Optimization Problems

Optimization problems involve finding the absolute maximum or minimum value of a quantity in a real-world context, such as maximizing area or minimizing cost. We can solve these problems efficiently using derivatives.

Steps to Solve Optimization Problems

1. Define Variables: Draw a diagram if helpful and assign variables to the unknown quantities.
2. Identify the Objective Function: Write an equation for the quantity you want to maximize or minimize.
3. Find the Constraints: Use the given information to write an equation relating your variables. Use this to substitute into the objective function so it depends on only one variable.
4. Determine the Domain: Find the practical limits for your variable.
5. Find Critical Points: Calculate the first derivative of your objective function and set it to zero.
6. Verify the Extrema: Use the First Derivative Test, Second Derivative Test, or check the endpoints of a closed interval to confirm whether your critical point is a maximum or minimum.

Example 1: Maximizing Area

Problem: Find the dimensions of a rectangle with a perimeter of $40$ that has the maximum area.

Solution: Let the length be $x$ and the width be $y$.

• Constraint: The perimeter is $40$, so $2x + 2y = 40$, which simplifies to $x + y = 20$, or $y = 20 - x$.
• Objective Function: We want to maximize Area, $A = x \cdot y$.
• Substitute $y$: $A(x) = x(20 - x) = 20x - x^2$.
• Critical Points: Take the derivative and set it to zero. $A'(x) = 20 - 2x$ $20 - 2x = 0 \implies x = 10$
• Verify: The second derivative is $A''(x) = -2$. Since $A'' < 0$, the graph is concave down, confirming a local maximum.
• Dimensions: If $x = 10$, then $y = 20 - 10 = 10$. The rectangle is a $10 \times 10$ square.

Example 2: Maximizing Volume

Problem: An open-top box is made from a $20 \times 30$ sheet by cutting equal squares from the corners and folding up the sides. Find the cut size for maximum volume.

Solution: Let $x$ be the side length of the square cut from each corner.

• Dimensions of the box: Height $= x$, Length $= 30 - 2x$, Width $= 20 - 2x$.
• Domain: Since length and width must be positive, $20 - 2x > 0 \implies 0 < x < 10$.
• Objective Function: Volume $V = \text{length} \times \text{width} \times \text{height}$. $V(x) = (30 - 2x)(20 - 2x)x = (600 - 100x + 4x^2)x = 4x^3 - 100x^2 + 600x$
• Critical Points: $V'(x) = 12x^2 - 200x + 600$ Set $V'(x) = 0$ and divide by $4$: $3x^2 - 50x + 150 = 0$
• Use the quadratic formula: $x = \frac{50 \pm \sqrt{(-50)^2 - 4(3)(150)}}{2(3)} = \frac{50 \pm \sqrt{2500 - 1800}}{6} = \frac{50 \pm \sqrt{700}}{6} = \frac{25 \pm 5\sqrt{7}}{3}$
• Since $\frac{25 + 5\sqrt{7}}{3} \approx 12.74$ is outside our domain ($x < 10$), the only valid critical point is: $x = \frac{25 - 5\sqrt{7}}{3} \approx 3.92$

Cutting squares of approximately $3.92$ units will yield the maximum volume.