$ contains an equality sign $(=)$ and the variable $x$ has an exponent of $1$ with a non-zero coefficient $2$. Therefore, it is a linear equation.
   • Option (d): $4=1+3$ contains an equality sign, but it contains no variable; it is purely a numerical identity, not an algebraic linear equation.

3. Conclusion.

   Option (c) is the only linear equation.

Solution

1. Isolate the variable term.

   Given the linear equation:

   $$\begin{array}{cc}5x-10& =10\end{array}$$

   Add $10$ to both sides:

   $$\begin{array}{cc}5x& =10+10\\ 5x& =20\end{array}$$

2. Solve for $x$.

   Divide both sides by $5$:

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

3. Verify and analyze distractors.

   Substitute $x=4$ into the original equation:

   $$\begin{array}{cc}5(4)-10& =20-10\\ & =10\end{array}$$

   The left-hand side matches the right-hand side.

   • Option (a) $0$: Occurs if one incorrectly subtracts $10-10=0$.
   • Option (b) $50$: Occurs if one incorrectly multiplies $10\times 5$.
   • Option (c) $4$: Correct value.
   • Option (d) $-4$: Result of a sign error.

Solution

1. Solve the linear inequality.

   Given:

   $$\begin{array}{cc}7x+4& <6x+6\end{array}$$

   Subtract $6x$ from both sides:

   $$\begin{array}{cc}7x-6x+4& <6\\ x+4& <6\end{array}$$

   Subtract $4$ from both sides:

   $$\begin{array}{cc}x& <6-4\\ x& <2\end{array}$$

2. Express the solution in interval notation.

   The inequality $x<2$ represents all real numbers strictly less than $2$. Since the inequality is strict $(<)$, the endpoint $2$ is not included (represented by a round parenthesis):

   $$\begin{array}{cc}x& \in (-\infty ,2)\end{array}$$

3. Analyze distractors.

   • Option (a) $(2,\infty )$: Represents $x>2$, reversing the inequality.
   • Option (b) $[2,\infty )$: Represents $x\ge 2$, reversing direction and improperly including the endpoint.
   • Option (c) $(-\infty ,2)$: Correct strict interval.
   • Option (d) $(-\infty ,2]$: Incorrectly includes the boundary point $2$ with a square bracket $(x\le 2)$.

Solution

1. Understand plane partitioning by a line.

   In coordinate geometry, any straight line $L$ drawn on the Cartesian plane partitions the points of the plane into three disjoint sets:

   1. Points lying on the line $L$ itself.
   2. Points lying on one side of $L$.
   3. Points lying on the opposite side of $L$.

   The regions on either side are each called a half-plane. Together, the line creates two half planes.

2. Consider the specific case of a vertical line.

   A vertical line has equation:

   $$\begin{array}{cc}x& =c\end{array}$$

   It divides the Cartesian plane into:

   1. The left half plane, defined by $x<c$.
   2. The right half plane, defined by $x>c$.

   Thus, the vertical line divides the plane into two half planes (the left half plane and the right half plane).

3. Analyze distractors.

   • Options (a) and (b): Each names only one of the two regions resulting from the division.
   • Option (c): Incorrect because a line partitions the plane rather than keeping it whole.
   • Option (d): Correct because the division results in two half planes.

Solution

1. State the definition.

   In the study of linear inequalities and linear programming: When the inequality symbol $(<,>,\le ,\text{ or }\ge )$ in a linear inequality such as:

   $$\begin{array}{cc}ax+by& <c\text{or}ax+by\le c\end{array}$$

   is replaced by an equality symbol $(=)$, the resulting equation:

   $$\begin{array}{cc}ax+by& =c\end{array}$$

   is specifically defined as the associated equation (also known as the boundary equation or corresponding equation).

2. Purpose of the associated equation.

   The graph of the associated equation is a straight line that acts as the boundary separating the plane into two half-planes.

3. Analyze distractors.

   • Option (a): While the resulting equation is linear, the specific terminology used in the curriculum for the equation derived from an inequality is the "associated equation".
   • Option (b): Exactly matches the formal textbook definition.
   • Option (c): Incorrect because the degree of the variables remains $1$, not $2$.
   • Option (d): Incorrect since option (b) is the correct term.

