Tasks 05-03 - Differentiation Rules & Tangent Lines

Section 05: Differential Calculus

A: Skills

Problem 1: Basic Differentiation Rules (x)

Differentiate the following functions with the power, constant, constant multiple and sum rules.

  1. \(f(x) = 5x^4 - 3x^2 + 7x - 2\)

  2. \(g(x) = 2\sqrt{x} + \frac{3}{x} - \frac{1}{x^3}\)

  3. \(h(x) = \frac{1}{2}x^{-4} + 4x^{1/3}\)

  4. \(k(x) = \frac{x^2}{4} - \frac{8}{\sqrt{x}} + \pi\)

First write every root and every fraction as a power of \(x\), then use the power rule term by term.

  1. Power rule and sum rule, the constant \(-2\) gives 0:

\[f'(x) = 5 \cdot 4x^3 - 3 \cdot 2x + 7 - 0 = 20x^3 - 6x + 7\]

  1. Rewrite with exponents: \(g(x) = 2x^{1/2} + 3x^{-1} - x^{-3}\).

\[g'(x) = 2 \cdot \frac{1}{2}x^{-1/2} + 3 \cdot (-1)x^{-2} - (-3)x^{-4} = x^{-1/2} - 3x^{-2} + 3x^{-4} = \frac{1}{\sqrt{x}} - \frac{3}{x^2} + \frac{3}{x^4}\]

  1. Power rule with the exponents \(-4\) and \(\frac{1}{3}\):

\[h'(x) = \frac{1}{2} \cdot (-4)x^{-5} + 4 \cdot \frac{1}{3}x^{-2/3} = -2x^{-5} + \frac{4}{3}x^{-2/3} = -\frac{2}{x^5} + \frac{4}{3\sqrt[3]{x^2}}\]

  1. Rewrite: \(k(x) = \frac{1}{4}x^2 - 8x^{-1/2} + \pi\). The number \(\pi\) is a constant, so its derivative is 0.

\[k'(x) = \frac{1}{4} \cdot 2x - 8 \cdot \left(-\frac{1}{2}\right)x^{-3/2} + 0 = \frac{x}{2} + 4x^{-3/2} = \frac{x}{2} + \frac{4}{x\sqrt{x}}\]

Typical mistake: \(\frac{d}{dx}[\pi] = 1\) or \(\frac{d}{dx}[\pi] = \pi\). \(\pi\) is a number like 7, not a variable.

Problem 2: Product Rule Practice (x)

Differentiate the following functions with the product rule.

  1. \(f(x) = (3x^2 + 1)(x - 4)\)

  2. \(g(x) = x^3(2x^2 - 5x + 1)\)

  3. Check your answer to a) by expanding \(f(x)\) first and then differentiating.

  4. \(k(x) = \sqrt{x}\,(x^2 - 6)\). Check your result by writing \(k(x)\) as a sum of powers of \(x\) first.

Product rule: \((u \cdot v)' = u' \cdot v + u \cdot v'\).

  1. \(u = 3x^2 + 1\), \(u' = 6x\) and \(v = x - 4\), \(v' = 1\):

\[f'(x) = 6x(x - 4) + (3x^2 + 1) \cdot 1 = 6x^2 - 24x + 3x^2 + 1 = 9x^2 - 24x + 1\]

  1. \(u = x^3\), \(u' = 3x^2\) and \(v = 2x^2 - 5x + 1\), \(v' = 4x - 5\):

\[g'(x) = 3x^2(2x^2 - 5x + 1) + x^3(4x - 5) = 6x^4 - 15x^3 + 3x^2 + 4x^4 - 5x^3 = 10x^4 - 20x^3 + 3x^2\]

  1. Expand first: \(f(x) = 3x^3 - 12x^2 + x - 4\). Power rule: \(f'(x) = 9x^2 - 24x + 1\) ✓, the same as in a).

  2. \(u = \sqrt{x} = x^{1/2}\), \(u' = \frac{1}{2\sqrt{x}}\) and \(v = x^2 - 6\), \(v' = 2x\):

\[k'(x) = \frac{x^2 - 6}{2\sqrt{x}} + \sqrt{x} \cdot 2x = \frac{x^2 - 6}{2\sqrt{x}} + \frac{4x^2}{2\sqrt{x}} = \frac{5x^2 - 6}{2\sqrt{x}}\]

For the second fraction, \(2x\sqrt{x} = \frac{2x \cdot 2x}{2\sqrt{x}}\), because \(\sqrt{x} \cdot \sqrt{x} = x\).

Check: \(k(x) = x^{1/2} \cdot x^2 - 6x^{1/2} = x^{5/2} - 6x^{1/2}\), so

\[k'(x) = \frac{5}{2}x^{3/2} - 3x^{-1/2} = \frac{5x^2}{2\sqrt{x}} - \frac{6}{2\sqrt{x}} = \frac{5x^2 - 6}{2\sqrt{x}} \quad \checkmark\]

Typical mistake: \((u \cdot v)' = u' \cdot v'\). For a) this gives \(6x \cdot 1 = 6x\), which is far from \(9x^2 - 24x + 1\).

Problem 3: Quotient Rule Applications (xx)

Differentiate the following functions with the quotient rule. Simplify the numerator as far as possible.

  1. \(f(x) = \frac{x^2 + 8}{x + 1}\)

  2. \(g(x) = \frac{4x + 1}{2x - 3}\)

  3. \(h(x) = \frac{x^3 - 1}{x^2 + 1}\)

  4. For \(g\) from b), a student writes \(g'(x) = \frac{4}{2} = 2\). Explain the mistake. Decide with your result from b) whether \(g\) increases or decreases.

