Optimization Problems
Using calculus to find maximum and minimum values in real-world situations
🎯 Optimization Problems
What is Optimization?
Optimization means finding the best solution to a problem - usually the maximum or minimum value of some quantity.
Real-World Examples
- Business: Maximize profit, minimize cost
- Engineering: Minimize material used, maximize strength
- Geometry: Maximize area with fixed perimeter, minimize surface area for given volume
- Physics: Minimize time, maximize distance
💡 Key Idea: Use calculus to find where the derivative equals zero, then verify it's actually a max or min!
The General Strategy
Step-by-Step Process
Step 1: Read carefully and identify what to maximize or minimize
Step 2: Draw a diagram (if applicable)
Step 3: Assign variables to all quantities
Step 4: Write the primary equation - the function to optimize
Step 5: Find constraint equations (relationships between variables)
Step 6: Reduce to one variable using constraints
Step 7: Find the derivative and critical points
Step 8: Verify it's a max/min (Second Derivative Test or endpoints)
Step 9: Answer the question (often needs more than just the critical point!)
Example 1: Basic Geometry
Problem: A farmer has 100 meters of fence to enclose a rectangular garden. What dimensions give the maximum area?
Step 1: Identify
Maximize: Area
Step 2: Draw
x
┌───┐
y │ │ y
└───┘
x
Step 3: Assign variables
Let = width, = length
Step 4: Primary equation
Area:
This is what we want to maximize.
Step 5: Constraint
Perimeter:
Step 6: Reduce to one variable
From constraint: , so
Substitute into area:
Now is a function of only!
Step 7: Find critical points
Set equal to zero:
meters
Step 8: Verify maximum
(concave down)
Since , this is a maximum ✓
Also check endpoints: or gives (no area)
Step 9: Answer the question
meters (width)
meters (length)
Answer: The garden should be a 25 × 25 meter square with maximum area of 625 square meters.
Example 2: Manufacturing/Cost
Problem: A box with a square base and open top must have a volume of 32 cubic feet. Find the dimensions that minimize the amount of material used.
Step 1: Identify
Minimize: Surface area (amount of material)
Step 2: Draw
┌─── x ───┐
│ │
h │ │
│ │
└─────────┘
x
Square base: , height:
Step 3: Variables
= side length of base = height
Step 4: Primary equation
Surface area (no top!):
(base plus four sides each)
Step 5: Constraint
Volume:
Step 6: Reduce to one variable
From constraint:
Substitute:
Step 7: Find critical points
Set equal to zero:
feet
Step 8: Verify minimum
✓
Positive → minimum ✓
Step 9: Find all dimensions
feet (base side length)
feet (height)
Answer: Base is 4 × 4 feet, height is 2 feet. Minimum surface area is 48 square feet.
Common Problem Types
Type 1: Fencing Problems
Given perimeter, maximize area (or vice versa)
Key: Use constraint to eliminate one variable
Type 2: Box/Container Problems
Minimize surface area for given volume
Key: Remember which surfaces are included!
Type 3: Distance Problems
Minimize distance from a point to a curve
Key: Can minimize distance² to avoid square roots
Type 4: Economics
Maximize revenue, profit; minimize cost
Key:
Type 5: Right Triangle Problems
Often use Pythagorean Theorem as constraint
Key: Draw the triangle and label everything
Important Tips
Domain Considerations
Always check the domain!
- Variables often must be positive:
- Constraints might limit the range
- Check endpoints if the domain is closed
Units
Pay attention to units and include them in your answer!
Answer the Actual Question
Sometimes the question asks for:
- The dimensions (not just )
- The maximum value (not just where it occurs)
- Multiple quantities
⚠️ Common Mistakes
Mistake 1: Not Reducing to One Variable
You MUST express the function in terms of a single variable before differentiating!
Mistake 2: Forgetting Constraints
Use ALL given information to create constraint equations.
Mistake 3: Not Verifying Max/Min
Use Second Derivative Test or check endpoints to confirm!
Mistake 4: Domain Errors
Make sure your answer makes physical sense (positive lengths, etc.)