Solution

1. Identify the components of the expression.

   The given mathematical statement is:

   $$\begin{array}{cc}3x+4& <0\end{array}$$

   It consists of an algebraic expression $3x+4$ and a constant $0$ connected by the strictly less than relational operator $(<)$.

2. Apply mathematical definitions.

   • An inequality is any mathematical statement stating that one expression is greater than, greater than or equal to, less than, or less than or equal to another expression using symbols $>,\ge ,<,\le$.
   • An equation asserts that two expressions are equal using $(=)$.
   • An identity is an equation that holds true for all permissible values of the variable.

   Since $3x+4<0$ contains the symbol $(<)$, it is an inequality.

3. Analyze distractors.

   • Option (a) equation: Incorrect because it lacks the equality symbol $(=)$.
   • Option (b) inequality: Correct formal classification.
   • Option (c) not inequality: False statement.
   • Option (d) identity: Incorrect because it is neither an equation nor universally true for all $x\in ℝ$.

Solution

1. Define corner point in linear programming.

   In linear inequalities and linear programming: A corner point of a feasible region is a point in the solution space where two boundary lines intersect.

2. Geometric correspondence.

   In geometry:

   • A point where two straight line segments or boundary lines meet to form an angle or corner of a polygon is called a vertex (plural: vertices).
   • Therefore, in linear programming literature and geometry, a corner point of the solution or feasible region is also called a vertex.

3. Analyze distractors.

   • Option (a) code: Unrelated mathematical or computing term.
   • Option (b) vertex: Correct standard mathematical synonym.
   • Option (c) curve: A continuous one-dimensional geometric object, not an isolated point.
   • Option (d) region: A two-dimensional set of points or area, not an intersection point.

Solution

1. Understand the test point method.

   A point $(x_0,y_0)=(0,0)$ is a solution to an inequality if substituting $x=0$ and $y=0$ into the inequality results in a true mathematical statement.

2. Test each option at $(0,0)$.

   • Option (a): $4x+5y>8$

   $$\begin{array}{cc}4(0)+5(0)& >8\\ 0+0& >8\\ 0& >8(\text{False})\end{array}$$

   • Option (b): $3x+y>6$

   $$\begin{array}{cc}3(0)+0& >6\\ 0+0& >6\\ 0& >6(\text{False})\end{array}$$

   • Option (c): $-2x+3y<6$

   $$\begin{array}{cc}-2(0)+3(0)& <6\\ 0+0& <6\\ 0& <6(\text{True})\end{array}$$

   • Option (d): $x+y>4$

   $$\begin{array}{cc}0+0& >4\\ 0& >4(\text{False})\end{array}$$

3. Conclusion.

   Since $0<6$ is true, the origin $(0,0)$ satisfies only the inequality in option (c).

Solution

1. Understand constraints and non-negativity.

   In a linear programming problem (LPP), the system contains:

   1. Problem constraints (linear inequalities describing physical/economic limitations).
   2. Non-negativity constraints:

   $$\begin{array}{cc}x& \ge 0\\ y& \ge 0\end{array}$$

   These non-negativity constraints restrict all valid solutions strictly to the first quadrant (including its boundary axes).

2. Define feasible region.

   The set of all points satisfying all the given system of linear inequalities together with the non-negativity constraints $(x\ge 0,y\ge 0)$ is termed the feasible region. Each point within this region is called a feasible solution.

3. Analyze distractors.

   • Option (a) objective region: Fictitious terminology; the objective function is the linear expression to be optimized, not a region.
   • Option (b) feasible region: Correct standard technical term.
   • Option (c) solution region: A general term for the graph of any inequality or system of inequalities without necessarily imposing the first-quadrant non-negativity condition.
   • Option (d) constraints region: Non-standard descriptive phrase, not the formal mathematical term.

Solution

1. Define optimization in linear programming.

   In linear programming: The primary goal is to find optimal decisions (values of variables $x$ and $y$) that maximize a benefit (such as profit or revenue) or minimize a cost (such as expenditure or loss).

