Tasks 05-04 - Chain Rule & Implicit Differentiation
Section 05: Differential Calculus
A: Skills
Problem 1: Basic Chain Rule (x)
Differentiate the following functions with the chain rule. Name the outer and the inner function first.
\(f(x) = (3x + 7)^6\)
\(g(x) = \sqrt{5x - 2}\)
\(h(x) = (x^2 - 4x + 1)^{10}\)
\(k(x) = \frac{1}{(2x + 3)^4}\)
Chain rule: \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\), the derivative of the outer function (evaluated at the inner function) times the derivative of the inner function.
- Outer: \(u^6\) with derivative \(6u^5\). Inner: \(3x + 7\) with derivative \(3\).
\[f'(x) = 6(3x + 7)^5 \cdot 3 = 18(3x + 7)^5\]
- Rewrite: \(g(x) = (5x - 2)^{1/2}\). Outer: \(u^{1/2}\) with derivative \(\frac{1}{2}u^{-1/2}\). Inner: \(5x - 2\) with derivative \(5\).
\[g'(x) = \frac{1}{2}(5x - 2)^{-1/2} \cdot 5 = \frac{5}{2\sqrt{5x - 2}}\]
- Outer: \(u^{10}\) with derivative \(10u^9\). Inner: \(x^2 - 4x + 1\) with derivative \(2x - 4\).
\[h'(x) = 10(x^2 - 4x + 1)^9 \cdot (2x - 4) = 20(x - 2)(x^2 - 4x + 1)^9\]
- Rewrite: \(k(x) = (2x + 3)^{-4}\). Outer: \(u^{-4}\) with derivative \(-4u^{-5}\). Inner: \(2x + 3\) with derivative \(2\).
\[k'(x) = -4(2x + 3)^{-5} \cdot 2 = -\frac{8}{(2x + 3)^5}\]
Typical mistake: forgetting the inner derivative, e.g. \(f'(x) = 6(3x + 7)^5\) in a). This result is too small by the factor 3.
Problem 2: Chain Rule with Product and Quotient Rule (xx)
Differentiate the following functions. Factor your result as far as possible.
\(f(x) = x^3(2x - 1)^4\)
\(g(x) = (x^2 + 1)^2(3x - 5)^3\)
\(h(x) = \frac{x^2}{(x + 1)^3}\)
Strategy from the lecture: the product (or quotient) rule first, then the chain rule for every composite factor.
- Product rule with \(u = x^3\), \(u' = 3x^2\) and \(v = (2x - 1)^4\), \(v' = 4(2x - 1)^3 \cdot 2 = 8(2x - 1)^3\) (chain rule):
\[f'(x) = 3x^2(2x - 1)^4 + x^3 \cdot 8(2x - 1)^3\]
Factor out \(x^2(2x - 1)^3\):
\[f'(x) = x^2(2x - 1)^3[3(2x - 1) + 8x] = x^2(2x - 1)^3(14x - 3)\]
- Product rule with \(u = (x^2 + 1)^2\), \(u' = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)\) and \(v = (3x - 5)^3\), \(v' = 3(3x - 5)^2 \cdot 3 = 9(3x - 5)^2\):
\[g'(x) = 4x(x^2 + 1)(3x - 5)^3 + (x^2 + 1)^2 \cdot 9(3x - 5)^2\]
Factor out \((x^2 + 1)(3x - 5)^2\):
\[g'(x) = (x^2 + 1)(3x - 5)^2[4x(3x - 5) + 9(x^2 + 1)] = (x^2 + 1)(3x - 5)^2(21x^2 - 20x + 9)\]
- Quotient rule with \(u = x^2\), \(u' = 2x\) and \(v = (x + 1)^3\), \(v' = 3(x + 1)^2 \cdot 1 = 3(x + 1)^2\):
\[h'(x) = \frac{2x(x + 1)^3 - x^2 \cdot 3(x + 1)^2}{(x + 1)^6}\]
Factor out \((x + 1)^2\) in the numerator and cancel:
\[h'(x) = \frac{(x + 1)^2[2x(x + 1) - 3x^2]}{(x + 1)^6} = \frac{2x - x^2}{(x + 1)^4} = \frac{x(2 - x)}{(x + 1)^4}\]
Typical mistake: \(v' = 4(2x - 1)^3\) in a), without the inner derivative 2. The product rule is set up correctly, but the chain rule step inside it is missing.