Quotient rule: \(\left(\frac{u}{v}\right)' = \frac{u' \cdot v - u \cdot v'}{v^2}\).

  1. \(u = x^2 + 8\), \(u' = 2x\) and \(v = x + 1\), \(v' = 1\):

\[f'(x) = \frac{2x(x + 1) - (x^2 + 8) \cdot 1}{(x + 1)^2} = \frac{2x^2 + 2x - x^2 - 8}{(x + 1)^2} = \frac{x^2 + 2x - 8}{(x + 1)^2} = \frac{(x + 4)(x - 2)}{(x + 1)^2}\]

  1. \(u = 4x + 1\), \(u' = 4\) and \(v = 2x - 3\), \(v' = 2\):

\[g'(x) = \frac{4(2x - 3) - (4x + 1) \cdot 2}{(2x - 3)^2} = \frac{8x - 12 - 8x - 2}{(2x - 3)^2} = \frac{-14}{(2x - 3)^2}\]

  1. \(u = x^3 - 1\), \(u' = 3x^2\) and \(v = x^2 + 1\), \(v' = 2x\):

\[h'(x) = \frac{3x^2(x^2 + 1) - (x^3 - 1) \cdot 2x}{(x^2 + 1)^2} = \frac{3x^4 + 3x^2 - 2x^4 + 2x}{(x^2 + 1)^2} = \frac{x^4 + 3x^2 + 2x}{(x^2 + 1)^2}\]

  1. The student divides the derivative of the numerator by the derivative of the denominator. But \(\left(\frac{u}{v}\right)' \neq \frac{u'}{v'}\): the full quotient rule is needed.

In b), the numerator of \(g'(x)\) is \(-14 < 0\) and the denominator \((2x - 3)^2\) is a square, positive for every \(x \neq 1.5\). So \(g'(x) < 0\) everywhere: \(g\) decreases (left and right of \(x = 1.5\), where it is not defined). The student’s positive value would even give the wrong direction.

Problem 4: Finding Tangent Lines (xx)

Determine the equation of the tangent line at the given point.

  1. \(f(x) = x^3 - 2x^2 + 1\) at \(x = 2\)

  2. \(g(x) = \frac{x + 1}{x - 1}\) at \(x = 3\)

  3. \(h(x) = (2x - 1)(x^2 + 3)\) at \(x = 1\) (use the product rule)

  4. Let \(k(x) = 2x^3 - 54x + 1\). Determine all points of the graph of \(k\) at which the tangent line is horizontal, and give the equations of these tangent lines.

Every time the same three steps: point \((a, f(a))\), slope \(f'(a)\), point-slope form \(y - f(a) = f'(a)(x - a)\).

  1. Point: \(f(2) = 8 - 8 + 1 = 1\), so \((2, 1)\).

Slope: \(f'(x) = 3x^2 - 4x\), \(f'(2) = 12 - 8 = 4\).

\[y - 1 = 4(x - 2) \iff y = 4x - 7\]

  1. Point: \(g(3) = \frac{4}{2} = 2\), so \((3, 2)\).

Slope with the quotient rule:

\[g'(x) = \frac{1 \cdot (x - 1) - (x + 1) \cdot 1}{(x - 1)^2} = \frac{-2}{(x - 1)^2}, \qquad g'(3) = \frac{-2}{4} = -\frac{1}{2}\]

\[y - 2 = -\frac{1}{2}(x - 3) \iff y = -\frac{1}{2}x + \frac{7}{2}\]

  1. Point: \(h(1) = 1 \cdot 4 = 4\), so \((1, 4)\).

Slope with the product rule, \(u = 2x - 1\), \(u' = 2\) and \(v = x^2 + 3\), \(v' = 2x\):

\[h'(x) = 2(x^2 + 3) + (2x - 1) \cdot 2x = 6x^2 - 2x + 6, \qquad h'(1) = 6 - 2 + 6 = 10\]

\[y - 4 = 10(x - 1) \iff y = 10x - 6\]

  1. A horizontal tangent line has slope 0:

\[k'(x) = 6x^2 - 54 = 0 \iff x^2 = 9 \iff x = 3 \text{ or } x = -3\]

  • \(k(3) = 54 - 162 + 1 = -107\): point \((3, -107)\), tangent line \(y = -107\)
  • \(k(-3) = -54 + 162 + 1 = 109\): point \((-3, 109)\), tangent line \(y = 109\)

Typical mistake: writing the tangent line as \(y = f'(a) \cdot x\), e.g. \(y = 4x\) in a). This line has the right slope but does not pass through \((2, 1)\).

Problem 5: Choosing the Rule (xx)

Before you differentiate, ask: can I simplify first? Decide for each function which rule gives the derivative fastest, then differentiate. State which of the five functions really needs the quotient rule.