2. Formal definition of the objective function.

   The linear function of the decision variables:

   $$\begin{array}{cc}z& =f(x,y)\\ & =ax+by\end{array}$$

   which is formulated to be maximized or minimized subject to a set of linear constraints is termed the objective function.

3. Analyze distractors.

   • Option (a) solution function: Incorrect, not a standard mathematical term in optimization.
   • Option (b) objective function: Exactly matches the formal definition.
   • Option (c) feasible function: Incorrect terminology.
   • Option (d) none of these: Incorrect since option (b) is the correct term.

Solution

Solution

1. Clear fractions by multiplying both sides by $3$.

   Multiply both sides of the equation by $3$:

   $$\begin{array}{cc}3\left(\frac{x+5}{3}\right)& =3(1-x)\\ x+5& =3-3x\end{array}$$

2. Collect like terms and solve for $x$.

   Add $3x$ to both sides:

   $$\begin{array}{cc}x+3x+5& =3\\ 4x+5& =3\end{array}$$

   Subtract $5$ from both sides:

   $$\begin{array}{cc}4x& =3-5\\ 4x& =-2\end{array}$$

   Divide both sides by $4$:

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

3. Check the solution.

   Substitute $x=-\frac{1}{2}$ into the original equation:

   $$\begin{array}{cc}\text{L.H.S.}& =\frac{(-\frac{1}{2})+5}{3}\\ & =\frac{\frac{-1+10}{2}}{3}\\ & =\frac{\frac{9}{2}}{3}\\ & =\frac{9}{2\times 3}\\ & =\frac{3}{2}\end{array}$$

   Evaluate the Right-Hand Side:

   $$\begin{array}{cc}\text{R.H.S.}& =1-(-\frac{1}{2})\\ & =1+\frac{1}{2}\\ & =\frac{3}{2}\end{array}$$

   Since $\text{L.H.S.}=\text{R.H.S.}=\frac{3}{2}$, the solution $x=-\frac{1}{2}$ is verified.

4. Representation on the real number line.

   On the real number line, the single discrete solution $x=-\frac{1}{2}=-0.5$ is represented by a solid dot midway between $-1$ and $0$.

   Solution: x = -1/2 on Real Linescale: 1 unit

Solution

1. Clear fractions using the least common multiple (L.C.M.).

   The denominators are $3$ and $2$. Their least common multiple is $\text{L.C.M.}(3,2)=6$. Multiply every term on both sides by $6$:

   $$\begin{array}{cc}6\left(\frac{2x+1}{3}\right)+6\left(\frac{1}{2}\right)& =6(1)-6\left(\frac{x-1}{3}\right)\\ 2(2x+1)+3(1)& =6-2(x-1)\end{array}$$

2. Expand parentheses and simplify both sides.

   $$\begin{array}{cc}4x+2+3& =6-2x+2\\ 4x+5& =8-2x\end{array}$$

3. Isolate the variable $x$.

   Add $2x$ to both sides:

   $$\begin{array}{cc}4x+2x+5& =8\\ 6x+5& =8\end{array}$$

   Subtract $5$ from both sides:

   $$\begin{array}{cc}6x& =8-5\\ 6x& =3\end{array}$$

   Divide both sides by $6$:

   $$\begin{array}{cc}x& =\frac{3}{6}\\ x& =\frac{1}{2}\end{array}$$

4. Check the solution.

   Substitute $x=\frac{1}{2}$ into the original equation:

   $$\begin{array}{cc}\text{L.H.S.}& =\frac{2\left(\frac{1}{2}\right)+1}{3}+\frac{1}{2}\\ & =\frac{1+1}{3}+\frac{1}{2}\\ & =\frac{2}{3}+\frac{1}{2}\\ & =\frac{4+3}{6}\\ & =\frac{7}{6}\end{array}$$

   Evaluate the Right-Hand Side:

   $$\begin{array}{cc}\text{R.H.S.}& =1-\frac{\frac{1}{2}-1}{3}\\ & =1-\frac{-\frac{1}{2}}{3}\\ & =1-(-\frac{1}{6})\\ & =1+\frac{1}{6}\\ & =\frac{7}{6}\end{array}$$

   Since $\text{L.H.S.}=\text{R.H.S.}=\frac{7}{6}$, the solution $x=\frac{1}{2}$ is confirmed.

