Difficulty: Intermediate | Prerequisites: First derivative, quotient rule, critical points, concavity (see companion notes)
Applied optimization uses derivatives to find where a real-world quantity (revenue, profit, cost) hits its maximum or minimum. For rational functions, you apply the quotient rule to get the derivative, set it to zero, factor, and use a sign chart to confirm whether you have a maximum or minimum. The denominator of a rational function is never zero (no real solutions), so there are no vertical asymptotes to worry about in these business models.
Quotient rule
The derivative rule for a function written as a fraction: if h(x) = f(x)/g(x), then h'(x) = [f'(x)·g(x) - f(x)·g'(x)] / [g(x)]². Think of it as "low d-high minus high d-low, over the square of what's below."
Critical point (critical value)
A value of x where f'(x) = 0 or f'(x) is undefined. These are the only places where a local maximum or minimum can occur. In simple terms, it is where the function's slope is flat or the slope breaks.
Relative maximum (local maximum)
A point where the function value is higher than at all nearby points. At a relative maximum, the function changes from increasing to decreasing, so f' goes from positive to negative.
Relative minimum (local minimum)
A point where the function value is lower than at all nearby points. The function changes from decreasing to increasing, so f' goes from negative to positive.
Rational function
A function that can be written as the ratio of two polynomials, such as R(t) = (63t - t²)/(t² + 63). These require the quotient rule for differentiation.
First derivative test
A method that uses a sign chart of f' to classify critical points. If f' changes from + to - at a critical point, that point is a relative maximum. If f' changes from - to +, it is a relative minimum. If there is no sign change, it is neither.
The revenue from the sale of a new kind of motorised skateboard, t weeks after its introduction, is given by:
R(t) = (63t - t²) / (t² + 63), measured in million dollars, where 0 ≤ t ≤ 63.
The question: when does the maximum revenue occur, and how much is it?
With f(t) = 63t - t² and g(t) = t² + 63:
f'(t) = 63 - 2t
g'(t) = 2t
Applying the quotient rule:
R'(t) = [(63 - 2t)(t² + 63) - (63t - t²)(2t)] / (t² + 63)²
Expanding the numerator:
(63 - 2t)(t² + 63) = 63t² + 63² - 2t³ - 126t
(63t - t²)(2t) = 126t² - 2t³
Numerator = 63t² + 3969 - 2t³ - 126t - 126t² + 2t³
Simplifying: -63t² - 126t + 3969 = -63(t² + 2t - 63)
Factoring: -63(t + 9)(t - 7)
So R'(t) = -63(t + 9)(t - 7) / (t² + 63)².
Set R'(t) = 0. The denominator (t² + 63)² is always positive, so only the numerator matters:
-63(t + 9)(t - 7) = 0, giving t = -9 or t = 7.
Since t represents weeks after introduction and 0 ≤ t ≤ 63, we discard t = -9. The only relevant critical point is t = 7.
Also check whether the denominator ever equals zero: t² + 63 = 0 gives t² = -63, which has no real solutions. There are no vertical asymptotes.
Test the sign of R'(t) on either side of t = 7 (within the domain):
At t = 1 (left of 7): numerator = -63(1 + 9)(1 - 7) = -63(10)(-6) = positive. R' > 0, so R is increasing.
At t = 10 (right of 7): numerator = -63(10 + 9)(10 - 7) = -63(19)(3) = negative. R' < 0, so R is decreasing.
Since R' changes from positive to negative at t = 7, this is a relative maximum.
Plug t = 7 back into R(t):
R(7) = (63(7) - 7²) / (7² + 63) = (441 - 49) / (49 + 63) = 392 / 112 = 3.5
The maximum revenue is 3.5 million dollars, occurring 7 weeks after introduction.
\text{Quotient Rule: } \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}R(t) = \frac{63t - t^2}{t^2 + 63}R'(t) = \frac{-63(t+9)(t-7)}{(t^2+63)^2}The factored form of R'(t) is where all the work pays off. Once the numerator is in factored form, reading the sign chart is straightforward.