Problem 3: Simplifying Before Differentiating (x)
Rewrite each function as a single power first, then differentiate it with the chain rule.
\(f(x) = \sqrt[3]{(4x - 1)^2}\)
\(g(x) = \frac{1}{\sqrt{x^2 + 1}}\)
\(h(x) = \frac{(x^2 + 3)^4}{\sqrt{x^2 + 3}}\)
Power laws: \(\sqrt[n]{a^m} = a^{m/n}\), \(\frac{1}{a^r} = a^{-r}\) and \(\frac{a^r}{a^s} = a^{r - s}\).
- \(f(x) = (4x - 1)^{2/3}\):
\[f'(x) = \frac{2}{3}(4x - 1)^{-1/3} \cdot 4 = \frac{8}{3\sqrt[3]{4x - 1}} \qquad (x \neq 0.25)\]
- \(g(x) = (x^2 + 1)^{-1/2}\):
\[g'(x) = -\frac{1}{2}(x^2 + 1)^{-3/2} \cdot 2x = -\frac{x}{(x^2 + 1)^{3/2}} = -\frac{x}{(x^2 + 1)\sqrt{x^2 + 1}}\]
- \(h(x) = (x^2 + 3)^{4 - 1/2} = (x^2 + 3)^{7/2}\):
\[h'(x) = \frac{7}{2}(x^2 + 3)^{5/2} \cdot 2x = 7x(x^2 + 3)^{5/2}\]
Typical mistake: \(\sqrt[3]{(4x - 1)^2} = (4x - 1)^{3/2}\). The root index goes into the denominator of the exponent: \(\sqrt[3]{a^2} = a^{2/3}\). In c), the quotient rule also works, but it takes much longer.
Problem 4: Derivatives of \(e^x\) and \(\ln(x)\) (xx)
Differentiate the following functions. For the logarithms, \(x\) is always chosen so that the argument is positive.
\(f(t) = 300e^{0.08t}\)
\(g(x) = e^{x^2 - 2x}\)
\(h(x) = \ln(5x - 3)\)
\(k(x) = \ln(x^2 + 2x + 5)\)
\(m(x) = (x + 3)e^{2x}\)
\(n(x) = x^2\ln(4x)\)
A student writes \(\frac{d}{dx}\left[e^{-3x}\right] = -3x \cdot e^{-3x - 1}\). Explain the mistake and give the correct derivative.
Rules: \(\left(e^{g(x)}\right)' = e^{g(x)} \cdot g'(x)\), in particular \(\left(e^{kx}\right)' = k \cdot e^{kx}\), and \(\left(\ln(g(x))\right)' = \frac{g'(x)}{g(x)}\).
- Inner function \(0.08t\) with derivative \(0.08\):
\[f'(t) = 300 \cdot 0.08e^{0.08t} = 24e^{0.08t}\]
- Inner function \(x^2 - 2x\) with derivative \(2x - 2\):
\[g'(x) = e^{x^2 - 2x} \cdot (2x - 2) = 2(x - 1)e^{x^2 - 2x}\]
- Inner function \(5x - 3\) with derivative \(5\) (defined for \(x > 0.6\)):
\[h'(x) = \frac{5}{5x - 3}\]
- Inner function \(x^2 + 2x + 5 = (x + 1)^2 + 4 > 0\) for every \(x\):
\[k'(x) = \frac{2x + 2}{x^2 + 2x + 5}\]
- Product rule with \(u = x + 3\), \(u' = 1\) and \(v = e^{2x}\), \(v' = 2e^{2x}\), then factor out \(e^{2x}\):
\[m'(x) = 1 \cdot e^{2x} + (x + 3) \cdot 2e^{2x} = e^{2x}(1 + 2x + 6) = e^{2x}(2x + 7)\]
- Product rule with \(u = x^2\), \(u' = 2x\) and \(v = \ln(4x)\), \(v' = \frac{4}{4x} = \frac{1}{x}\) (defined for \(x > 0\)):
\[n'(x) = 2x\ln(4x) + x^2 \cdot \frac{1}{x} = 2x\ln(4x) + x = x\left(2\ln(4x) + 1\right)\]
- The student uses the power rule “bring the exponent down and subtract 1”. But in \(e^{-3x}\) the variable is in the exponent, so the power rule does not apply. With \(\left(e^{kx}\right)' = k \cdot e^{kx}\) and \(k = -3\):
\[\frac{d}{dx}\left[e^{-3x}\right] = -3e^{-3x}\]
Typical mistake: in f), \(\left(\ln(4x)\right)' = \frac{1}{4x}\). The inner derivative 4 is missing. Correct is \(\frac{4}{4x} = \frac{1}{x}\).