5. Representation on the real number line.

   On the real line, the unique solution $x=\frac{1}{2}=0.5$ is plotted as a single solid dot midway between $0$ and $1$.

   Solution: x = 1/2 on Real Linescale: 1 unit

Solution

1. Isolate the variable term.

   Subtract $7$ from both sides of the inequality:

   $$\begin{array}{cc}3x+7-7& <16-7\\ 3x& <9\end{array}$$

2. Solve for $x$.

   Divide both sides by $3$ (since $3>0$, the inequality direction is preserved):

   $$\begin{array}{cc}\frac{3x}{3}& <\frac{9}{3}\\ x& <3\end{array}$$

3. Express in set-builder and interval notation.

   • Set-builder notation:

   $$\{x|x<3,x\in ℝ\}$$

   • Interval notation:

   $$(-\infty ,3)$$

4. Representation on the real number line.

   Because $x<3$ is a strict inequality, an open circle (hollow ring) is placed at $x=3$ to show that $3$ is excluded from the solution set. A highlighted ray extends indefinitely to the left toward $-\infty$.

   Inequality: x < 3 on Real Linescale: 1 unit

Solution

1. Expand parentheses on both sides.

   Distribute $5$ on the left-hand side and simplify the right-hand side:

   $$\begin{array}{cc}5(x)-5(3)& \ge 26x-10x-4\\ 5x-15& \ge 16x-4\end{array}$$

2. Collect the variable terms on the left.

   Subtract $16x$ from both sides:

   $$\begin{array}{cc}5x-16x-15& \ge -4\\ -11x-15& \ge -4\end{array}$$

3. Collect constant terms on the right.

   Add $15$ to both sides:

   $$\begin{array}{cc}-11x& \ge -4+15\\ -11x& \ge 11\end{array}$$

4. Divide by the negative coefficient and reverse the inequality sign.

   Dividing both sides by the negative number $-11$ reverses the inequality sign from $\ge$ to $\le$:

   $$\begin{array}{cc}\frac{-11x}{-11}& \le \frac{11}{-11}\\ x& \le -1\end{array}$$

5. Express in set-builder and interval notation.

   • Set-builder notation:

   $$\{x|x\le -1,x\in ℝ\}$$

   • Interval notation:

   $$(-\infty ,-1]$$

6. Representation on the real number line.

   Because $x\le -1$ includes the endpoint $x=-1$ (non-strict inequality), a solid dot (closed circle) is placed at $x=-1$. A highlighted ray extends indefinitely to the left toward $-\infty$.

   Inequality: x <= -1 on Real Linescale: 1 unit

Solution

Solution

1. Formulate associated equations and find coordinate intercepts.

   Given the system of inequalities:

   $$\begin{array}{cc}3x-4y& \le 12\text{— (1)}\\ 3x+2y& \ge 3\text{— (2)}\end{array}$$

   1. For Inequality (1): The associated (boundary) equation is:

   $$\begin{array}{cc}3x-4y& =12\end{array}$$

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

   $$\begin{array}{cc}3x-4(0)& =12\\ 3x& =12\\ x& =\frac{12}{3}\\ x& =4\end{array}$$

   Thus, the boundary line passes through $A(4,0)$.

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

   $$\begin{array}{cc}3(0)-4y& =12\\ -4y& =12\\ y& =\frac{12}{-4}\\ y& =-3\end{array}$$

   Thus, the boundary line passes through $B(0,-3)$.

   Plot the points $A(4,0)$ and $B(0,-3)$, and join them with a solid straight line (since the inequality includes $\le$).