  1. \(f(x) = (3x - 2)(3x + 2)\)

  2. \(g(x) = \frac{3x^4 + 8x}{x}\), \(x \neq 0\)

  3. \(h(x) = \frac{7}{2x^3}\)

  4. \(k(x) = \frac{x^2 - 1}{x + 3}\)

  5. \(m(x) = 4x\sqrt{x}\)

Function Fastest way Derivative
a) \((3x - 2)(3x + 2)\) difference of squares: \(9x^2 - 4\), then power rule \(f'(x) = 18x\)
b) \(\frac{3x^4 + 8x}{x}\) divide each term by \(x\): \(3x^3 + 8\), then power rule \(g'(x) = 9x^2\)
c) \(\frac{7}{2x^3}\) write as \(\frac{7}{2}x^{-3}\), constant multiple and power rule \(h'(x) = -\frac{21}{2}x^{-4} = -\frac{21}{2x^4}\)
d) \(\frac{x^2 - 1}{x + 3}\) quotient rule (nothing cancels) \(k'(x) = \frac{x^2 + 6x + 1}{(x + 3)^2}\)
e) \(4x\sqrt{x}\) \(x \cdot x^{1/2} = x^{3/2}\), so \(4x^{3/2}\), then power rule \(m'(x) = 6x^{1/2} = 6\sqrt{x}\)
  1. in detail: \(u = x^2 - 1\), \(u' = 2x\) and \(v = x + 3\), \(v' = 1\):

\[k'(x) = \frac{2x(x + 3) - (x^2 - 1)}{(x + 3)^2} = \frac{2x^2 + 6x - x^2 + 1}{(x + 3)^2} = \frac{x^2 + 6x + 1}{(x + 3)^2}\]

Only d) really needs the quotient rule. The numerator \(x^2 - 1 = (x - 1)(x + 1)\) has no factor \(x + 3\), so nothing cancels.

The other rules give the same results, only with more work. For c), the quotient rule gives \(\frac{0 \cdot 2x^3 - 7 \cdot 6x^2}{(2x^3)^2} = \frac{-42x^2}{4x^6} = -\frac{21}{2x^4}\) ✓.

B: Set Up & Check

Problem 6: From Text to Derivatives (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

  1. “A dance school earns \(R(x) = 90x - 0.25x^2\) EUR per month with \(x\) course participants.” The manager wants to show “the tangent line of \(R\) at 100 participants” in a presentation. Write down what you need for this tangent line and give its equation in point-slope form.

    Check: Does your line give 6,500 EUR at 100 participants?

  2. “A photo studio makes a weekly profit of \(P(x) = -2x^2 + 180x - 2400\) EUR with \(x\) photo shoots.” The owner says: “This week we had 30 shoots. With 32 shoots next week, we will earn about 1,320 EUR, because our marginal profit at 30 shoots is 60 EUR per shoot.” Write the owner’s estimate as a linear approximation.

    Check: Compute \(P(30)\), \(P'(30)\) and the exact profit \(P(32)\). How large is the error of the estimate?

  3. “A cheese dairy sells cheese wheels to restaurants. Its monthly revenue for \(x\) wheels is \(R(x) = 120x - 0.4x^2\) EUR, and the price per wheel is \(\frac{R(x)}{x}\).” The sales manager says: “With every further wheel we sell, our price per wheel falls by 0.40 EUR.” Write this statement with a derivative.

    Check: Compute the derivative of \(\frac{R(x)}{x}\) once with the quotient rule and once after simplifying the fraction. Do both give \(-0.4\)?

  4. “A company builds tiny houses (small, fully equipped homes on a trailer). Its monthly profit for \(x\) houses is \(P(x) = -0.02x^3 + 1.5x^2 - 24x - 100\) thousand EUR.” An analyst says: “At 40 houses per month, the tangent line of the profit curve is horizontal.” Write this statement with the derivative.

    Check: Compute \(P'(40)\) and \(P(40)\). Is 40 the only output with a horizontal tangent line?

  5. “A new cinema in a shopping centre has a monthly revenue of \(R(t) = 0.25t^2 + 2t + 30\) thousand EUR in month \(t\) after its opening.” A report says: “In month 4, the tangent line of the revenue curve rises at an angle of about 76° against the \(t\)-axis.” The angle \(\alpha\) between a line with slope \(m\) and the \(t\)-axis is \(\alpha = \arctan(m)\). Write the report’s statement as an equation.

    Check: Compute \(R'(4)\) and the angle. The indicator D (degrees) must show on your calculator.

    Hint: SHIFT + tan gives \(\arctan\), just as SHIFT + sin gives \(\sin^{-1}\).

  1. A tangent line needs two things: the point \((100, R(100))\) and the slope \(R'(100)\).

    \[y - R(100) = R'(100) \cdot (x - 100)\]

    With \(R(100) = 9000 - 2500 = 6{,}500\) and \(R'(x) = 90 - 0.5x\), \(R'(100) = 40\):

    \[y - 6500 = 40(x - 100) \iff y = 40x + 2500\]

    Check: at \(x = 100\), the line gives \(4000 + 2500 = 6{,}500\) EUR ✓. The tangent line touches the graph of \(R\) at 100 participants.

    Typical mistake: writing \(y = 40x\). This line has the right slope, but at 100 participants it gives 4,000 EUR instead of 6,500 EUR: it misses the point of the graph.

  2. The estimate uses the tangent line at 30 shoots:

    \[P(32) \approx P(30) + P'(30) \cdot (32 - 30) = 1{,}320\]

    Check: \(P(30) = -1800 + 5400 - 2400 = 1{,}200\) EUR and \(P'(x) = -4x + 180\), so \(P'(30) = 60\) EUR per shoot. The estimate is \(1200 + 60 \cdot 2 = 1{,}320\) EUR ✓. Exactly, \(P(32) = -2048 + 5760 - 2400 = 1{,}312\) EUR. The error is \(1320 - 1312 = 8\) EUR.