Problem 5: First, Second and Third Derivative (xx)
Note: The second derivative \(f''\) is the derivative of \(f'\), and the third derivative \(f'''\) is the derivative of \(f''\).
Determine the first, second and third derivative of \(f(x) = (1 - 2x)^5\).
Determine the first, second and third derivative of \(g(x) = (2x + 1)e^{-x}\). Factor out \(e^{-x}\) in every step. Determine the \(x\)-value at which the tangent line of \(g\) is horizontal.
Determine the first, second and third derivative of \(h(x) = \ln(3x + 2)\), \(x > -\frac{2}{3}\).
- Chain rule with the inner derivative \(-2\) in every step:
\[f'(x) = 5(1 - 2x)^4 \cdot (-2) = -10(1 - 2x)^4\]
\[f''(x) = -40(1 - 2x)^3 \cdot (-2) = 80(1 - 2x)^3\]
\[f'''(x) = 240(1 - 2x)^2 \cdot (-2) = -480(1 - 2x)^2\]
- Product rule with \(u = 2x + 1\), \(u' = 2\) and \(v = e^{-x}\), \(v' = -e^{-x}\):
\[g'(x) = 2e^{-x} - (2x + 1)e^{-x} = e^{-x}(2 - 2x - 1) = e^{-x}(1 - 2x)\]
Now \(u = 1 - 2x\), \(u' = -2\) and \(v = e^{-x}\), \(v' = -e^{-x}\):
\[g''(x) = -2e^{-x} - (1 - 2x)e^{-x} = e^{-x}(-2 - 1 + 2x) = e^{-x}(2x - 3)\]
\[g'''(x) = 2e^{-x} - (2x - 3)e^{-x} = e^{-x}(2 - 2x + 3) = e^{-x}(5 - 2x)\]
Horizontal tangent line: \(g'(x) = 0\). Since \(e^{-x} > 0\) for every \(x\), only the bracket can become zero:
\[1 - 2x = 0 \iff x = 0.5\]
- Rewrite the first derivative as a power, then use the chain rule with the inner derivative \(3\):
\[h'(x) = \frac{3}{3x + 2} = 3(3x + 2)^{-1}\]
\[h''(x) = -3(3x + 2)^{-2} \cdot 3 = -\frac{9}{(3x + 2)^2}\]
\[h'''(x) = 18(3x + 2)^{-3} \cdot 3 = \frac{54}{(3x + 2)^3}\]
Typical mistake: \(\left(e^{-x}\right)' = e^{-x}\). The inner function is \(-x\), so the inner derivative \(-1\) gives \(\left(e^{-x}\right)' = -e^{-x}\). With the wrong sign, \(g'(x)\) becomes \(e^{-x}(2x + 3)\), and every further derivative is wrong too.
B: Set Up & Check
Problem 6: From Text to Rates (xx)
Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.
“A pet insurance company has 50,000 customers today, and the number of its customers grows by 2,000 per month.” Let \(C(t)\) be the number of customers in thousands, \(t\) months from today. Write both statements with \(C\) and its derivative.
Check: If the rate stays the same, how many customers does the company have after 6 months? Which unit does your derivative have?
“A company sells an online booking system to doctors’ practices. With \(C\) thousand practices as customers, its monthly revenue is \(R(C) = 4C^{0.75}\) thousand EUR. The number of customers grows by 1,200 practices per month.” Revenue depends on the customer base, and the customer base depends on time. Write the rate at which the monthly revenue grows as a product of two rates.
Check: Does your formula give 1,800 EUR per month at the moment when 16,000 practices use the system?
Hint: The fourth root of 16 is 2, because \(2^4 = 16\). So \(16^{0.25} = 2\).
“A company rents out bouncy castles (inflatable play castles) for children’s parties. The owner keeps the monthly rental revenue at 4,800 EUR: when the price \(p\) per rental changes, the number \(q\) of rentals per month changes so that price times quantity stays the same.” Write this as an equation and determine \(\frac{dq}{dp}\) by implicit differentiation.
Check: Compute \(q\) and \(\frac{dq}{dp}\) at a price of 240 EUR per rental. Do you get the same slope if you first solve the equation for \(q\)?