   2. For Inequality (2): The associated (boundary) equation is:

   $$\begin{array}{cc}3x+2y& =3\end{array}$$

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

   $$\begin{array}{cc}3x+2(0)& =3\\ 3x& =3\\ x& =\frac{3}{3}\\ x& =1\end{array}$$

   Thus, the boundary line passes through $C(1,0)$.

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

   $$\begin{array}{cc}3(0)+2y& =3\\ 2y& =3\\ y& =\frac{3}{2}\end{array}$$

   Thus, the boundary line passes through $D(0,\frac{3}{2})$ (or $D(0,1.5)$).

   Plot the points $C(1,0)$ and $D(0,\frac{3}{2})$, and join them with a solid straight line (since the inequality includes $\ge$).

   ---

2. Determine the solution region of each inequality using test point $(0,0)$.

   • Test Inequality (1) at origin $(0,0)$:

   $$\begin{array}{cc}3(0)-4(0)& \le 12\\ 0-0& \le 12\\ 0& \le 12(\text{True})\end{array}$$

   Since the statement is true, the solution region for $3x-4y\le 12$ is the closed half-plane containing the origin $(0,0)$ (the region above and to the left of the line $3x-4y=12$).

   • Test Inequality (2) at origin $(0,0)$:

   $$\begin{array}{cc}3(0)+2(0)& \ge 3\\ 0+0& \ge 3\\ 0& \ge 3(\text{False})\end{array}$$

   Since the statement is false, the solution region for $3x+2y\ge 3$ is the closed half-plane NOT containing the origin $(0,0)$ (the region above and to the right of the line $3x+2y=3$).

   ---

3. Find the point of intersection of the boundary lines.

   Solve the system of linear equations simultaneously:

   $$\begin{array}{cc}3x-4y& =12\text{— (1)}\\ 3x+2y& =3\text{— (2)}\end{array}$$

   Subtract equation (1) from equation (2):

   $$\begin{array}{cc}(3x+2y)-(3x-4y)& =3-12\\ 3x+2y-3x+4y& =-9\\ 6y& =-9\\ y& =-\frac{9}{6}\\ y& =-\frac{3}{2}\end{array}$$

   Substitute $y=-\frac{3}{2}$ into equation (2):

   $$\begin{array}{cc}3x+2(-\frac{3}{2})& =3\\ 3x-3& =3\\ 3x& =3+3\\ 3x& =6\\ x& =\frac{6}{3}\\ x& =2\end{array}$$

   Hence, the two boundary lines intersect at the point:

   $$\begin{array}{c}P(2,-\frac{3}{2})\end{array}$$

   ---

4. Conclusion: Solution Region.

   The solution region of the given system of inequalities is the common (overlapping) shaded region satisfying both:

   1. The closed half-plane of $3x-4y\le 12$ towards the origin.
   2. The closed half-plane of $3x+2y\ge 3$ away from the origin.

   This forms an unbounded wedge-shaped region with vertex at $P(2,-\frac{3}{2})$, including all points lying on both solid boundary lines within the region.

Solution

1. Formulate associated equations and find coordinate intercepts.

   Given the system of inequalities:

   $$\begin{array}{cc}2x+y& \le 4\text{— (1)}\\ x+2y& \le 6\text{— (2)}\end{array}$$

   1. For Inequality (1): The associated (boundary) equation is:

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

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

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

   Thus, the boundary line passes through $A(2,0)$.

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

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

   Thus, the boundary line passes through $B(0,4)$.

   Plot the points $A(2,0)$ and $B(0,4)$, and draw a solid straight line through them.

   2. For Inequality (2): The associated (boundary) equation is:

   $$\begin{array}{cc}x+2y& =6\end{array}$$

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

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

   Thus, the boundary line passes through $C(6,0)$.

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

   $$\begin{array}{cc}0+2y& =6\\ 2y& =6\\ y& =\frac{6}{2}\\ y& =3\end{array}$$

   Thus, the boundary line passes through $D(0,3)$.

   Plot the points $C(6,0)$ and $D(0,3)$, and draw a solid straight line through them.

   ---

2. Determine the solution region of each inequality using test point $(0,0)$.

   • Test Inequality (1) at origin $(0,0)$:

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

   Since the statement is true, the solution region for $2x+y\le 4$ is the closed half-plane containing the origin $(0,0)$ (the region on and below the line $2x+y=4$).

