Week 11, Session 1: Optimization Problems II and Behavior of Implicit Relations

Session Focus

Today you extend optimization to problems in three dimensions and in business, then apply the tools of Unit 5 to curves defined implicitly: where a curve has horizontal or vertical tangents, and whether a point with a horizontal tangent is a local high or low point of the curve (CED Topics 5.11 and 5.12).

Learning Objectives

By the end of this session you should be able to:

• Solve optimization problems involving surface area, volume, revenue, cost, and profit.
• Find horizontal and vertical tangents of an implicitly defined curve.
• Use \(\dfrac{d^2y}{dx^2}\) to classify a horizontal tangent on an implicit curve as a local maximum or minimum of \(y\).

Key Language

surface area, profit, marginal revenue, marginal cost, implicit relation, local extremum of \(y\) on a curve

Teacher Explanation

Optimization in three dimensions follows the same procedure as Session 40. The formulas are the new part: a closed cylinder has volume \(V = \pi r^2 h\) and surface area \(S = 2\pi r^2 + 2\pi r h\).

In business problems, profit is revenue minus cost: \(P(x) = R(x) - C(x)\). Setting \(P'(x) = 0\) gives \(R'(x) = C'(x)\): profit is maximized where marginal revenue equals marginal cost. This is a standard result in economics, and it follows directly from the derivative.

Behavior of implicit relations. For a curve given implicitly, \(\dfrac{dy}{dx}\) usually depends on both \(x\) and \(y\).

• A horizontal tangent needs \(\dfrac{dy}{dx} = 0\) (numerator 0, denominator not 0) at a point on the curve.
• A vertical tangent needs the denominator of \(\dfrac{dy}{dx}\) to be 0 (numerator not 0) at a point on the curve.

At a point with a horizontal tangent, the sign of \(\dfrac{d^2y}{dx^2}\) works like the Second Derivative Test: negative means the curve has a local maximum in \(y\) there, positive means a local minimum.

Core Rule

For implicit curves, always solve the tangent condition together with the original equation. A point that makes \(\dfrac{dy}{dx} = 0\) but is not on the curve is meaningless.

Worked Example 1: The cheapest can

A closed cylindrical can must hold \(16\pi\) cubic centimeters. What radius and height minimize the surface area?

Constraint: \(\pi r^2 h = 16\pi\), so \(h = \dfrac{16}{r^2}\). Objective: \(S = 2\pi r^2 + 2\pi r\left(\dfrac{16}{r^2}\right) = 2\pi r^2 + \dfrac{32\pi}{r}\), for \(r > 0\).

\(S'(r) = 4\pi r - \dfrac{32\pi}{r^2} = 0\) gives \(r^3 = 8\), so \(r = 2\). \(S'\) changes from negative to positive at \(r = 2\), the only critical point, so it is the absolute minimum. Then \(h = \dfrac{16}{4} = 4\). The best can is 2 cm in radius and 4 cm tall: its height equals its diameter.

Worked Example 2: Maximum profit

A company sells \(x\) items at price \(p = 100 - 2x\) dollars each. The cost of making \(x\) items is \(C(x) = 20x + 50\) dollars.

\(R(x) = x(100 - 2x)\), so \(P(x) = 100x - 2x^2 - 20x - 50 = 80x - 2x^2 - 50\). \(P'(x) = 80 - 4x = 0\) at \(x = 20\), and \(P''(x) = -4 < 0\). The maximum profit is \(P(20) = 1600 - 800 - 50 = 750\) dollars. Check: marginal revenue \(R'(20) = 100 - 80 = 20\) equals marginal cost \(C'(20) = 20\).

Worked Example 3: An implicit curve

For \(x^2 + xy + y^2 = 3\), Session 23 gave \(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}\) and horizontal tangents at \((1, -2)\) and \((-1, 2)\).

Differentiate \(\dfrac{dy}{dx}\) again with the quotient rule. At a point where \(\dfrac{dy}{dx} = 0\) and \(2x + y = 0\), most terms vanish, leaving \(\dfrac{d^2y}{dx^2} = -\dfrac{2}{x + 2y}\).

At \((1, -2)\): \(x + 2y = -3\), so \(\dfrac{d^2y}{dx^2} = \dfrac{2}{3} > 0\). The curve has a local minimum in \(y\) there.

At \((-1, 2)\): \(x + 2y = 3\), so \(\dfrac{d^2y}{dx^2} = -\dfrac{2}{3} < 0\). The curve has a local maximum in \(y\) there.

Common Errors to Avoid

• Forgetting the top and bottom of a closed can in the surface area.
• Maximizing revenue when the question asks for profit.
• Solving \(\dfrac{dy}{dx} = 0\) without substituting back into the curve's equation.

Independent Practice

• A rectangle has its base on the \(x\)-axis and its upper corners on \(y = 12 - x^2\). Find its maximum area.
• An open-top cylindrical tank must hold \(27\pi\) cubic meters. Find the radius that minimizes the material used.
• Find all points on \(x^2 - xy + y^2 = 12\) where the tangent line is horizontal.

Error-Log Reflection

When you finish, write two sentences in your notebook. The first begins with "The step I must watch most carefully is..." The second begins with "The clue that tells me to use this method is..." Keep these sentences in one running error log for the whole course. Before every unit assessment, reread the log; it is a record of your own decision-making, and it is the fastest review tool you will have.

References

• College Board. (2020). AP Calculus AB and BC course and exam description (Effective Fall 2020). https://apcentral.collegeboard.org/media/pdf/ap-calculus-ab-and-bc-course-and-exam-description.pdf
• Strang, G., & Herman, E. (2016). Calculus volume 1. OpenStax. https://openstax.org/details/books/calculus-volume-1
• Stewart, J., Clegg, D., & Watson, S. (2021). Calculus: Early transcendentals (9th ed.). Cengage Learning.