“A bank has 250 cash machines (ATMs) in a region. Because more and more people pay by card, it reduces their number with a continuous decline of 4 % per year.” Write the number \(N(t)\) of cash machines \(t\) years from now and determine \(N'(t)\) with its unit.
Check: By how many machines per year does the number fall today? Is \(N'(t) = -0.04 \cdot N(t)\)?
The number today is a value of \(C\), the growth per month is a rate, i.e. a derivative:
\[C(0) = 50, \qquad C'(t) = 2\]
Unit of \(C'(t)\): thousand customers per month.
Check: with a constant rate, \(C(6) = 50 + 6 \cdot 2 = 62\), i.e. 62,000 customers after 6 months.
Typical mistake: writing \(C(t) = 2\) or \(C = 2{,}000\). “Grows by 2,000 per month” describes how fast the number changes, not the number itself. A second mistake is \(C'(t) = 2{,}000\) while \(C\) counts thousands of customers.
Chain rule in Leibniz notation, as for related rates:
\[\frac{dR}{dt} = \frac{dR}{dC} \cdot \frac{dC}{dt} = 3C^{-0.25} \cdot 1.2\]
with \(\frac{dR}{dC} = 4 \cdot 0.75 C^{-0.25} = 3C^{-0.25}\) and \(\frac{dC}{dt} = 1.2\) thousand practices per month.
Check: at \(C = 16\), \(\frac{dR}{dC} = \frac{3}{16^{0.25}} = \frac{3}{2} = 1.5\) thousand EUR per thousand practices, so \(\frac{dR}{dt} = 1.5 \cdot 1.2 = 1.8\) ✓. The monthly revenue grows by about 1,800 EUR per month.
Typical mistake: \(\frac{dR}{dt} = R'(16) = 1.5\). This is the revenue per additional thousand practices. It still has to be multiplied by the number of new practices per month.
Price times quantity stays at 4,800 EUR:
\[p \cdot q = 4800\]
\(q\) depends on \(p\), so differentiate both sides with respect to \(p\) (product rule on the left):
\[q + p \cdot \frac{dq}{dp} = 0 \iff \frac{dq}{dp} = -\frac{q}{p}\]
Check: at \(p = 240\), \(q = \frac{4800}{240} = 20\) rentals and \(\frac{dq}{dp} = -\frac{20}{240} = -\frac{1}{12} \approx -0.083\) rentals per EUR: about one rental fewer for every 12 EUR more. Solving first gives \(q = 4800p^{-1}\) and \(\frac{dq}{dp} = -4800p^{-2} = -\frac{4800}{57600} = -\frac{1}{12}\) ✓.
Typical mistake: differentiating \(p \cdot q\) as if \(q\) were a constant (\(\frac{d}{dp}[p \cdot q] = q\)). Then \(q = 0\), which makes no sense. \(q\) changes with \(p\), so the product rule is needed.
Continuous decline with the rate \(k = 0.04\) per year (lecture 04-03):
\[N(t) = 250e^{-0.04t}, \qquad N'(t) = 250 \cdot (-0.04)e^{-0.04t} = -10e^{-0.04t}\]
Unit of \(N'(t)\): cash machines per year.
Check: \(N'(0) = -10\), so today the number falls by about 10 machines per year. And \(-0.04 \cdot N(t) = -0.04 \cdot 250e^{-0.04t} = -10e^{-0.04t} = N'(t)\) ✓: in every year, the decline is 4 % of the current number.
Typical mistake: \(N(t) = 250 \cdot 0.96^t\). This is a decline of 4 % once per year, not a continuous one. A second mistake is \(N'(t) = 250e^{-0.04t}\): the inner derivative \(-0.04\) is missing.
Problem 7: One Paragraph, All Conditions (xx)
An online shop sells second-hand children’s clothes. Today it has 40,000 customers, and the number of customers grows continuously by 3 % per month. The monthly revenue of the shop depends on the number \(U\) of customers in thousands: it is \(R = 2.4U - 0.02U^2\) thousand EUR. The owners want to know how fast the monthly revenue grows 10 months from now, at which number of customers the revenue stops growing, and when the shop will have 50,000 customers.
List every condition and every function in this paragraph as a mathematical statement. Do not solve anything.
Check: An assistant writes the growth condition as \(U'(t) = 3\). Explain what is wrong with it.
