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

  1. Business: Maximize profit, minimize cost
  2. Engineering: Minimize material used, maximize strength
  3. Geometry: Maximize area with fixed perimeter, minimize surface area for given volume
  4. 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 xx = width, yy = length


Step 4: Primary equation

Area: A=xyA = xy

This is what we want to maximize.


Step 5: Constraint

Perimeter: 2x+2y=1002x + 2y = 100


Step 6: Reduce to one variable

From constraint: 2y=1002x2y = 100 - 2x, so y=50xy = 50 - x

Substitute into area: A(x)=x(50x)=50xx2A(x) = x(50 - x) = 50x - x^2

Now AA is a function of xx only!


Step 7: Find critical points

A(x)=502xA'(x) = 50 - 2x

Set equal to zero: 502x=050 - 2x = 0

x=25x = 25 meters


Step 8: Verify maximum

A(x)=2<0A''(x) = -2 < 0 (concave down)

Since A<0A'' < 0, this is a maximum

Also check endpoints: x=0x = 0 or x=50x = 50 gives A=0A = 0 (no area)


Step 9: Answer the question

x=25x = 25 meters (width)

y=5025=25y = 50 - 25 = 25 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: x×xx \times x, height: hh


Step 3: Variables

xx = side length of base hh = height


Step 4: Primary equation

Surface area (no top!): S=x2+4xhS = x^2 + 4xh

(base x2x^2 plus four sides xhxh each)


Step 5: Constraint

Volume: x2h=32x^2h = 32


Step 6: Reduce to one variable

From constraint: h=32x2h = \frac{32}{x^2}

Substitute: S(x)=x2+4x(32x2)=x2+128xS(x) = x^2 + 4x\left(\frac{32}{x^2}\right) = x^2 + \frac{128}{x}


Step 7: Find critical points

S(x)=2x128x2S'(x) = 2x - \frac{128}{x^2}

Set equal to zero: 2x128x2=02x - \frac{128}{x^2} = 0

2x=128x22x = \frac{128}{x^2}

2x3=1282x^3 = 128

x3=64x^3 = 64

x=4x = 4 feet


Step 8: Verify minimum

S(x)=2+256x3S''(x) = 2 + \frac{256}{x^3}

S(4)=2+25664=2+4=6>0S''(4) = 2 + \frac{256}{64} = 2 + 4 = 6 > 0

Positive → minimum


Step 9: Find all dimensions

x=4x = 4 feet (base side length)

h=3242=3216=2h = \frac{32}{4^2} = \frac{32}{16} = 2 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: Profit=RevenueCost\text{Profit} = \text{Revenue} - \text{Cost}

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: x>0x > 0
  • 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 xx)
  • 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 xx.


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 d=f(x)d = \sqrt{f(x)}, minimize d2=f(x)d^2 = f(x) 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 [a,b][a, b]:

  1. Find critical points in (a,b)(a, b)
  2. Evaluate function at critical points
  3. Evaluate function at endpoints aa and bb
  4. 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

  1. Start with simple problems (single constraint, basic geometry)
  2. Draw everything - visual representation helps enormously
  3. Label clearly - know what each variable represents
  4. Check endpoints - don't forget the domain!
  5. Verify your answer - does it make physical sense?
  6. 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 xx and yy


Step 3: Primary equation

Product: P=xyP = xy


Step 4: Constraint

Sum is 20: x+y=20x + y = 20


Step 5: Reduce to one variable

From constraint: y=20xy = 20 - x

Substitute: P(x)=x(20x)=20xx2P(x) = x(20 - x) = 20x - x^2


Step 6: Find domain

Since both numbers must be positive: x>0x > 0 and y=20x>0y = 20 - x > 0

So 0<x<200 < x < 20


Step 7: Find critical points

P(x)=202xP'(x) = 20 - 2x

Set equal to zero: 202x=020 - 2x = 0

x=10x = 10


Step 8: Verify maximum

P(x)=2<0P''(x) = -2 < 0 (concave down)

This is a maximum

Check endpoints:

