Solve The Following Problems Show Your Complete Solution

Let's tackle some challenging problems with detailed, step-by-step solutions. This approach allows for a thorough understanding of the problem-solving process, making it easier to apply these techniques to future challenges.

Problem 1: Calculus - Optimization

Problem Statement: A rectangular garden is to be fenced off. You have 500 feet of fencing material. What dimensions should the garden have if you want to maximize its area? What is the maximum area?

Solution:

1. Define Variables:

   • Let l represent the length of the rectangular garden.
   • Let w represent the width of the rectangular garden.
   • Let A represent the area of the rectangular garden.

2. Formulate Equations:

   • Perimeter (Constraint): The total fencing material is 500 feet, so the perimeter of the rectangle is 2l + 2w = 500.
   • Area (Objective Function): We want to maximize the area, which is given by A = l * w.

3. Express the Objective Function in Terms of One Variable:

   • Solve the perimeter equation for one variable (let's solve for l):

     2l = 500 - 2w l = 250 - w

   • Substitute this expression for l into the area equation:

     A = (250 - w) * w A = 250w - w²

4. Find the Critical Points:

   • Take the derivative of the area function with respect to w:

     dA/dw = 250 - 2w

   • Set the derivative equal to zero and solve for w:

     250 - 2w = 0 2w = 250 w = 125

5. Determine if the Critical Point is a Maximum:

   • Take the second derivative of the area function with respect to w:

     d²A/dw² = -2

   • Since the second derivative is negative, the critical point w = 125 corresponds to a maximum area.

6. Find the Length:

   • Substitute w = 125 back into the equation l = 250 - w:

     l = 250 - 125 l = 125

7. Calculate the Maximum Area:

   • Substitute l = 125 and w = 125 into the area equation A = l * w:

     A = 125 * 125 A = 15625

8. Answer:

   • The dimensions that maximize the area of the garden are 125 feet by 125 feet (a square).
   • The maximum area of the garden is 15625 square feet.

Problem 2: Discrete Mathematics - Combinatorics

Problem Statement: How many different committees of 5 people can be formed from a group of 10 people?

Solution:

This problem involves combinations because the order in which the people are chosen for the committee does not matter. We use the combination formula:

nCr = n! / (r! * (n-r)!)

where:

• n is the total number of items (in this case, 10 people).
• r is the number of items to choose (in this case, 5 people).
• ! denotes the factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1).

1. Identify n and r:

   • n = 10 (total number of people)
   • r = 5 (number of people to choose for the committee)

2. Apply the Combination Formula:

   10C5 = 10! / (5! * (10-5)!) 10C5 = 10! / (5! * 5!)

3. Calculate the Factorials:

   • 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3,628,800
   • 5! = 5 * 4 * 3 * 2 * 1 = 120

4. Substitute and Simplify:

   10C5 = 3,628,800 / (120 * 120) 10C5 = 3,628,800 / 14,400 10C5 = 252

5. Answer:

   There are 252 different committees of 5 people that can be formed from a group of 10 people.

Problem 3: Linear Algebra - Solving Systems of Equations

Problem Statement: Solve the following system of linear equations:

2x + y - z = 8 -3x - y + 2z = -11 -2x + y + 2z = -3

Solution:

We can use several methods to solve this system, such as Gaussian elimination, matrix inversion, or Cramer's rule. Let's use Gaussian elimination.

1. Write the Augmented Matrix:

   The augmented matrix representing the system of equations is:

   [ 2  1 -1 |  8 ]
   [-3 -1  2 | -11 ]
   [-2  1  2 | -3 ]
   

2. Perform Row Operations to Obtain Row-Echelon Form:

   • Step 1: Eliminate x in the second and third rows.