- Number of customers: \(U(t) = 40e^{0.03t}\) (\(U\) in thousand customers, \(t\) in months from today), so \(U(0) = 40\)
- Continuous growth by 3 % per month: \(U'(t) = 0.03 \cdot U(t)\)
- Revenue: \(R(U) = 2.4U - 0.02U^2\) (thousand EUR per month)
- Rate of change of the revenue (chain rule): \(\frac{dR}{dt} = R'(U) \cdot U'(t) = (2.4 - 0.04U) \cdot U'(t)\)
- Wanted: \(\frac{dR}{dt}\) at \(t = 10\)
- The revenue stops growing: \(R'(U) = 0\), i.e. \(2.4 - 0.04U = 0\)
- 50,000 customers: \(U(t) = 50\)
Check: “grows continuously by 3 % per month” means 3 % of the current number per month, not 3 thousand customers. The correct condition is \(U'(t) = 0.03 \cdot U(t)\). Today this is \(0.03 \cdot 40 = 1.2\) thousand customers per month, and the rate grows with \(U\).
Typical mistake: \(U(t) = 40 + 3t\) (a fixed number per month) or \(U(t) = 40 \cdot 1.03^t\) (3 % once per month). The paragraph says “continuously”, so the base is \(e\): \(U(t) = 40e^{0.03t}\).
C: Applied
Problem 8: Staff and Robots in an Office Cleaning Company (xx)
A company cleans office buildings. For the weekly cleaning of its customers’ offices, it combines \(L\) hours of work by its staff with \(K\) hours of work by cleaning robots (machines that vacuum and wipe the floors on their own). All combinations that get the weekly work done satisfy
\[\sqrt{L} \cdot K = 600\]
At the moment, the company uses 100 staff hours and 60 robot hours per week.
Check that the current combination satisfies the equation. Solve the equation for \(K\) and determine \(\frac{dK}{dL}\) with the power rule.
Determine \(\frac{dK}{dL}\) by implicit differentiation and show that \(\frac{dK}{dL} = -\frac{K}{2L}\). Compare both results at the current combination.
(for verification: \(\frac{dK}{dL} = -0.3\) at \(L = 100\), \(K = 60\))
Interpret the value \(-0.3\). A staff hour costs the company 30 EUR, a robot hour 80 EUR. Decide whether the company saves money if it replaces a few robot hours with staff hours, or a few staff hours with robot hours.
Two employees leave the company. In the coming weeks, the staff hours fall by 5 hours per week. Determine how fast the robot hours must grow at the moment, so that the weekly work still gets done.
Estimate with \(\frac{dK}{dL}\) how many robot hours the company needs if it cuts the staff hours from 100 to 90 per week. Compare with the exact value (round to two decimal places).
a) \(\sqrt{100} \cdot 60 = 10 \cdot 60 = 600\) ✓
Solve for \(K\): \(K = \frac{600}{\sqrt{L}} = 600L^{-1/2}\). Power rule:
\[\frac{dK}{dL} = 600 \cdot \left(-\frac{1}{2}\right)L^{-3/2} = -300L^{-3/2} = -\frac{300}{L\sqrt{L}}\]
At \(L = 100\): \(\frac{dK}{dL} = -\frac{300}{100 \cdot 10} = -0.3\).
b) \(K\) depends on \(L\), so every term with \(K\) gets the factor \(\frac{dK}{dL}\). Differentiate both sides with respect to \(L\), with the product rule on the left:
\[\frac{1}{2}L^{-1/2} \cdot K + L^{1/2} \cdot \frac{dK}{dL} = 0\]
\[\frac{dK}{dL} = -\frac{\frac{1}{2}L^{-1/2} \cdot K}{L^{1/2}} = -\frac{K}{2L} \quad \checkmark\]
At \(L = 100\), \(K = 60\): \(\frac{dK}{dL} = -\frac{60}{200} = -0.3\), the same as in a). (With \(K = 600L^{-1/2}\), the formula \(-\frac{K}{2L}\) becomes \(-300L^{-3/2}\) in general.)
c) At the current combination, one more staff hour replaces about 0.3 robot hours, and one staff hour less needs about 0.3 more robot hours. The negative sign shows the trade-off: more of one input, less of the other.
- Replace robot hours with staff: one more staff hour costs 30 EUR and saves about \(0.3 \cdot 80 = 24\) EUR of robot time. The costs rise by about 6 EUR.
