Navigating the complexities of master's level mathematics can be challenging. For those grappling with intricate problems, a reliable discrete math assignment solver can provide the guidance and clarity needed to excel. At MathsAssignmentHelp.com, experts are available to help tackle even the most demanding assignments. In this blog, we’ll delve into two advanced math problems, providing comprehensive solutions to enhance your understanding and skills.

Problem 1: Proving a Property of Graphs

Question:

Prove that in any simple graph $G$ with $n$ vertices and $e$ edges, the following inequality holds:

e≤n(n−1)2e \leq \frac{n(n-1)}{2}e≤2n(n−1)​

where $e$ is the number of edges and $n$ is the number of vertices.

Solution:

This problem involves understanding basic properties of graphs. The inequality we are asked to prove is known as the edge bound of a simple graph, which states that the maximum number of edges in a simple graph is n(n−1)2\frac{n(n-1)}{2}2n(n−1)​. Here’s a step-by-step approach to prove this:

1. Define a Simple Graph:

   A simple graph is an undirected graph with no loops or multiple edges between the same pair of vertices. Each edge connects two distinct vertices.

2. Count Possible Edges:

   To find the maximum number of edges in a simple graph, consider that each vertex can connect to every other vertex exactly once.

   • For a graph with $n$ vertices, each vertex has $(n-1)$ potential connections (edges).

   • The total number of connections (edges) in the graph, considering each connection is bidirectional, can be represented as the combination (n2)\binom{n}{2}(2n​):

     (n2)=n(n−1)2\binom{n}{2} = \frac{n(n-1)}{2}(2n​)=2n(n−1)​

   • This formula calculates the number of ways to choose 2 vertices out of $n$ to form an edge.

3. Maximum Edge Count:

   Since every edge connects two distinct vertices and each pair of vertices is connected by at most one edge, the maximum number of edges in a simple graph is:

   emax=n(n−1)2e_{\text{max}} = \frac{n(n-1)}{2}emax​=2n(n−1)​

Problem 2: Solving a Linear Programming Problem

Question:

Maximize the objective function:

Z=3x1+2x2Z = 3x_1 + 2x_2Z=3x1​+2x2​

subject to the constraints:

2x1+x2≤82x_1 + x_2 \leq 82x1​+x2​≤8 x1+2x2≤6x_1 + 2x_2 \leq 6x1​+2x2​≤6 x1≥0,x2≥0x_1 \geq 0, \quad x_2 \geq 0x1​≥0,x2​≥0

Solution:

Linear programming problems are pivotal in optimization and operations research. Here’s a step-by-step solution to this problem using the graphical method.

1. Graph the Constraints:

   Convert each inequality constraint into an equality to find the boundary lines of the feasible region.

   • For 2x1+x2=82x_1 + x_2 = 82x1​+x2​=8:

     • When x1=0x_1 = 0x1​=0, x2=8x_2 = 8x2​=8.
     • When x2=0x_2 = 0x2​=0, x1=4x_1 = 4x1​=4.

     Plot these points and draw the line.

   • For x1+2x2=6x_1 + 2x_2 = 6x1​+2x2​=6:

     • When x1=0x_1 = 0x1​=0, x2=3x_2 = 3x2​=3.
     • When x2=0x_2 = 0x2​=0, x1=6x_1 = 6x1​=6.

     Plot these points and draw the line.

2. Determine the Feasible Region:

   The feasible region is the area where all the constraints overlap. It is bounded by the lines plotted above and the axes x1≥0x_1 \geq 0x1​≥0 and x2≥0x_2 \geq 0x2​≥0. Shade this region on the graph.

3. Find the Corner Points:

   Identify the corner points (vertices) of the feasible region. These are typically where the constraints intersect or touch the axes.

   • Intersection of 2x1+x2=82x_1 + x_2 = 82x1​+x2​=8 and x1+2x2=6x_1 + 2x_2 = 6x1​+2x2​=6:

     Solving these equations simultaneously:

     2x1+x2=8x1+2x2=6\begin{aligned} 2x_1 + x_2 &= 8 \\ x_1 + 2x_2 &= 6 \end{aligned}2x1​+x2​x1​+2x2​​=8=6​

     Multiply the second equation by 2:

     2x1+4x2=122x_1 + 4x_2 = 122x1​+4x2​=12

     Subtract the first equation from this result:

     3x2=4⇒x2=433x_2 = 4 \quad \Rightarrow \quad x_2 = \frac{4}{3}3x2​=4⇒x2​=34​

     Substitute x2=43x_2 = \frac{4}{3}x2​=34​ back into x1+2x2=6x_1 + 2x_2 = 6x1​+2x2​=6:

     x1+2(43)=6⇒x1=103x_1 + 2 \left(\frac{4}{3}ight) = 6 \quad \Rightarrow \quad x_1 = \frac{10}{3}x1​+2(34​)=6⇒x1​=310​

     So, the intersection point is (103,43)\left(\frac{10}{3}, \frac{4}{3}ight)(310​,34​).

   • Other corner points are $(0,0)$, $(0,3)$, and $(4,0)$.

4. Evaluate the Objective Function:

   Calculate the objective function Z=3x1+2x2Z = 3x_1 + 2x_2Z=3x1​+2x2​ at each corner point:

   • At $(0,0)$:

     $$Z=3\times 0+2\times 0=0$$

   • At $(0,3)$:

     $$Z=3\times 0+2\times 3=6$$

   • At $(4,0)$:

     $$Z=3\times 4+2\times 0=12$$

   • At (103,43)\left(\frac{10}{3}, \frac{4}{3}ight)(310​,34​):

     Z=3×103+2×43=10+83=383≈12.67Z = 3 \times \frac{10}{3} + 2 \times \frac{4}{3} = 10 + \frac{8}{3} = \frac{38}{3} \approx 12.67Z=3×310​+2×34​=10+38​=338​≈12.67

Final Thoughts

Master's level mathematics requires not only understanding advanced concepts but also applying them effectively. The problems discussed above highlight key techniques in graph theory and optimization. Whether you're struggling with similar problems or need help with other complex topics, a discrete math assignment solver at MathsAssignmentHelp.com is available to assist. With expert guidance, you can master these challenging problems and excel in your studies.