Premed · Premed · Calculus 1
Lecture 16: Optimization Problems
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Translate word problems into mathematical optimization problems
- Identify the objective function and constraint(s)
- Reduce optimization problems to single-variable calculus problems
- Find absolute maxima and minima in applied contexts
- Verify that a critical point is indeed a maximum or minimum
Lecture Content
I. General Strategy for Optimization
Optimization problems follow a systematic nine-step strategy. Step 1: Understand the problem by reading carefully, drawing a diagram, and identifying what is to be maximized or minimized. Step 2: Introduce variables by assigning symbols to all relevant quantities. Step 3: Write the objective function, expressing the quantity to optimize as a function of the variables. Step 4: Identify constraints -- equations that relate the variables, such as a fixed perimeter or fixed volume. Step 5: Reduce to one variable by using the constraint to eliminate extra variables from the objective function. Step 6: Find the domain, determining the physically meaningful interval for the remaining variable. Step 7: Find the critical numbers by setting the derivative equal to zero and solving. Step 8: Determine the absolute extremum using the Closed Interval Method, the First Derivative Test, or the Second Derivative Test. Step 9: Answer the question by substituting back and stating the result with appropriate units.
II. Example — Maximizing Area with Fixed Perimeter
A farmer has 400 m of fencing and wants to enclose the largest possible rectangular area. Let x and y be the side lengths. The constraint is 2x + 2y = 400, giving y = 200 - x. The objective is to maximize A = x * y = x(200 - x) = 200x - x^2, with domain 0 <= x <= 200 (both dimensions must be non-negative). Setting A'(x) = 200 - 2x = 0 gives x = 100. Since A''(x) = -2 < 0, this is indeed a maximum. The optimal dimensions are x = y = 100 m (a square), yielding a maximum area of 10,000 m^2.
III. Example — Minimizing Material for a Box
An open-top rectangular box with a square base must have a volume of 32,000 cm^3. We want to find the dimensions that minimize the surface area (amount of material used). Let x be the side length of the base and h the height. The volume constraint gives x^2 * h = 32,000, so h = 32,000/x^2. The surface area is S = x^2 + 4xh = x^2 + 128,000/x, with domain x > 0.
Setting S'(x) = 2x - 128,000/x^2 = 0 gives 2x = 128,000/x^2, so x^3 = 64,000 and x = 40 cm. The height is then h = 32,000/1600 = 20 cm. Since S''(x) = 2 + 256,000/x^3 > 0 for all x > 0, this critical point is confirmed as a minimum. The minimum surface area is 1600 + 3200 = 4800 cm^2.
<image>A three-dimensional diagram of the open-top box with square base. The base is labeled x by x, and the height is labeled h. Next to it, the surface area formula S = x^2 + 4xh is shown with the base area (one square) and four side panels (four rectangles) "unfolded" into a flat net. The constraint V = x^2 h = 32,000 is displayed, with an arrow showing the substitution h = 32,000/x^2. Title: "Open-top box optimization."</image>
IV. Example — Closest Point on a Curve
To find the point on the parabola y = x^2 closest to the point (0, 1), note that the distance is D = sqrt(x^2 + (x^2 - 1)^2). Since minimizing D is equivalent to minimizing D^2, define f(x) = x^2 + (x^2 - 1)^2 = x^4 - x^2 + 1. Setting f'(x) = 4x^3 - 2x = 2x(2x^2 - 1) = 0 gives x = 0 or x = +/- 1/sqrt(2). Evaluating, f(0) = 1 and f(+/- 1/sqrt(2)) = 1/4 - 1/2 + 1 = 3/4. The minimum distance occurs at x = +/- 1/sqrt(2) with y = 1/2, so the closest points are (+/- 1/sqrt(2), 1/2).
V. Example — Maximizing Revenue or Profit
A company sells x units at price p = 200 - 0.5x dollars per unit. Revenue is R(x) = x * p = 200x - 0.5x^2. Setting R'(x) = 200 - x = 0 gives x = 200, and the maximum revenue is R(200) = 20,000 dollars. If a cost function C(x) is also given, profit P(x) = R(x) - C(x) would be maximized instead.
VI. Optimization with Trigonometry
Some optimization problems require trigonometric parametrization. For example, consider a window in the shape of a rectangle topped by a semicircle, with a total perimeter of 12 m. Let w be the width (also the diameter of the semicircle) and h the height of the rectangular part. The perimeter constraint is w + 2h + (piw/2) = 12, which determines h in terms of w. The total area A = wh + (pi/8)*w^2 can then be expressed as a function of w alone, differentiated, and set equal to zero to find the optimal dimensions.
<image>A diagram of the Norman window (rectangle topped by semicircle). The width w and rectangular height h are labeled. The semicircle sits on top with diameter w. The perimeter equation P = w + 2h + piw/2 = 12 is shown. The total area A = wh + (pi/8)w^2 is shown as the sum of the rectangular area (shaded light blue) and the semicircular area (shaded light green). Title: "Norman Window Optimization."</image>
VII. Tips and Common Mistakes
Several pitfalls must be avoided. Always verify the answer by confirming that the critical point is a maximum or minimum, using the Second Derivative Test, First Derivative Test, or endpoint checking. Check endpoints, since the absolute extremum might occur at the boundary of the domain. Do not forget the constraint -- the most common error is failing to use the constraint to reduce to one variable. Draw a picture, as a good diagram prevents most setup errors. Always include units in the final answer, and re-read the question to make sure you answer what was actually asked -- whether it is the dimensions, the maximum value, or both.