- Replace staff hours with robots: one staff hour less saves 30 EUR, the 0.3 extra robot hours cost about 24 EUR. The costs fall by about 6 EUR.
So at the current combination, the company saves money if it replaces a few staff hours with robot hours.
d) Both \(L\) and \(K\) change with time \(t\). Differentiate \(L^{1/2} \cdot K = 600\) with respect to \(t\) (product rule and chain rule):
\[\frac{1}{2}L^{-1/2} \cdot \frac{dL}{dt} \cdot K + L^{1/2} \cdot \frac{dK}{dt} = 0 \iff \frac{dK}{dt} = -\frac{K}{2L} \cdot \frac{dL}{dt}\]
With \(L = 100\), \(K = 60\) and \(\frac{dL}{dt} = -5\) hours per week:
\[\frac{dK}{dt} = -0.3 \cdot (-5) = 1.5\]
At the moment, the robot hours must grow by about 1.5 hours per week.
e) Linear approximation with \(\Delta L = 90 - 100 = -10\):
\[\Delta K \approx \frac{dK}{dL} \cdot \Delta L = -0.3 \cdot (-10) = 3, \qquad K \approx 60 + 3 = 63\]
Exact: \(K = \frac{600}{\sqrt{90}} \approx 63.25\) robot hours. The estimate is about 0.25 hours too low. The error is small, because 90 is close to 100.
Problem 9: Parcel Lockers in a City (xx)
A logistics company runs parcel lockers in a city (automatic stations with lockable compartments where customers pick up their parcels). With \(n\) lockers, its yearly revenue is
\[R(n) = 120\ln(0.1n + 1)\]
thousand EUR, and its yearly costs are \(C(n) = 1.5n + 20\) thousand EUR. The company runs 40 lockers today and installs 15 new lockers per year. Round to two decimal places.
Determine the marginal revenue with the chain rule and show that \(R'(n) = \frac{120}{n + 10}\). Compute \(R'(40)\) and interpret it.
Give the profit \(P(n)\). Write the relationship between \(\frac{dP}{dt}\) and \(\frac{dn}{dt}\) and compute how fast the profit changes today and at 110 lockers.
(for verification: 13,500 EUR per year today)
Determine the number of lockers at which the marginal revenue equals the marginal cost. Explain with the values from b) why the profit is highest there, and compute the highest profit.
Determine when the company reaches this number of lockers with its plan. Explain why the profit grows more and more slowly until then.
The city introduces a fee of 500 EUR per locker and year for the use of public space. Determine the new best number of lockers and decide what the company should do with its plan.
a) Chain rule with the inner function \(0.1n + 1\) and its derivative \(0.1\):
\[R'(n) = 120 \cdot \frac{0.1}{0.1n + 1} = \frac{12}{0.1n + 1} = \frac{120}{n + 10} \quad \checkmark\]
(numerator and denominator multiplied by 10). \(R'(40) = \frac{120}{50} = 2.4\) thousand EUR per locker: at 40 lockers, one more locker brings about 2,400 EUR more revenue per year.
b) \(P(n) = R(n) - C(n) = 120\ln(0.1n + 1) - 1.5n - 20\) with the marginal profit \(P'(n) = \frac{120}{n + 10} - 1.5\).
\(n\) changes with time, so by the chain rule
\[\frac{dP}{dt} = \frac{dP}{dn} \cdot \frac{dn}{dt} = \left(\frac{120}{n + 10} - 1.5\right) \cdot \frac{dn}{dt}, \qquad \frac{dn}{dt} = 15 \text{ lockers per year}\]
- Today (\(n = 40\)): \(\frac{dP}{dt} = (2.4 - 1.5) \cdot 15 = 13.5\). The profit grows by about 13,500 EUR per year.