    Typical mistake: \(P(32) \approx 1200 + 60 = 1{,}260\) EUR. The marginal profit is per shoot, and there are \(32 - 30 = 2\) more shoots, so it must be multiplied by 2.

  3. Price per wheel \(p(x) = \frac{R(x)}{x}\). “Falls by 0.40 EUR with every further wheel” means a constant negative slope:

    \[p'(x) = \left(\frac{R(x)}{x}\right)' = -0.4 \quad \text{for every } x\]

    Check with the quotient rule, \(u = R(x)\) with \(u' = 120 - 0.8x\) and \(v = x\) with \(v' = 1\):

    \[p'(x) = \frac{(120 - 0.8x) \cdot x - (120x - 0.4x^2) \cdot 1}{x^2} = \frac{120x - 0.8x^2 - 120x + 0.4x^2}{x^2} = \frac{-0.4x^2}{x^2} = -0.4 \quad \checkmark\]

    After simplifying: \(p(x) = \frac{120x - 0.4x^2}{x} = 120 - 0.4x\), so \(p'(x) = -0.4\) ✓. The manager is right.

    Typical mistake: taking \(R'(x) = 120 - 0.8x\) as the change of the price. \(R'(x)\) is the marginal revenue, the change of the total revenue. The price per wheel is \(\frac{R(x)}{x}\), and \(\left(\frac{u}{v}\right)' \neq \frac{u'}{v'}\).

  4. Horizontal tangent line means slope 0:

    \[P'(40) = 0\]

    Check: \(P'(x) = -0.06x^2 + 3x - 24\), so \(P'(40) = -96 + 120 - 24 = 0\) ✓, and \(P(40) = -1280 + 2400 - 960 - 100 = 60\) thousand EUR. But 40 is not the only output: \(P'(x) = 0 \iff x^2 - 50x + 400 = 0 \iff x = 10\) or \(x = 40\). At 10 houses, \(P(10) = -20 + 150 - 240 - 100 = -210\) thousand EUR: the tangent line is horizontal there, too, but the company makes a large loss. A horizontal tangent line alone does not tell you that the profit is high.

    Typical mistake: writing \(P(40) = 0\). This would say that the profit is zero at 40 houses (a break-even point), not that the graph is flat there.

  5. The slope of the tangent line in month 4 is \(R'(4)\):

    \[\arctan(R'(4)) \approx 76°\]

    Check: \(R'(t) = 0.5t + 2\), so \(R'(4) = 4\) and \(\alpha = \arctan(4) \approx 75.96° \approx 76°\) ✓.

    Typical mistake: using the value \(R(4) = 4 + 8 + 30 = 42\) instead of the slope: \(\arctan(42) \approx 88.64°\). A second mistake is the indicator R: then the calculator shows \(\arctan(4) \approx 1.3258\), an angle in radians.

Problem 7: One Paragraph, All Conditions (xx)

A small workshop builds wooden play kitchens for children. Its weekly total costs \(C\) in EUR are a polynomial of third degree in the number \(x\) of play kitchens, and its fixed costs are 400 EUR per week. At an output of 10 play kitchens, the marginal cost is 35 EUR per play kitchen. The tangent line of the cost curve at 20 play kitchens is the line \(y = 38x + 360\). The workshop sells every play kitchen for 80 EUR. The owner looks for the output at which the marginal revenue equals the marginal cost, and wants to know the average cost per play kitchen at 20 play kitchens.

List every condition and every function in this paragraph as a mathematical statement. Do not solve anything.

Check: An assistant writes only one condition for the tangent line: \(C'(20) = 38\). Name the missing condition and explain why a tangent line gives two equations.

  1. Cost function: \(C(x) = ax^3 + bx^2 + cx + d\) (\(x\) in play kitchens, \(C\) in EUR per week), so \(C'(x) = 3ax^2 + 2bx + c\)
  2. Fixed costs: \(C(0) = 400\), i.e. \(d = 400\)
  3. Marginal cost at 10 play kitchens: \(C'(10) = 35\)
  4. The tangent line touches the graph at \(x = 20\): \(C(20) = 38 \cdot 20 + 360 = 1{,}120\)
  5. The slope of the tangent line is the slope of \(C\) at \(x = 20\): \(C'(20) = 38\)
  6. Revenue: \(R(x) = 80x\), so \(R'(x) = 80\)
  7. Profit: \(P(x) = R(x) - C(x)\)
  8. Output wanted: \(R'(x) = C'(x)\), i.e. \(C'(x) = 80\)
  9. Average cost at 20 play kitchens: \(\frac{C(20)}{20}\)

Check: the missing condition is \(C(20) = 1{,}120\). The tangent line and the graph have the same point at \(x = 20\) (same value) and the same slope there. The slope gives \(C'(20) = 38\), the common point gives \(C(20) = 1{,}120\). Together with conditions 2 and 3, these are four equations for the four unknowns \(a\), \(b\), \(c\), \(d\).

Typical mistake: writing \(C(20) = 38\). 38 is the slope of the tangent line, not a cost. The value of \(C\) at 20 comes from inserting \(x = 20\) into the equation of the tangent line.

C: Applied

Problem 8: Profit of a Sunglasses Brand (xx)

A company designs sunglasses and sells them online. Its monthly profit is

\[P(x) = -0.2x^2 + 12x - 50\]

thousand EUR, where \(x\) is the number of sunglasses sold per month, in thousands.