   • Test Inequality (2) at origin $(0,0)$:

   $$\begin{array}{cc}(0)+2(0)& \le 6\\ 0+0& \le 6\\ 0& \le 6(\text{True})\end{array}$$

   Since the statement is true, the solution region for $x+2y\le 6$ is the closed half-plane containing the origin $(0,0)$ (the region on and below the line $x+2y=6$).

   ---

3. Find the point of intersection of the boundary lines.

   Solve the system of equations simultaneously:

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

   Multiply equation (1) by $2$:

   $$\begin{array}{cc}2(2x+y)& =2(4)\\ 4x+2y& =8\text{— (3)}\end{array}$$

   Subtract equation (2) from equation (3):

   $$\begin{array}{cc}(4x+2y)-(x+2y)& =8-6\\ 3x& =2\\ x& =\frac{2}{3}\end{array}$$

   Substitute $x=\frac{2}{3}$ into equation (1):

   $$\begin{array}{cc}2\left(\frac{2}{3}\right)+y& =4\\ \frac{4}{3}+y& =4\\ y& =4-\frac{4}{3}\\ y& =\frac{12-4}{3}\\ y& =\frac{8}{3}\end{array}$$

   Hence, the two boundary lines intersect at the point:

   $$\begin{array}{c}E(\frac{2}{3},\frac{8}{3})\end{array}$$

   ---

4. Conclusion: Solution Region.

   The solution region of the given system of inequalities is the common (overlapping) shaded region that satisfies both half-planes containing the origin $(0,0)$. Because there are no non-negativity constraints ($x\ge 0,y\ge 0$) specified, this solution region extends infinitely into the negative directions, bounded by the rays of the two boundary lines meeting at the corner point $E(\frac{2}{3},\frac{8}{3})$.

Solution

1. Formulate the associated equation and find intercepts.

   Given constraints:

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

   For inequality (1), 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 $A(4,0)$ lies on the boundary line.

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

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

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

   ---

2. Determine the solution half-plane using test point $(0,0)$.

   Substitute the origin $(0,0)$ into inequality (1):

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

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

   ---

3. Identify the feasible region.

   The non-negativity constraints:

   $$\begin{array}{cc}x& \ge 0\\ y& \ge 0\end{array}$$

   restrict the solution region exclusively to the first quadrant. Thus, the feasible region is the closed triangular region bounded by the coordinate axes and the line segment joining $A(4,0)$ and $B(0,4)$.

   ---

4. Determine the corner points (vertices) of the feasible region.

   The corner points bounding the feasible region are:

   1. The origin:

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

   2. The $x$-intercept:

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

   3. The $y$-intercept:

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

   ---

5. Evaluate the objective function $g(x,y)=x+4y$ at each corner point.

   Evaluate $g(x,y)$ at each vertex:

   • At $O(0,0)$:

   $$\begin{array}{cc}g(0,0)& =0+4(0)\\ & =0+0\\ & =0\end{array}$$

   • At $A(4,0)$:

   $$\begin{array}{cc}g(4,0)& =4+4(0)\\ & =4+0\\ & =4\end{array}$$

   • At $B(0,4)$:

   $$\begin{array}{cc}g(0,4)& =0+4(4)\\ & =0+16\\ & =16\end{array}$$

   Summary table:

   Corner Point $(x,y)$	Objective Function $g(x,y)=x+4y$	Value
   $O(0,0)$	$0+4(0)$	$0$
   $A(4,0)$	$4+4(0)$	$4$
   $B(0,4)$	$0+4(4)$	$16$

   ---

6. Conclude the optimal solution.

   Comparing all values, the maximum value of the objective function is $16$, which occurs at the corner point $B(0,4)$.

Solution

1. Formulate associated equations and find intercepts.

   Given constraints:

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

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

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

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

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

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

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

   $$\begin{array}{cc}0+3y& =3\\ 3y& =3\\ y& =1\end{array}$$

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

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

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

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

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

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

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

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

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

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2. Determine the solution half-planes using test point $(0,0)$.