Mistake 5: Incomplete Answers
If the problem asks for dimensions, give ALL dimensions, not just .
Special Techniques
Technique 1: Implicit Differentiation
If the constraint is hard to solve, use implicit differentiation (in related rates style).
Technique 2: Eliminating Square Roots
To minimize distance , minimize instead (same critical points, easier derivative).
Technique 3: Symmetry
Often the optimal solution has symmetric dimensions (like a square instead of rectangle).
Closed Interval Method for Optimization
If the domain is a closed interval :
- Find critical points in
- Evaluate function at critical points
- Evaluate function at endpoints and
- Compare all values
📝 Problem-Solving Checklist
- [ ] Read problem carefully
- [ ] Draw a diagram
- [ ] Label all variables
- [ ] Write equation to optimize
- [ ] Find constraint(s)
- [ ] Reduce to one variable
- [ ] Find derivative and critical points
- [ ] Verify max/min (Second Derivative Test or endpoints)
- [ ] Calculate all requested values
- [ ] Check answer makes sense
- [ ] Include units!
Practice Tips
- Start with simple problems (single constraint, basic geometry)
- Draw everything - visual representation helps enormously
- Label clearly - know what each variable represents
- Check endpoints - don't forget the domain!
- Verify your answer - does it make physical sense?
- Practice word problems - these take time to master
📚 Practice Problems
1Problem 1easy
❓ Question:
Find two positive numbers whose sum is 20 and whose product is as large as possible.
💡 Show Solution
Step 1: Identify
Maximize: Product of two numbers
Step 2: Assign variables
Let the numbers be and
Step 3: Primary equation
Product:
Step 4: Constraint
Sum is 20:
Step 5: Reduce to one variable
From constraint:
Substitute:
Step 6: Find domain
Since both numbers must be positive: and
So
Step 7: Find critical points
Set equal to zero:
Step 8: Verify maximum
(concave down)
This is a maximum ✓
Check endpoints:
- At :
- At :
- At :
Maximum is clearly at ✓
Step 9: Find both numbers
Answer: The two numbers are 10 and 10 with maximum product 100.
2Problem 2hard
❓ Question:
A rectangular poster is to have 50 square inches of printed material with 2-inch margins on top and bottom and 1-inch margins on each side. What dimensions minimize the total area of the poster?
💡 Show Solution
Step 1: Draw a diagram
┌─────────────────┐
│ 1" 1" │ ← 2" margin top
│ ┌───────────┐ │
│ │ printed │ │
│ │ area │ │
│ └───────────┘ │
│ │ ← 2" margin bottom
└─────────────────┘
Step 2: Assign variables
Let = width of printed area
Let = height of printed area
Step 3: Constraint
Printed area:
Step 4: Primary equation (total area)
Total width:
Total height:
Total area:
Step 5: Reduce to one variable
From constraint:
Substitute:
Step 6: Find critical points
Set equal to zero:
(positive solution)
Step 7: Verify minimum
✓
Minimum confirmed ✓
Step 8: Find dimensions
Printed width: inches
Printed height: inches
Total poster dimensions:
Total width: inches
Total height: inches
Answer: The poster should be 7 inches wide by 14 inches tall.
3Problem 3expert
❓ Question:
A cylindrical can is to hold 1000 cubic centimeters of juice. Find the dimensions that will minimize the cost of metal to make the can.
💡 Show Solution
Step 1: Draw
┌───────┐
│ │
h │ │ ← height
│ │
└───────┘
r (radius)
Step 2: Identify
Minimize: Surface area (amount of metal)
Step 3: Assign variables
= radius of base
= height
Step 4: Constraint
Volume:
Step 5: Primary equation
Surface area (top + bottom + side):
Step 6: Reduce to one variable
From constraint:
Substitute:
Step 7: Find critical points
Set equal to zero:
cm
Step 8: Verify minimum
Since , we have for all ✓
This is a minimum ✓
Step 9: Find height
Computing: cm
Note: (interesting!)
Answer:
- Radius: cm
- Height: cm
The optimal can has height equal to the diameter!
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