  1. Compute the profit at 25,000 sunglasses per month.

  2. Determine the marginal profit \(P'(x)\) and compute \(P'(25)\). Interpret this value for the company.

(for verification: \(P'(x) = 12 - 0.4x\))

  1. Next month, the company expects to sell 26,000 sunglasses. Estimate the profit with the linear approximation at \(x = 25\). Compute the exact profit and the error of the estimate in EUR and in percent of the exact profit (round to two decimal places).

  2. The marketing department uses the same linear approximation to estimate the profit at 30,000 sunglasses. Compute the estimate, the exact profit and the error. Explain why the error is larger than in c) and why the estimate is too high.

  3. Determine the output at which the marginal profit is 0. Show with the vertex of \(P\) that the profit is maximal there and compute the maximum profit. Explain what this means for the estimate from d).

a) \(P(25) = -0.2 \cdot 625 + 300 - 50 = -125 + 300 - 50 = 125\). The profit is 125,000 EUR per month.

b) Power rule: \(P'(x) = -0.4x + 12\), so \(P'(25) = -10 + 12 = 2\).

At 25,000 sunglasses, the profit rises by about 2 thousand EUR per further thousand sunglasses: each further pair adds about 2 EUR to the profit. The marginal profit is positive, so selling more still pays off.

c) Linear approximation \(P(x) \approx P(a) + P'(a)(x - a)\) with \(a = 25\):

\[P(26) \approx P(25) + P'(25) \cdot (26 - 25) = 125 + 2 \cdot 1 = 127\]

Estimate: 127,000 EUR. Exact:

\[P(26) = -0.2 \cdot 676 + 312 - 50 = -135.2 + 312 - 50 = 126.8\]

The exact profit is 126,800 EUR. Error: \(127 - 126.8 = 0.2\) thousand EUR, i.e. 200 EUR, which is \(\frac{0.2}{126.8} \cdot 100\,\% \approx 0.16\,\%\) of the exact profit. Near \(x = 25\), the linear approximation is very good.

d) Estimate with the same tangent line:

\[P(30) \approx 125 + 2 \cdot (30 - 25) = 135\]

Exact: \(P(30) = -0.2 \cdot 900 + 360 - 50 = -180 + 360 - 50 = 130\). Error: \(135 - 130 = 5\) thousand EUR, i.e. 5,000 EUR, about \(\frac{5}{130} \cdot 100\,\% \approx 3.85\,\%\).

The linear approximation keeps the slope \(P'(25) = 2\) for the whole way from 25 to 30. But the marginal profit \(P'(x) = 12 - 0.4x\) falls as \(x\) grows: \(P'(26) = 1.6\), \(P'(28) = 0.8\), \(P'(30) = 0\). The profit grows more and more slowly, so the tangent line rises faster than the profit and the estimate is too high. The farther \(x\) is from 25, the longer the tangent line goes the wrong way, and the error grows.

e) \(P'(x) = 12 - 0.4x = 0 \iff x = 30\).

Vertex of \(P(x) = -0.2x^2 + 12x - 50\): \(a = -0.2 < 0\), so the parabola opens downward and its vertex is a maximum.

\[x_V = -\frac{12}{2 \cdot (-0.2)} = 30 \quad \checkmark\]

The maximum profit is \(P(30) = 130\) thousand EUR, i.e. 130,000 EUR per month at 30,000 sunglasses.

The estimate 135,000 EUR from d) is larger than the maximum profit the company can reach at all. A linear approximation is only reliable close to the point where the tangent line touches the graph.

Problem 9: Handmade Sneakers from a Small Manufactory (xx)

A small manufactory makes sneakers by hand and sells them in its own shop. Its weekly total costs for \(x\) pairs are

\[C(x) = 0.1x^3 - 3x^2 + 50x + 900 \text{ EUR}, \qquad 0 \leq x \leq 40\]

where 40 pairs per week is its capacity (the maximum possible output). To sell \(x\) pairs per week, the manufactory can ask the price \(p(x) = 130 - 2x\) EUR per pair.

  1. Give the revenue \(R(x)\) and the profit \(P(x)\). Determine the marginal cost, the marginal revenue and the marginal profit.

(for verification: \(P'(x) = -0.3x^2 + 2x + 80\))

  1. Determine the output at which the marginal revenue equals the marginal cost. Check that the marginal profit is 0 there.

  2. Compare the marginal revenue and the marginal cost at 10 pairs and at 30 pairs. Explain why the profit is highest at the output from b). Calculate the price and the maximum profit.

(for verification: 300 EUR per week)

  1. The landlord raises the rent of the workshop by 500 EUR per week. Explain without new calculations why the best output does not change. Compute the new highest profit and explain what it means for the manufactory.

a) Revenue = number of pairs · price:

\[R(x) = x(130 - 2x) = 130x - 2x^2\]

\[P(x) = R(x) - C(x) = 130x - 2x^2 - 0.1x^3 + 3x^2 - 50x - 900 = -0.1x^3 + x^2 + 80x - 900\]

Marginal functions with the power rule:

  • Marginal cost: \(C'(x) = 0.3x^2 - 6x + 50\)
  • Marginal revenue: \(R'(x) = 130 - 4x\)
  • Marginal profit: \(P'(x) = -0.3x^2 + 2x + 80\) (the same as \(R'(x) - C'(x)\) ✓)

b) \(MR = MC\):

\[130 - 4x = 0.3x^2 - 6x + 50 \iff 0 = 0.3x^2 - 2x - 80 \;\big|\; \cdot 10 \iff 3x^2 - 20x - 800 = 0\]

Quadratic formula with \(\Delta = (-20)^2 - 4 \cdot 3 \cdot (-800) = 400 + 9600 = 10{,}000\) and \(\sqrt{10{,}000} = 100\):

\[x_{1,2} = \frac{20 \pm 100}{6}, \qquad x_1 = 20, \quad x_2 = -\frac{40}{3}\]

\(x_2 < 0\) is no output. So \(MR = MC\) at 20 pairs per week.