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

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

   Since this is false, the solution region for $x+3y\ge 3$ is the closed half-plane not containing the origin $(0,0)$ (the region above and away from line 1).

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

   $$\begin{array}{cc}(0)+(0)& \ge 2\\ 0& \ge 2(\text{False})\end{array}$$

   Since this is false, the solution region for $x+y\ge 2$ is the closed half-plane not containing the origin $(0,0)$ (the region above and away from line 2).

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3. Determine the feasible region.

   The non-negativity constraints $x\ge 0$ and $y\ge 0$ restrict the feasible region to the first quadrant. Since both inequality conditions require points above their respective boundary lines, the feasible region is an unbounded polygonal region extending infinitely in the first quadrant. Although the region is unbounded, a minimum exists: both coefficients of $f(x,y)=3x+5y$ are positive and $x,y\ge 0$ on the region, so $f$ is bounded below there and attains its minimum at a corner point.

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

   1. Point on the positive $x$-axis: Between $A(3,0)$ and $C(2,0)$, testing both constraints:

   • At $x=3$: $3+3(0)=3\ge 3$ and $3+0=3\ge 2$ (both satisfied).
   • At $x=2$: $2+3(0)=2\ge ̸3$ (not in feasible region). Therefore, the vertex on the $x$-axis is $A(3,0)$.

   2. Point of intersection $E$ of the two boundary lines: Solve simultaneously:

   $$\begin{array}{cc}x+3y& =3\text{— (i)}\\ x+y& =2\text{— (ii)}\end{array}$$

   Subtracting equation (ii) from equation (i):

   $$\begin{array}{cc}(x+3y)-(x+y)& =3-2\\ 2y& =1\\ y& =\frac{1}{2}\end{array}$$

   Substitute $y=\frac{1}{2}$ into equation (ii):

   $$\begin{array}{cc}x+\frac{1}{2}& =2\\ x& =2-\frac{1}{2}\\ x& =\frac{3}{2}\end{array}$$

   Thus, the intersection point is $E(\frac{3}{2},\frac{1}{2})$ or $E(1.5,0.5)$.

   3. Point on the positive $y$-axis: Between $B(0,1)$ and $D(0,2)$, testing both constraints:

   • At $y=2$: $0+3(2)=6\ge 3$ and $0+2=2\ge 2$ (both satisfied).
   • At $y=1$: $0+1=1\ge ̸2$ (not in feasible region). Therefore, the vertex on the $y$-axis is $D(0,2)$.

   The corner points bounding the feasible region are:

   $$\begin{array}{c}A(3,0),E(\frac{3}{2},\frac{1}{2}),D(0,2)\end{array}$$

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

   • At $A(3,0)$:

   $$\begin{array}{cc}f(3,0)& =3(3)+5(0)\\ & =9+0\\ & =9\end{array}$$

   • At $E(\frac{3}{2},\frac{1}{2})$:

   $$\begin{array}{cc}f(\frac{3}{2},\frac{1}{2})& =3\left(\frac{3}{2}\right)+5\left(\frac{1}{2}\right)\\ & =\frac{9}{2}+\frac{5}{2}\\ & =\frac{14}{2}\\ & =7\end{array}$$

   • At $D(0,2)$:

   $$\begin{array}{cc}f(0,2)& =3(0)+5(2)\\ & =0+10\\ & =10\end{array}$$

   Summary table:

   Corner Point $(x,y)$	Objective Function $f(x,y)=3x+5y$	Value
   $A(3,0)$	$3(3)+5(0)$	$9$
   $E(\frac{3}{2},\frac{1}{2})$	$3\left(\frac{3}{2}\right)+5\left(\frac{1}{2}\right)$	$7$
   $D(0,2)$	$3(0)+5(2)$	$10$

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6. Conclude the optimal solution.

   Comparing all evaluated values, the minimum value of the objective function is $7$, which occurs at the corner point $E(\frac{3}{2},\frac{1}{2})$ (or $(1.5,0.5)$).