     • Multiply the first row by 3/2 and add it to the second row: R2 = R2 + (3/2)R1

       [ 2  1 -1 |  8 ]
       [ 0  1/2  1/2 |  1 ]
       [-2  1  2 | -3 ]
       

     • Add the first row to the third row: R3 = R3 + R1

       [ 2  1 -1 |  8 ]
       [ 0  1/2  1/2 |  1 ]
       [ 0  2  1 |  5 ]
       

   • Step 2: Eliminate y in the third row.

     • Multiply the second row by -4 and add it to the third row: R3 = R3 - 4R2

       [ 2  1 -1 |  8 ]
       [ 0  1/2  1/2 |  1 ]
       [ 0  0 -1 |  1 ]
       

3. Convert Back to Equations:

   The row-echelon form matrix represents the following system of equations:

   2x + y - z = 8 (1/2)y + (1/2)z = 1 -z = 1

4. Solve Using Back-Substitution:

   • From the third equation: z = -1
   • Substitute z = -1 into the second equation: (1/2)y + (1/2)(-1) = 1 (1/2)y - (1/2) = 1 (1/2)y = 3/2 y = 3
   • Substitute y = 3 and z = -1 into the first equation: 2x + 3 - (-1) = 8 2x + 4 = 8 2x = 4 x = 2

5. Answer:

   The solution to the system of equations is x = 2, y = 3, and z = -1. We can write this as an ordered triple: (2, 3, -1).

Problem 4: Probability - Conditional Probability

Problem Statement: A bag contains 3 red balls and 5 blue balls. Two balls are drawn at random without replacement. What is the probability that the first ball is red and the second ball is blue?

Solution:

This is a conditional probability problem. We need to find the probability of drawing a red ball first and then drawing a blue ball second.

1. Define Events:

   • Let R1 be the event that the first ball drawn is red.
   • Let B2 be the event that the second ball drawn is blue.

2. Calculate the Probability of the First Event (R1):

   • There are 3 red balls and 8 total balls.
   • P(R1) = (Number of red balls) / (Total number of balls) = 3/8

3. Calculate the Conditional Probability of the Second Event (B2 given R1):

   • After drawing a red ball, there are now 2 red balls and 5 blue balls left in the bag, for a total of 7 balls.
   • P(B2 | R1) = (Number of blue balls) / (Total number of balls remaining) = 5/7

4. Calculate the Probability of Both Events Occurring:

   • The probability of drawing a red ball first and then a blue ball second is the product of the individual probabilities: P(R1 and B2) = P(R1) * P(B2 | R1) = (3/8) * (5/7)

5. Simplify:

   • P(R1 and B2) = 15/56

6. Answer:

   The probability that the first ball is red and the second ball is blue is 15/56.

Problem 5: Abstract Algebra - Group Theory

Problem Statement: Show that the set of integers Z under the operation of addition (+) forms a group.

Solution:

To show that the set of integers Z under addition forms a group, we need to verify the four group axioms: closure, associativity, identity, and inverse.

1. Closure:

   • For any two integers a and b in Z, their sum (a + b) is also an integer in Z. Therefore, the set of integers is closed under addition.
   • Example: 3 + 5 = 8. Since 3, 5, and 8 are all integers, the closure property holds.

2. Associativity:

   • For any three integers a, b, and c in Z, the order in which we perform the addition does not affect the result: (a + b) + c = a + (b + c). This is a fundamental property of addition.
   • Example: (2 + 4) + 1 = 6 + 1 = 7 and 2 + (4 + 1) = 2 + 5 = 7. Therefore, associativity holds.

3. Identity Element:

   • There exists an identity element e in Z such that for any integer a in Z, a + e = e + a = a. The identity element for addition is 0.
   • For any integer a, a + 0 = 0 + a = a.
   • Example: 7 + 0 = 0 + 7 = 7.

4. Inverse Element:

   • For every integer a in Z, there exists an inverse element a^-1 in Z such that a + a^-1 = a^-1 + a = e (the identity element). In this case, the inverse of a is -a.
   • For any integer a, a + (-a) = (-a) + a = 0.
   • Example: The inverse of 5 is -5 because 5 + (-5) = (-5) + 5 = 0.