Check: \(P'(20) = -0.3 \cdot 400 + 40 + 80 = -120 + 40 + 80 = 0\) ✓.

c) Marginal values:

Output \(MR\) \(MC\) Comparison
10 pairs \(130 - 40 = 90\) \(30 - 60 + 50 = 20\) \(MR > MC\)
30 pairs \(130 - 120 = 10\) \(270 - 180 + 50 = 140\) \(MR < MC\)

At 10 pairs, a further pair brings about 90 EUR more revenue and costs only about 20 EUR: the profit still grows, so the manufactory should make more. At 30 pairs, a further pair brings only about 10 EUR and costs about 140 EUR: the profit falls, so it should make fewer.

This holds on the whole interval. With the zeros from b), \(MR - MC = -0.3x^2 + 2x + 80 = -0.1(3x + 40)(x - 20)\). For \(0 \leq x < 20\), the factor \(x - 20\) is negative, so \(MR - MC > 0\). For \(x > 20\), it is positive, so \(MR - MC < 0\). The profit increases up to 20 pairs and decreases afterwards: it is highest at 20 pairs per week.

Price: \(p(20) = 130 - 40 = 90\) EUR per pair.

Maximum profit: \(P(20) = -0.1 \cdot 8000 + 400 + 1600 - 900 = -800 + 400 + 1600 - 900 = 300\) EUR per week.

d) The rent belongs to the fixed costs. The new costs are \(C_1(x) = C(x) + 500\), and the derivative of the constant 500 is 0: \(C_1'(x) = C'(x)\). The marginal cost and the marginal revenue do not change, so \(MR = MC\) still holds at 20 pairs, and the comparison from c) stays the same. The best output is still 20 pairs.

New highest profit: \(P_1(20) = P(20) - 500 = 300 - 500 = -200\) EUR per week.

The manufactory now makes a loss at every output. With 20 pairs per week, the loss is smallest (200 EUR per week). With this demand, it cannot make a profit anymore. It must lower its costs, for example with a cheaper workshop, or find customers who pay more.

Problem 10: Sales of a New Football Shirt (xxx)

A football club presents its new home shirt for the next season. The club’s online shop sells

\[S(t) = \frac{900t}{t^2 + 36}\]

hundred shirts per week, \(t\) weeks after the presentation (\(0 \leq t \leq 30\)).

  1. Show with the quotient rule that

\[S'(t) = \frac{900(36 - t^2)}{(t^2 + 36)^2}\]

  1. Compute \(S(2)\), \(S'(2)\), \(S(18)\) and \(S'(18)\) and interpret the four values with units. Compare week 2 with week 18.

    (for verification: \(S(2) = S(18) = 45\))

  2. A small change \(\Delta t\) changes the weekly sales by about \(\Delta S \approx S'(t) \cdot \Delta t\). Estimate the change of the weekly sales from \(t = 2\) to \(t = 2.5\) and from \(t = 18\) to \(t = 18.5\). Compare with the exact changes (round to two decimal places). Explain at which of the two times the weekly sales are more sensitive to time.

  3. Determine the time at which the rate of change of the weekly sales is zero. Explain with the sign of \(S'(t)\) before and after this time that the weekly sales are highest there, and compute the highest weekly sales.

    Hint: The denominator \((t^2 + 36)^2\) is always positive, so the numerator alone decides the sign of \(S'(t)\).

  4. The shop hires extra packing staff as long as it sells at least 6,000 shirts per week. Determine this period and its length.

    Hint: The denominator is positive, so you can multiply both sides by it without changing the sign.

a) \(u = 900t\), \(u' = 900\) and \(v = t^2 + 36\), \(v' = 2t\):

\[S'(t) = \frac{900(t^2 + 36) - 900t \cdot 2t}{(t^2 + 36)^2} = \frac{900t^2 + 32400 - 1800t^2}{(t^2 + 36)^2} = \frac{32400 - 900t^2}{(t^2 + 36)^2} = \frac{900(36 - t^2)}{(t^2 + 36)^2} \quad \checkmark\]

b) Values (\(S\) in hundred shirts per week, \(S'\) in hundred shirts per week, per week):

  • \(S(2) = \frac{1800}{40} = 45\): in week 2, the shop sells about 4,500 shirts per week.
  • \(S'(2) = \frac{900 \cdot 32}{40^2} = \frac{28800}{1600} = 18\): the weekly sales grow fast, by about 1,800 shirts per week within one week.
  • \(S(18) = \frac{16200}{360} = 45\): in week 18, the shop again sells about 4,500 shirts per week.
  • \(S'(18) = \frac{900 \cdot (36 - 324)}{360^2} = \frac{-259200}{129600} = -2\): the weekly sales fall slowly, by about 200 shirts per week within one week.