- At \(n = 110\): \(\frac{dP}{dt} = \left(\frac{120}{120} - 1.5\right) \cdot 15 = -0.5 \cdot 15 = -7.5\). The profit falls by about 7,500 EUR per year.
c) Marginal cost \(C'(n) = 1.5\). \(MR = MC\):
\[\frac{120}{n + 10} = 1.5 \iff n + 10 = 80 \iff n = 70\]
Comparison as in lecture 05-03: at 40 lockers, \(MR = 2.4 > 1.5 = MC\), so a further locker adds more revenue than cost and the profit still grows. At 110 lockers, \(MR = 1 < 1.5 = MC\), so a further locker lowers the profit. On the whole range, \(MR = \frac{120}{n + 10}\) gets smaller as \(n\) grows (the same number divided by a larger number). So \(MR > MC\) below 70 lockers and \(MR < MC\) above 70 lockers: the profit rises up to 70 lockers and falls afterwards. It is highest at 70 lockers:
\[P(70) = 120\ln(8) - 105 - 20 \approx 249.53 - 125 = 124.53\]
The highest profit is about 124,530 EUR per year (today, at 40 lockers: \(P(40) = 120\ln(5) - 80 \approx 113.13\)).
d) \(40 + 15t = 70 \iff t = 2\): in 2 years.
Until then, \(\frac{dP}{dt} = \left(\frac{120}{n + 10} - 1.5\right) \cdot 15\) gets smaller with every new locker, because \(\frac{120}{n + 10}\) falls: each new locker brings less additional revenue than the one before (the lockers compete for the same customers), but costs the same 1,500 EUR per year. At 70 lockers, \(\frac{dP}{dt} = 0\).
e) The fee raises the costs to \(C_1(n) = 2n + 20\), so \(MC = 2\):
\[\frac{120}{n + 10} = 2 \iff n + 10 = 60 \iff n = 50\]
With the same comparison as in c), the profit is now highest at 50 lockers. The company should install only 10 more lockers (at 15 per year, within about 8 months) and then stop its plan. Each locker above 50 would cost more than it brings.
D: FSP Problem
Work on this problem as in the exam: write down every step. The points show what each part is worth.
FSP Problem: Fibre-Optic Internet in a Rural District [20 pts. total]
The public utility company of a rural district (a company owned by the district that supplies water and energy) builds a fibre-optic network (very fast internet through glass fibre cables). \(t\) months after the start of sales,
\[A(t) = 90 - 80e^{-0.2t}, \qquad t \geq 0\]
hundred households have booked a fibre-optic connection. Round all results to two decimal places.
Part a)
Determine the limit of \(A(t)\) for \(t \to \infty\) and explain what it means for the company. Compute \(A(0)\) and \(A(12)\) and interpret both values. [3 pts.]
Limit: \(e^{-0.2t} \to 0\) for \(t \to \infty\), so
\[\lim_{t \to \infty} A(t) = 90 - 0 = 90\]
Since \(80e^{-0.2t} > 0\) for every \(t\), \(A(t) < 90\): \(A\) approaches 90 from below and never reaches it.
Meaning: in the long run, almost 9,000 households book a connection, but never more. This is the market limit of the district, for example the number of households the network can reach.
Values: \(A(0) = 90 - 80e^{0} = 90 - 80 = 10\): at the start of sales, 1,000 households have already booked (for example by pre-order). \(A(12) = 90 - 80e^{-2.4} \approx 90 - 7.26 = 82.74\): after one year, about 8,274 households have booked.
Grading: 1 Pt. limit 90 with \(e^{-0.2t} \to 0\), 1 Pt. approached from below and meaning as a market limit, 1 Pt. \(A(0) = 10\) and \(A(12) \approx 82.74\) with meaning.
Part b)
Determine the first, second and third derivative of \(A\). [3 pts.]
Chain rule with \(\left(e^{kt}\right)' = k \cdot e^{kt}\) and \(k = -0.2\). The constant 90 has the derivative 0:
\[A'(t) = -80 \cdot (-0.2)e^{-0.2t} = 16e^{-0.2t}\]
\[A''(t) = 16 \cdot (-0.2)e^{-0.2t} = -3.2e^{-0.2t}\]
\[A'''(t) = -3.2 \cdot (-0.2)e^{-0.2t} = 0.64e^{-0.2t}\]
Grading: 1 Pt. each derivative.
Part c)
Show that \(A'(t) = 0.2 \cdot (90 - A(t))\). Compute \(A'(0)\) and interpret it with units. Explain with the formula why the bookings grow more and more slowly. [3 pts.]
Show: \(90 - A(t) = 90 - 90 + 80e^{-0.2t} = 80e^{-0.2t}\), so
\[0.2 \cdot (90 - A(t)) = 0.2 \cdot 80e^{-0.2t} = 16e^{-0.2t} = A'(t) \quad \checkmark\]
Rate at the start: \(A'(0) = 16\) hundred households per month. At the start of sales, about 1,600 households per month book a connection.
