Calculus I - Week 9

Optimization

Optimization is one of the most powerful applications of calculus: finding the maximum or minimum value of a function subject to constraints. Engineers minimize material cost, physicists minimize energy, and businesses maximize profit — all using the same derivative technique.

Learning Objectives

- Identify the objective function and the constraint in a word problem. - Express the objective function in one variable using the constraint. - Find critical points of the objective function. - Use the first or second derivative test to classify the critical point. - Check endpoints when the domain is bounded. - Interpret the solution in context and verify it is reasonable.

1. The Optimization Strategy

Step 1: Identify what you want to maximize or minimize (the objective function). Step 2: Identify the constraint — the equation linking the variables. Step 3: Use the constraint to write the objective in one variable. Step 4: Differentiate and find critical points. Step 5: Verify it is a max or min (first/second derivative test or endpoints). Step 6: Answer the original question.

2. Example: Maximum Area with Fixed Perimeter

Find the rectangle of maximum area with perimeter 40 m. Let length = l, width = w. Constraint: 2l + 2w = 40, so l + w = 20. Objective: A = l * w. Substitute: w = 20 - l. A(l) = l(20 - l) = 20l - l^2. A'(l) = 20 - 2l = 0 → l = 10. A''(l) = -2 < 0 → maximum. Dimensions: 10 × 10 (a square). Area = 100 m^2.

3. Example: Minimum Material for a Box

A box with square base and open top must hold 32 m^3. Minimize the surface area. Let base side = x, height = h. Volume: x^2 * h = 32, so h = 32/x^2. Surface area: S = x^2 + 4xh = x^2 + 4x(32/x^2) = x^2 + 128/x. S'(x) = 2x - 128/x^2 = 0 2x^3 = 128 x^3 = 64 → x = 4. h = 32/16 = 2. S''(x) = 2 + 256/x^3 > 0 → minimum.

4. Example: Closest Point on a Parabola

Find the point on y = x^2 closest to the point (0, 1). Distance from (x, x^2) to (0, 1): D = sqrt(x^2 + (x^2 - 1)^2) Minimize D^2 = x^2 + (x^2 - 1)^2 (same minimum). Let f = x^2 + x^4 - 2x^2 + 1 = x^4 - x^2 + 1. f' = 4x^3 - 2x = 2x(2x^2 - 1) = 0. x = 0 or x = ±1/sqrt(2). f''(0) = -2 < 0 → local max (discard for distance min). f''(±1/sqrt(2)) = 4 > 0 → local min. Points: (±1/sqrt(2), 1/2). Distance = sqrt(3)/2.

5. Example: Revenue Maximization

A company sells x units at price p = 50 - 0.5x. The cost is C = 10x + 200. Revenue: R = xp = x(50 - 0.5x) = 50x - 0.5x^2. Profit: P = R - C = 50x - 0.5x^2 - 10x - 200 = 40x - 0.5x^2 - 200. P'(x) = 40 - x = 0 → x = 40. P''(x) = -1 < 0 → maximum profit. P(40) = 40(40) - 0.5(1600) - 200 = 1600 - 800 - 200 = 600.

6. Example: Minimum Fence Cost

A farmer wants to enclose 600 m^2 with 3 sides of fence (one side is a wall). Find the minimum total fence length. Let width = w, length = l. Area: l*w = 600, so l = 600/w. Fence (3 sides): F = l + 2w = 600/w + 2w. F' = -600/w^2 + 2 = 0 2w^2 = 600 w^2 = 300 w = 10*sqrt(3). l = 600/(10*sqrt(3)) = 60/sqrt(3) = 20*sqrt(3). Min fence: 20*sqrt(3) + 20*sqrt(3) = 40*sqrt(3) ≈ 69.3 m.

7. The Closed Interval Method

If the domain is a closed interval [a, b], the absolute extreme values occur at: - Critical points inside (a, b), or - Endpoints a or b. Evaluate the objective function at all these points and compare. This is called the closed interval method and is guaranteed by the Extreme Value Theorem.

8. Open Domain Optimization

When the domain is open (e.g., x > 0), there are no endpoints to check. If there is only one critical point: - If f'' < 0 there, it is the absolute max. - If f'' > 0 there, it is the absolute min. This is the most common situation in applied optimization: a physical constraint forces the domain to be positive (length, area, volume cannot be negative).

9. Physics Application: Projectile Range

A projectile launched at speed v and angle theta has horizontal range: R = v^2 * sin(2*theta) / g To maximize range: dR/d(theta) = v^2 * 2*cos(2*theta)/g = 0. cos(2*theta) = 0 → 2*theta = pi/2 → theta = pi/4 = 45 degrees. A 45° launch angle gives maximum range on flat ground.

10. Common Mistakes

- Writing the objective function in two variables and forgetting to use the constraint. - Finding a critical point but not checking whether it is a max or min. - Ignoring endpoints on a closed interval. - Answering with the x-value when the problem asks for the maximum value. - Forgetting that physical domains have restrictions (x > 0, 0 < theta < pi/2, etc.).