This type of problem models any situation where a quantity rises, peaks, and then falls. Product revenue after launch follows this pattern: sales ramp up, hit a ceiling as the market saturates, then decline. The same mathematics applies to drug concentration in the bloodstream over time, or the spread of a new technology through a population.
Students often forget to check the domain when discarding critical points. In this problem, t = -9 is thrown out because t ≥ 0, but you must state why. On an exam, silently dropping a solution without justification can cost marks.
A common algebra error is expanding the quotient rule numerator incorrectly. Students frequently lose a negative sign when distributing. Write out each product separately before combining.
Students sometimes try to set the denominator equal to zero to find asymptotes, find no real solutions, and then get confused. No real solutions simply means the function has no vertical asymptotes, which is fine.
Forgetting to state the answer in context. The question asks "when" and "how much," so the answer must include both the time (t = 7 weeks) and the revenue value (3.5 million dollars), with units.
⚠️ The quotient rule with full algebraic simplification is a staple exam question. Expect to spend several minutes on the algebra. Practice expanding and factoring the numerator until it is automatic.
⚠️ Applied optimization problems always require a written interpretation: state the time, the quantity, and the units. "The maximum revenue of $3.5 million occurs at t = 7 weeks" is the expected form.
⚠️ You may be asked to verify that a critical point is a maximum (not a minimum) using either the first derivative test (sign chart) or the second derivative test. Know both methods.
⚠️ Checking whether the denominator has real zeros (and therefore vertical asymptotes) is a common sub-question. Be ready to show that t² + 63 = 0 has no real solutions because t² cannot equal -63.
True or false: the quotient rule states (f/g)' = (f'g + fg') / g². (False. The numerator uses subtraction: f'g - fg'.)
Fill in the blank: a critical point where f' changes from positive to negative is a relative ____. (Maximum.)
True or false: if t² + 63 = 0 has no real solutions, then R(t) has no vertical asymptotes. (True.)
In the skateboard revenue problem, the maximum revenue occurs at t = ____ weeks and equals ____ million dollars. (7; 3.5.)
True or false: t = -9 is a valid critical point in this problem. (False. The domain restricts t ≥ 0.)
Q: The profit from a new product t months after launch is P(t) = (50t - t²) / (t² + 50), in thousands of dollars. When does the maximum profit occur?
A: Apply the quotient rule. P'(t) = [(50 - 2t)(t² + 50) - (50t - t²)(2t)] / (t² + 50)². Expand and simplify the numerator: 50t² + 2500 - 2t³ - 100t - 100t² + 2t³ = -50t² - 100t + 2500 = -50(t² + 2t - 50). Factor using the quadratic formula: t = [-2 ± √(4 + 200)] / 2 = [-2 ± √204] / 2. The positive root is approximately t ≈ 6.14 months. Verify with a sign chart that P' changes from + to - at this value.
Q: For R(t) = (63t - t²) / (t² + 63), explain why t² + 63 = 0 has no real solutions.
A: Rearranging gives t² = -63. Since t² is always non-negative (the square of any real number is ≥ 0), it can never equal a negative number. Therefore the equation has no real solutions, and the denominator is never zero.
Q: Suppose a revenue function has R'(t) = -10(t - 3)(t + 5) / (t² + 4)². Find the critical points in the domain t ≥ 0 and classify them.
A: Setting the numerator to zero: t = 3 or t = -5. Since t ≥ 0, discard t = -5. At t = 3, test: R'(1) = -10(1 - 3)(1 + 5) / (positive) = -10(-2)(6) / (positive) = positive, and R'(5) = -10(5 - 3)(5 + 5) / (positive) = -10(2)(10) / (positive) = negative. R' changes from + to -, so t = 3 is a relative maximum.
This connects to concavity and the second derivative (the companion study notes): you could also confirm the maximum at t = 7 by showing R''(7) < 0, which means the function is concave down at that point and therefore has a local maximum.
It also connects to curve sketching: once you know the domain, critical points, intervals of increase/decrease, concavity, and any asymptotes, you can draw the full shape of R(t) from t = 0 onward.
In the broader course, applied optimization leads into marginal analysis (cost, revenue, and profit functions) and elasticity of demand, both of which rely on the same derivative-based reasoning.
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