  • At x=0x = 0: P=0P = 0
  • At x=20x = 20: P=0P = 0
  • At x=10x = 10: P=10(10)=100P = 10(10) = 100

Maximum is clearly at x=10x = 10


Step 9: Find both numbers

x=10x = 10

y=2010=10y = 20 - 10 = 10

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 xx = width of printed area

Let yy = height of printed area


Step 3: Constraint

Printed area: xy=50xy = 50


Step 4: Primary equation (total area)

Total width: x+1+1=x+2x + 1 + 1 = x + 2

Total height: y+2+2=y+4y + 2 + 2 = y + 4

Total area: A=(x+2)(y+4)A = (x + 2)(y + 4)


Step 5: Reduce to one variable

From constraint: y=50xy = \frac{50}{x}

Substitute: A(x)=(x+2)(50x+4)A(x) = (x + 2)\left(\frac{50}{x} + 4\right)

=(x+2)(50+4xx)= (x + 2)\left(\frac{50 + 4x}{x}\right)

=(x+2)(50+4x)x= \frac{(x + 2)(50 + 4x)}{x}

=50x+4x2+100+8xx= \frac{50x + 4x^2 + 100 + 8x}{x}

=4x2+58x+100x= \frac{4x^2 + 58x + 100}{x}

=4x+58+100x= 4x + 58 + \frac{100}{x}


Step 6: Find critical points

A(x)=4100x2A'(x) = 4 - \frac{100}{x^2}

Set equal to zero: 4100x2=04 - \frac{100}{x^2} = 0

4=100x24 = \frac{100}{x^2}

4x2=1004x^2 = 100

x2=25x^2 = 25

x=5x = 5 (positive solution)


Step 7: Verify minimum

A(x)=200x3A''(x) = \frac{200}{x^3}

A(5)=200125=1.6>0A''(5) = \frac{200}{125} = 1.6 > 0

Minimum confirmed


Step 8: Find dimensions

Printed width: x=5x = 5 inches

Printed height: y=505=10y = \frac{50}{5} = 10 inches

Total poster dimensions:

Total width: 5+2=75 + 2 = 7 inches

Total height: 10+4=1410 + 4 = 14 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

rr = radius of base

hh = height


Step 4: Constraint

Volume: πr2h=1000\pi r^2 h = 1000


Step 5: Primary equation

Surface area (top + bottom + side): S=2πr2+2πrhS = 2\pi r^2 + 2\pi rh


Step 6: Reduce to one variable

From constraint: h=1000πr2h = \frac{1000}{\pi r^2}

Substitute: S(r)=2πr2+2πr(1000πr2)S(r) = 2\pi r^2 + 2\pi r\left(\frac{1000}{\pi r^2}\right)

=2πr2+2000r= 2\pi r^2 + \frac{2000}{r}


Step 7: Find critical points

S(r)=4πr2000r2S'(r) = 4\pi r - \frac{2000}{r^2}

Set equal to zero: 4πr2000r2=04\pi r - \frac{2000}{r^2} = 0

4πr=2000r24\pi r = \frac{2000}{r^2}

4πr3=20004\pi r^3 = 2000

r3=20004π=500πr^3 = \frac{2000}{4\pi} = \frac{500}{\pi}

r=500π35.42r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42 cm


Step 8: Verify minimum

S(r)=4π+4000r3S''(r) = 4\pi + \frac{4000}{r^3}

Since r>0r > 0, we have S(r)>0S''(r) > 0 for all rr

This is a minimum


Step 9: Find height

h=1000πr2=1000π(500π)2/3h = \frac{1000}{\pi r^2} = \frac{1000}{\pi \left(\frac{500}{\pi}\right)^{2/3}}

Computing: h=2500π310.84h = 2\sqrt[3]{\frac{500}{\pi}} \approx 10.84 cm

Note: h=2rh = 2r (interesting!)


Answer:

  • Radius: r=500π35.42r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42 cm
  • Height: h=2500π310.84h = 2\sqrt[3]{\frac{500}{\pi}} \approx 10.84 cm

The optimal can has height equal to the diameter!