Conclusion:

Since the set of integers Z under addition satisfies all four group axioms (closure, associativity, identity, and inverse), we can conclude that (Z, +) forms a group. Furthermore, since addition is commutative (a + b = b + a for all integers a and b), (Z, +) is an abelian group or commutative group.

Problem 6: Differential Equations - Solving First-Order Linear Equations

Problem Statement: Solve the following first-order linear differential equation:

dy/dx + (2/x)y = x

Solution:

1. Identify the Standard Form:

   The given differential equation is already in the standard form of a first-order linear differential equation:

   dy/dx + P(x)y = Q(x)

   Where P(x) = 2/x and Q(x) = x.

2. Find the Integrating Factor:

   The integrating factor, denoted by μ(x), is given by:

   μ(x) = e^{∫P(x) dx}

   In this case:

   μ(x) = e^{∫(2/x) dx} = e^2ln|x| = e^ln(x²) = x²

3. Multiply the Differential Equation by the Integrating Factor:

   Multiply both sides of the original differential equation by the integrating factor x²:

   x²(dy/dx) + x²(2/x)y = x²(x) x²(dy/dx) + 2xy = x³

4. Recognize the Left-Hand Side as a Derivative:

   The left-hand side of the equation is now the derivative of the product of y and the integrating factor:

   d/dx (x²y) = x³

5. Integrate Both Sides with Respect to x:

   ∫ d/dx (x²y) dx = ∫ x³ dx x²y = (x⁴)/4 + C where C is the constant of integration.

6. Solve for y:

   Divide both sides by x² to isolate y:

   y = (x²)/4 + C/x²

7. Answer:

   The general solution to the first-order linear differential equation is:

   y(x) = (x²)/4 + C/x²

Problem 7: Number Theory - Modular Arithmetic

Problem Statement: Find the remainder when 3¹⁰⁰ is divided by 7.

Solution:

We will use the properties of modular arithmetic to simplify the calculation.

1. Find a Pattern in the Powers of 3 Modulo 7:

   Calculate the first few powers of 3 modulo 7:

   • 3¹ ≡ 3 (mod 7)
   • 3² ≡ 9 ≡ 2 (mod 7)
   • 3³ ≡ 3 * 2 ≡ 6 (mod 7)
   • 3⁴ ≡ 3 * 6 ≡ 18 ≡ 4 (mod 7)
   • 3⁵ ≡ 3 * 4 ≡ 12 ≡ 5 (mod 7)
   • 3⁶ ≡ 3 * 5 ≡ 15 ≡ 1 (mod 7)

2. Identify the Cycle:

   We found that 3⁶ ≡ 1 (mod 7). This means the powers of 3 modulo 7 repeat in a cycle of length 6.

3. Divide the Exponent by the Cycle Length:

   Divide 100 by 6:

   100 = 6 * 16 + 4

4. Use the Remainder to Find the Equivalent Power:

   Since 100 = 6 * 16 + 4, we have:

   3¹⁰⁰ ≡ 3^{(6*16 + 4)} ≡ (3⁶)¹⁶ * 3⁴ (mod 7)

5. Simplify Using Modular Arithmetic:

   Since 3⁶ ≡ 1 (mod 7):

   3¹⁰⁰ ≡ (1)¹⁶ * 3⁴ (mod 7) 3¹⁰⁰ ≡ 1 * 3⁴ (mod 7) 3¹⁰⁰ ≡ 3⁴ (mod 7)

6. Calculate the Final Remainder:

   We already calculated 3⁴ ≡ 4 (mod 7).

   Therefore, 3¹⁰⁰ ≡ 4 (mod 7)

7. Answer:

   The remainder when 3¹⁰⁰ is divided by 7 is 4.

These examples cover a range of mathematical topics and demonstrate a systematic approach to problem-solving. Remember to clearly define variables, formulate equations, use appropriate formulas, and carefully interpret the results. Practice is key to improving your problem-solving skills.