Both weeks have the same weekly sales. In week 2, the sales are still rising quickly after the presentation. In week 18, they are slowly falling.

c) Estimates with \(\Delta t = 0.5\):

  • From \(t = 2\): \(\Delta S \approx 18 \cdot 0.5 = 9\) hundred shirts per week (about 900 shirts more)
  • From \(t = 18\): \(\Delta S \approx -2 \cdot 0.5 = -1\) hundred shirts per week (about 100 shirts fewer)

Exact changes:

  • \(S(2.5) = \frac{2250}{6.25 + 36} = \frac{2250}{42.25} \approx 53.25\), so \(S(2.5) - S(2) \approx 8.25\)
  • \(S(18.5) = \frac{16650}{342.25 + 36} = \frac{16650}{378.25} \approx 44.02\), so \(S(18.5) - S(18) \approx -0.98\)

Both estimates are close to the exact changes. In week 2, half a week changes the weekly sales by about 900 shirts, in week 18 only by about 100 shirts. Shortly after the presentation, the weekly sales are about nine times as sensitive to time (\(|S'(2)| = 18\) against \(|S'(18)| = 2\)). A stock plan that is a few days off matters much more then.

d) \(S'(t) = 0 \iff 900(36 - t^2) = 0 \iff t^2 = 36 \iff t = 6\) (the solution \(t = -6\) is not in \([0; 30]\)).

The denominator is positive, so the sign of \(S'(t)\) is the sign of \(36 - t^2\):

  • For \(0 \leq t < 6\): \(t^2 < 36\), so \(S'(t) > 0\). The weekly sales increase.
  • For \(t > 6\): \(t^2 > 36\), so \(S'(t) < 0\). The weekly sales decrease.

The weekly sales rise until week 6 and fall afterwards, so they are highest at \(t = 6\):

\[S(6) = \frac{5400}{36 + 36} = \frac{5400}{72} = 75\]

The highest weekly sales are 7,500 shirts per week, 6 weeks after the presentation.

e) 6,000 shirts are 60 hundred shirts. The denominator \(t^2 + 36\) is positive, so multiplying by it keeps the direction of the inequality:

\[\frac{900t}{t^2 + 36} \geq 60 \iff 900t \geq 60t^2 + 2160 \iff 0 \geq 60t^2 - 900t + 2160 \;\big|\; : 60 \iff t^2 - 15t + 36 \leq 0\]

Zeros with the quadratic formula: \(\Delta = 225 - 144 = 81\), so \(t = \frac{15 \pm 9}{2}\), i.e. \(t_1 = 3\) and \(t_2 = 12\).

The parabola \(t^2 - 15t + 36\) opens upward, so it is \(\leq 0\) between its zeros: \(3 \leq t \leq 12\).

The shop needs the extra staff from week 3 to week 12 after the presentation, for 9 weeks. Check: \(S(3) = \frac{2700}{45} = 60\) ✓ and \(S(12) = \frac{10800}{180} = 60\) ✓.

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: Online Advertising of a Solar Installer [20 pts. total]

A regional company installs solar panels on the roofs of private homes. It advertises online. With a monthly advertising budget of \(a\) thousand EUR, it receives

\[N(a) = \frac{300a + 500}{a + 5}, \qquad 0 \leq a \leq 40\]

requests for a quote (a household asks for a price offer) per month. Round all results to two decimal places.

Part a)

Compute \(N(0)\) and interpret it. Investigate the asymptotic behavior of \(N\) for \(a \to \infty\) and explain its meaning for the company. [4 pts.]

Without advertising: \(N(0) = \frac{500}{5} = 100\). Without online advertising, the company still receives 100 requests per month (for example from recommendations and its website).

Asymptotic behavior: numerator and denominator both have degree 1. The horizontal asymptote is the ratio of the leading coefficients:

\[\lim_{a \to \infty} \frac{300a + 500}{a + 5} = \frac{300}{1} = 300, \qquad \text{horizontal asymptote } y = 300\]

From below: for every \(a \geq 0\), \(300a + 500 < 300a + 1500 = 300(a + 5)\), so \(N(a) < 300\). \(N\) approaches 300 from below and never reaches it.

Meaning: however much the company spends on advertising, it receives fewer than 300 requests per month. The region has only a limited number of households that want solar panels, so more and more money brings fewer and fewer additional requests.

Grading: 1 Pt. \(N(0) = 100\) with meaning, 1 Pt. degree comparison and limit 300, 1 Pt. \(N(a) < 300\) (approached from below), 1 Pt. meaning for the company.

Part b)

The management wants 250 requests per month. Determine the monthly advertising budget it needs. [2 pts.]

Equation: \(a + 5 > 0\), so multiply both sides by \(a + 5\):

\[\frac{300a + 500}{a + 5} = 250 \iff 300a + 500 = 250a + 1250 \iff 50a = 750 \iff a = 15\]

The company needs an advertising budget of 15,000 EUR per month. Check: \(N(15) = \frac{5000}{20} = 250\) ✓

Grading: 1 Pt. equation multiplied by \(a + 5\), 1 Pt. \(a = 15\), i.e. 15,000 EUR per month.

Part c)

Show with the quotient rule that \(N'(a) = \frac{1000}{(a + 5)^2}\). Explain why a higher budget always brings more requests. [4 pts.]

Parts: \(u = 300a + 500\), \(u' = 300\) and \(v = a + 5\), \(v' = 1\).

Quotient rule:

\[N'(a) = \frac{300(a + 5) - (300a + 500) \cdot 1}{(a + 5)^2} = \frac{300a + 1500 - 300a - 500}{(a + 5)^2} = \frac{1000}{(a + 5)^2} \quad \checkmark\]

Sign: the numerator 1000 is positive, and the denominator \((a + 5)^2\) is a square, positive for every \(a \geq 0\). So \(N'(a) > 0\) for every budget: \(N\) increases, and a higher budget always brings more requests.

