Lecture on Optimization Problems

Introduction

• Optimization problems aim to find dimensions or situations that maximize or minimize certain quantities.
• Common goals include maximizing area, minimizing fencing, maximizing profit, or minimizing cost.
• Key idea: identify maximum and minimum values of a function.
• At maximum/minimum points, the slope of the tangent line is zero.
• Process: Take the first derivative, set it to zero, solve for variables.

Problem 1: Two Numbers with Maximum Product

• Problem Statement: Find two numbers whose sum is 60 and whose product is maximum.
• Equations:
  • Sum: x + y = 60
  • Product: P = x * y
• Objective and Constraint:
  • Sum is constraint (fixed value: 60).
  • Product is objective function (to be maximized).
• Solution Steps:
  1. Express y in terms of x: y = 60 - x
  2. Substitute in product equation: P = x(60 - x)
  3. Simplify and differentiate: P' = 60 - 2x
  4. Set P' = 0 and solve for x: x = 30
  5. Find y: y = 30
  6. Calculate maximum product: 30 * 30 = 900
• Verification:
  • Table to show other combinations yield less product.

Problem 2: Two Numbers with Minimum Product

• Problem Statement: Find two numbers whose difference is 40 and product is minimum.
• Equations:
  • Difference: y - x = 40
  • Product: P = x * y
• Objective and Constraint:
  • Difference is constraint.
  • Product is objective function (to minimize).
• Solution Steps:
  1. Express y in terms of x: y = 40 + x
  2. Substitute in product equation: P = x(40 + x)
  3. Simplify and differentiate: P' = 40 + 2x
  4. Set P' = 0 and solve for x: x = -20
  5. Find y: y = 20
  6. Calculate minimum product: -20 * 20 = -400
• Verification:
  • Table to show other combinations yield higher product.

Problem 3: Two Numbers with Minimum Sum

• Problem Statement: Find two positive numbers whose product is 400 and sum is minimized.
• Equations:
  • Product: x * y = 400
  • Sum: S = x + y
• Objective and Constraint:
  • Product is constraint.
  • Sum is objective function (to minimize).
• Solution Steps:
  1. Express y in terms of x: y = 400/x
  2. Substitute in sum equation: S = x + 400/x
  3. Differentiate: S' = 1 - 400/x^2
  4. Set S' = 0 and solve for x: x = 20
  5. Find y: y = 20
  6. Calculate minimum sum: 20 + 20 = 40*

Problem 4: Point with Minimum Sum of Number and Reciprocal

• Problem Statement: Find positive number x where sum of x and its reciprocal is minimum.
• Equation:
  • Sum: S = x + 1/x
• Solution Steps:
  1. Differentiate: S' = 1 - 1/x^2
  2. Set S' = 0 and solve for x: x = 1
  3. Verify minimum sum: 1 + 1/1 = 2

Problem 5: Rectangle with Maximum Area

• Problem Statement: Find dimensions of a rectangle with perimeter of 200 feet that has maximum area.
• Equations:
  • Perimeter: 2L + 2W = 200
  • Area: A = L * W
• Objective and Constraint:
  • Perimeter is constraint.
  • Area is objective function (to maximize).
• Solution Steps:
  1. Express W in terms of L: W = 100 - L
  2. Substitute in area equation: A = L(100 - L)
  3. Differentiate: A' = 100 - 2L
  4. Set A' = 0 and solve for L: L = 50
  5. Calculate W: W = 50
  6. Calculate maximum area: 50 * 50 = 2500 square feet

Problem 6: Field Along River with Maximum Area

• Problem Statement: A farmer with 600 feet of fencing wants a rectangular field along a river with largest area.
• Equation and Objective:
  • Perimeter: x + 2y = 600
  • Maximize area: A = x * y
• Solution Steps:
  1. Express x in terms of y: x = 600 - 2y
  2. Substitute in area equation: A = (600 - 2y) * y
  3. Differentiate: A' = 600 - 4y
  4. Set A' = 0 and solve for y: y = 150
  5. Calculate x: x = 300
  6. Maximum area: 300 * 150 = 45,000 square feet*

Problem 7: Field Along River with Fixed Area

• Problem Statement: A farmer needs a field of 10,000 square feet with least fencing along a river.
• Equation and Objective:
  • Area: x * y = 10,000
  • Minimize perimeter: P = x + 2y
• Solution Steps:
  1. Express y in terms of x: y = 10,000/x
  2. Substitute in perimeter equation: P = x + 20,000/x
  3. Differentiate: P' = 1 - 20,000/x^2
  4. Set P' = 0 and solve for x: x ≈ 141.42
  5. Calculate y: y ≈ 70.75
  6. Minimum fencing: P ≈ 282.8 feet*

Problem 8: Field Divided into Three Pens

• Problem Statement: A farmer divides a rectangular area into three pens using 1600 feet of fencing.
• Equation and Objective:
  • Perimeter: 2x + 4y = 1600
  • Maximize total area: A = x * y
• Solution Steps:
  1. Express x in terms of y: x = 800 - 2y
  2. Substitute in area equation: A = (800 - 2y) * y
  3. Differentiate: A' = 800 - 4y
  4. Set A' = 0 and solve for y: y = 200
  5. Calculate x: x = 400
  6. Maximum area: 400 * 200 = 80,000 square feet*

Problem 9: Closest Point on a Line to the Origin

• Problem Statement: Find point on the line y = 3x + 5 closest to the origin.
• Distance Formula:
  • d = sqrt(x^2 + y^2)
• Solution Steps:
  1. Express y in terms of x: y = 3x + 5
  2. Substitute in distance equation: d = sqrt(x^2 + (3x + 5)^2)
  3. Differentiate and simplify.
  4. Solve for x: x = -1.5
  5. Calculate y: y = 0.5
  6. Closest point: (-1.5, 0.5)

Problem 10: Closest Point on a Line to Another Point

• Problem Statement: Find point on line y = 4 - x closest to (7, 6).
• Solution Steps:
  1. Use distance formula and substitute y.
  2. Differentiate, find x, solve: x = 2.5
  3. Calculate y: y = 1.5
  4. Closest point: (2.5, 1.5)

Problem 11: Maximum Slope on a Curve

• Problem Statement: Find point on curve y = 6x^2 - x^3 + 10 with highest slope.
• Solution Steps:
  1. First derivative represents slope.
  2. Maximize this slope by setting second derivative to zero.
  3. Solve for x: x = 2
  4. Calculate y: y = 26
  5. Point with highest slope: (2, 26)
  6. Maximum slope value: 12

Problem 12: Rectangle in Semi-circle

• Problem Statement: A rectangle inscribed in a semi-circle with radius 10 cm.
• Equation and Objective:
  • Find dimensions that maximize area.
• Solution Steps:
  1. Use circle equation x^2 + y^2 = r^2 to express y.
  2. Substitute in area equation.
  3. Differentiate, solve for x, calculate y.
  4. Maximize area: Length is 10√2 cm, width is 5√2 cm.
  5. Maximum area: 100 square cm.

These notes summarize the key concepts and problem-solving strategies for various optimization problems taught in the lecture. They include problem statements, equations, objectives, solution steps, and verification methods. This structured approach aids in understanding how to tackle optimization problems in real-life scenarios.