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 $x$ = width, $y$ = length

Step 4: Primary equation

Area: $A = xy$

This is what we want to maximize.

Step 5: Constraint

Perimeter: $2x + 2y = 100$

Step 6: Reduce to one variable

From constraint: $2y = 100 - 2x$, so $y = 50 - x$

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

Now $A$ is a function of $x$ only!

Step 7: Find critical points

$A'(x) = 50 - 2x$

Set equal to zero: $50 - 2x = 0$

$x = 25$ meters

Step 8: Verify maximum

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

Since $A'' < 0$, this is a maximum ✓

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

Step 9: Answer the question

$x = 25$ meters (width)

$y = 50 - 25 = 25$ 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: $x \times x$, height: $h$

Step 3: Variables

$x$ = side length of base $h$ = height

Step 4: Primary equation

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

(base $x^2$ plus four sides $xh$ each)

Step 5: Constraint

Volume: $x^2h = 32$

Step 6: Reduce to one variable

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

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

Step 7: Find critical points

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

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

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

$2x^3 = 128$

$x^3 = 64$

$x = 4$ feet

Step 8: Verify minimum

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

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

Positive → minimum ✓

Step 9: Find all dimensions

$x = 4$ feet (base side length)

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

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: $x > 0$
• 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 $x$)
• 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 $x$.

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

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

Step 3: Primary equation

Product: $P = xy$

Step 4: Constraint

Sum is 20: $x + y = 20$

Step 5: Reduce to one variable

From constraint: $y = 20 - x$

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

Step 6: Find domain

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

So $0 < x < 20$

Step 7: Find critical points

$P'(x) = 20 - 2x$

Set equal to zero: $20 - 2x = 0$

$x = 10$

Step 8: Verify maximum

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

This is a maximum ✓

Check endpoints:

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

Maximum is clearly at $x = 10$ ✓

Step 9: Find both numbers

$x = 10$

$y = 20 - 10 = 10$

Answer: The two numbers are 10 and 10 with maximum product 100.