Optimization Problems

What is Optimization?

Optimization: Finding maximum or minimum value of a quantity

Real-world examples:

• Maximize profit
• Minimize cost
• Maximize area with fixed perimeter
• Minimize time or distance

Algebraic approach: Use equations and inequalities

Optimizing Quadratic Functions

Quadratic f(x) = ax² + bx + c

Vertex at: x = -b/(2a)

If a > 0: Minimum at vertex

If a < 0: Maximum at vertex

Example 1: Find Maximum

f(x) = -x² + 6x - 5

a = -1 < 0, so maximum exists

Vertex:

x = -6/(2(-1)) = -6/(-2) = 3

Maximum value:

f(3) = -(3)² + 6(3) - 5
     = -9 + 18 - 5
     = 4

Maximum: 4 at x = 3

Example 2: Find Minimum

g(x) = 2x² - 8x + 11

a = 2 > 0, so minimum exists

Vertex:

x = -(-8)/(2·2) = 8/4 = 2

Minimum value:

g(2) = 2(2)² - 8(2) + 11
     = 8 - 16 + 11
     = 3

Minimum: 3 at x = 2

Maximum Area Problems

Common setup: Fixed perimeter, maximize area

Example: Rectangular Garden

Problem: 100 feet of fencing for rectangular garden. Maximize area.**

Let: x = width, y = length

Perimeter: 2x + 2y = 100

Solve for y:

y = 50 - x

Area:

A = xy = x(50 - x)
A = 50x - x²

This is quadratic! a = -1 < 0, so maximum exists

Vertex:

x = -50/(2(-1)) = 25

When x = 25:

y = 50 - 25 = 25

Maximum area:

A = 25 · 25 = 625 square feet

Answer: 25 × 25 square garden (square shape!)

Example: Rectangular Area Against Wall

Problem: Use 60 m fencing for garden against wall (one side not needed). Maximize area.**

Let: x = width, y = length parallel to wall

Fencing equation: x + 2y + x = 60 (two widths + one length)

Simplify: 2x + y = 60

Solve for y: y = 60 - 2x

Area:

A = xy = x(60 - 2x)
A = 60x - 2x²

Vertex:

x = -60/(2(-2)) = -60/(-4) = 15

When x = 15:

y = 60 - 2(15) = 30

Maximum area:

A = 15 · 30 = 450 m²

Answer: 15 m × 30 m**

Product Optimization

Two numbers with constraint

Example: Maximum Product

Problem: Two positive numbers sum to 40. Maximize their product.**

Let: x and y be the numbers

Constraint: x + y = 40, so y = 40 - x

Product:

P = xy = x(40 - x)
P = 40x - x²

Vertex:

x = -40/(2(-1)) = 20

When x = 20: y = 20

Maximum product:

P = 20 · 20 = 400

Answer: Both numbers are 20 (equal split maximizes product)**

Revenue and Profit Optimization

Business applications

Example: Ticket Pricing

Problem: Theater has 500 seats. At $10/ticket, sells out. Each $1 increase decreases sales by 50 tickets. Maximize revenue.**

Let: x = number of $1 increases

Price: 10 + x

Tickets sold: 500 - 50x

Revenue:

R = (price)(tickets)
R = (10 + x)(500 - 50x)
R = 5000 + 500x - 500x - 50x²
R = 5000 - 50x²

Wait, that's wrong. Let me recalculate:

R = (10 + x)(500 - 50x)
R = 5000 - 500x + 500x - 50x²
R = 5000 - 50x²

Hmm, still getting same answer. Let me expand properly:

R = 10(500) + 10(-50x) + x(500) + x(-50x)
R = 5000 - 500x + 500x - 50x²
R = 5000 - 50x²

That suggests x = 0 is optimal, which doesn't seem right. Let me reconsider the problem.

Actually, I think the middle terms should be:

R = (10 + x)(500 - 50x)
R = 5000 - 500x + 500x - 50x²

The -500x + 500x cancel, leaving:

R = 5000 - 50x²

Maximum at x = 0 (no increase)

This suggests current price is optimal.

Let me use a different example:

Example: Better Revenue Problem

Problem: At $20, sell 100 items. Each $2 decrease increases sales by 10 items. Maximize revenue.**

Let: x = number of $2 decreases

Price: 20 - 2x

Quantity: 100 + 10x

Revenue:

R = (20 - 2x)(100 + 10x)
R = 2000 + 200x - 200x - 20x²
R = 2000 - 20x²

Same issue! The linear terms cancel.

Let me try once more carefully:

R = 20(100) + 20(10x) - 2x(100) - 2x(10x)
R = 2000 + 200x - 200x - 20x²
R = 2000 - 20x²

Maximum at x = 0.

Actually this makes sense - if the linear terms exactly cancel, the parabola is symmetric about x=0, so current price IS optimal.

Let me use a problem where they don't cancel:

Example: Revenue with Asymmetric Effect

Problem: At $15, sell 80 items. Each $1 increase decreases sales by 4 items. Maximize revenue.**

Let: x = number of $1 increases

Price: 15 + x

Quantity: 80 - 4x

Revenue:

R = (15 + x)(80 - 4x)
R = 1200 - 60x + 80x - 4x²
R = 1200 + 20x - 4x²

Vertex:

x = -20/(2(-4)) = -20/(-8) = 2.5

Price: $15 + $2.50 = $17.50

Quantity: 80 - 4(2.5) = 70

Maximum revenue:

R = 17.50 × 70 = $1,225

Answer: Charge $17.50**

Minimizing Cost or Distance

Similar approach

Example: Minimize Perimeter

Problem: Rectangular area of 200 m². Minimize perimeter (fencing).**

Let: x = width, y = length

Constraint: xy = 200, so y = 200/x

Perimeter:

P = 2x + 2y = 2x + 2(200/x)
P = 2x + 400/x

This is not a simple quadratic. To find minimum, we'd need calculus OR try completing a transformation.

Alternative approach - recognize xy = 200 is fixed product.

For fixed product, perimeter minimized when x = y (square)

If x = y and xy = 200:

x² = 200
x = √200 = 10√2 ≈ 14.14 m

Minimum perimeter:

P = 4x = 4(10√2) = 40√2 ≈ 56.57 m

Answer: Square with side 10√2 m**

Problem-Solving Strategy

Steps:

1. Identify what to maximize/minimize
2. Define variables
3. Write constraint equations
4. Express target quantity as function of one variable
5. Find vertex (or use other method)
6. Verify it's max or min (check a < 0 or a > 0)
7. Calculate optimum value
8. Answer in context

Common Patterns

Fixed perimeter → maximize area: Square shape

Fixed area → minimize perimeter: Square shape

Two numbers with fixed sum → maximize product: Equal numbers

Revenue problems: Often quadratic, find vertex

Real-World Applications

Business: Pricing, production levels

Engineering: Material usage, efficiency

Agriculture: Land allocation

Transportation: Route optimization (simplified)

Construction: Design specifications

Practice