Explanation: \(90 - A(t)\) is the number of households (in hundreds) that have not booked yet. The growth rate is 20 % of this remaining number. The more households have booked, the smaller the remaining number, so the bookings grow more and more slowly.
Grading: 1 Pt. show-that, 1 Pt. \(A'(0) = 16\) with unit, 1 Pt. explanation with the remaining households.
Part d)
The monthly costs of operating the network (support and maintenance) depend on the number \(A\) of households (in hundreds) that have booked a connection:
\[C(A) = 0.01A^2 + 0.5A + 30 \text{ thousand EUR}\]
Determine how fast the monthly costs grow at the moment when 6,500 households have booked. [4 pts.]
Related rates: \(C\) depends on \(A\), and \(A\) depends on \(t\). Chain rule:
\[\frac{dC}{dt} = C'(A) \cdot \frac{dA}{dt}, \qquad C'(A) = 0.02A + 0.5\]
At 6,500 households, \(A = 65\): \(C'(65) = 1.3 + 0.5 = 1.8\) thousand EUR per hundred households.
Rate of \(A\) at this moment, with c): \(A' = 0.2 \cdot (90 - 65) = 0.2 \cdot 25 = 5\) hundred households per month. (Or with the time: \(90 - 80e^{-0.2t} = 65 \iff e^{-0.2t} = \frac{25}{80}\), and \(A' = 16 \cdot \frac{25}{80} = 5\).)
\[\frac{dC}{dt} = 1.8 \cdot 5 = 9\]
At this moment, the monthly costs grow by about 9,000 EUR per month.
Grading: 1 Pt. \(\frac{dC}{dt} = C'(A) \cdot \frac{dA}{dt}\), 1 Pt. \(C'(65) = 1.8\), 1 Pt. \(A' = 5\) at 6,500 households, 1 Pt. result 9,000 EUR per month with meaning.
Part e)
The utility company ends its sales campaign as soon as fewer than 200 households per month book a connection. Determine this time and the number of households that have booked by then. [4 pts.]
Condition: 200 households are 2 hundred households, so the campaign ends when \(A'(t) = 2\). After that, \(A'(t) = 16e^{-0.2t}\) keeps falling, because \(e^{-0.2t}\) gets smaller as \(t\) grows.
Equation:
\[16e^{-0.2t} = 2 \iff e^{-0.2t} = \frac{1}{8} = 0.125\]
Logarithm:
\[-0.2t = \ln(0.125) \iff t = \frac{\ln(0.125)}{-0.2} = 5\ln(8) \approx 10.40\]
The campaign ends about 10.40 months after the start of sales, i.e. in the 11th month.
Households: \(A(10.40) = 90 - 80 \cdot 0.125 = 90 - 10 = 80\), i.e. about 8,000 households. (With c): \(A' = 2 \iff 0.2 \cdot (90 - A) = 2 \iff A = 80\) ✓.)
Grading: 1 Pt. condition \(A'(t) = 2\) (200 households = 2 hundred), 1 Pt. \(e^{-0.2t} = 0.125\), 1 Pt. \(t = 5\ln(8) \approx 10.40\) months, 1 Pt. 8,000 households.
Part f)
Compute the average number of new bookings per month in the first year (the slope of the secant from \(t = 0\) to \(t = 12\)). Compare it with \(A'(0)\) and \(A'(12)\) and explain the order of the three rates. [3 pts.]
Average rate (slope of the secant): with the values from a),
\[\frac{A(12) - A(0)}{12 - 0} = \frac{82.74 - 10}{12} \approx 6.06\]
hundred households per month. In the first year, on average about 606 households per month book a connection.
Rates at the ends (slopes of the tangents): \(A'(0) = 16\) from c), and \(A'(12) = 16e^{-2.4} \approx 1.45\). So
\[A'(12) \approx 1.45 \; < \; 6.06 \; < \; 16 = A'(0)\]
Explanation: by b), \(A''(t) = -3.2e^{-0.2t} < 0\) for every \(t\), so the booking rate \(A'\) falls all the time. At the start of sales, about 1,600 households per month book. After one year, only about 145 per month book. The average of the first year has to lie between the highest and the lowest rate of that year.
Grading: 1 Pt. secant slope \(\approx 6.06\) with unit, 1 Pt. \(A'(12) \approx 1.45\) and the order of the three rates, 1 Pt. explanation with the falling rate (\(A'' < 0\)).