Grading: 1 Pt. \(u\), \(u'\), \(v\), \(v'\), 1 Pt. quotient rule set up correctly, 1 Pt. simplified to \(\frac{1000}{(a + 5)^2}\), 1 Pt. \(N'(a) > 0\) with reason.

Part d)

The company currently spends 5,000 EUR per month on advertising. Determine the tangent of \(N\) at \(a = 5\) and explain its meaning. [4 pts.]

Point: \(N(5) = \frac{2000}{10} = 200\), so \((5, 200)\).

Slope: \(N'(5) = \frac{1000}{10^2} = 10\).

Point-slope form:

\[y - 200 = 10(a - 5) \iff y = 10a + 150\]

Meaning: close to the current budget of 5,000 EUR, the tangent line describes the requests as a straight line. Every further 1,000 EUR of advertising brings about 10 more requests per month, and 1,000 EUR less about 10 fewer. For budgets near 5,000 EUR, the company can plan with this simple line instead of \(N\).

Grading: 1 Pt. \(N(5) = 200\), 1 Pt. \(N'(5) = 10\), 1 Pt. \(y = 10a + 150\), 1 Pt. meaning.

Part e)

A line with slope \(m\) meets the \(a\)-axis at the angle \(\alpha = \arctan(m)\). Compute the angle of intersection \(\alpha\) between this tangent and the \(a\)-axis. Compute the angle for the tangent at the budget from b) as well, and explain what the smaller angle means for the company. [3 pts.]

Hint: The indicator D must show. SHIFT + tan gives \(\arctan\), just as SHIFT + sin gives \(\sin^{-1}\).

At 5,000 EUR: \(\alpha_1 = \arctan(10) \approx 84.29°\).

At 15,000 EUR: \(N'(15) = \frac{1000}{20^2} = \frac{1000}{400} = 2.5\), so \(\alpha_2 = \arctan(2.5) \approx 68.20°\).

Meaning: the tangent at 15,000 EUR is much flatter. There, a further 1,000 EUR of advertising brings only about 2.5 more requests per month instead of about 10. The additional effect of advertising becomes smaller as the budget grows, as the asymptote from a) suggests.

Grading: 1 Pt. \(\alpha_1 \approx 84.29°\), 1 Pt. \(N'(15) = 2.5\) and \(\alpha_2 \approx 68.20°\), 1 Pt. meaning.

Part f)

Estimate \(N(6)\) with the tangent from d) and give the error. A trainee uses the same tangent to find the budget for 250 requests per month. Compute the trainee’s budget, compare it with b) and explain the difference. [3 pts.]

Estimate: \(N(6) \approx 10 \cdot 6 + 150 = 210\). Exact: \(N(6) = \frac{2300}{11} \approx 209.09\). Error: \(210 - 209.09 = 0.91\) requests, very small.

Trainee: \(10a + 150 = 250 \iff a = 10\), i.e. 10,000 EUR per month. The correct budget from b) is 15,000 EUR: the trainee plans 5,000 EUR too little. With 10,000 EUR, the company would only get \(N(10) = \frac{3500}{15} \approx 233.33\) requests.

Reason: \(N'(a) = \frac{1000}{(a + 5)^2}\) gets smaller as \(a\) grows, because the denominator grows. The requests rise more and more slowly, but the tangent keeps the slope 10. Near \(a = 5\) this hardly matters, far away the tangent lies clearly above the graph of \(N\) and reaches 250 too early.

Grading: 1 Pt. estimate 210 and error about 0.91, 1 Pt. trainee’s budget 10,000 EUR compared with 15,000 EUR, 1 Pt. reason with the falling slope.

Calculator Practice

Problem 11: Calculator: Derivative Values (x)

Use the d/dx template (SHIFT + ∫) to compute the derivative at the given point. Round to four decimal places.

  1. \(f(x) = x^4 - 5x^2 + 3x\) at \(x = 2\)

  2. \(g(x) = \frac{x^2 + 3}{2x + 1}\) at \(x = 1\)

  3. \(h(x) = 5\sqrt{x} - 2x\) at \(x = 4\)

  4. Check your result from a) by hand with the power rule.

Press SHIFT + ∫, enter the function in the template, move the cursor to the \(x\)-value field, enter the value and press =.

  1. \(f'(2) = 15\)

  2. \(g'(1) = -\frac{2}{9} \approx -0.2222\)

  3. \(h'(4) = -0.75\)

  4. Power rule: \(f'(x) = 4x^3 - 10x + 3\), so \(f'(2) = 32 - 20 + 3 = 15\) ✓

Problem 12: Calculator: Rate of Change (xx)

A company’s monthly revenue \(t\) months after the launch of a new product is

\[R(t) = -0.1t^3 + 2.4t^2 + 24t + 150\]

thousand EUR, \(0 \leq t \leq 30\).

  1. Compute \(R'(5)\), \(R'(10)\) and \(R'(24)\) with the d/dx template and give the unit.

  2. Decide for each of the three months whether the revenue increases or decreases, and by about how much per month.

  3. The manager says: “In month 20, the revenue stops growing.” Check the statement with the d/dx template. The calculator only gives values, not the derivative function. Determine \(R'(t)\) by hand and show that \(t = 20\) is the only time in \([0; 30]\) with \(R'(t) = 0\).

  1. \(R'(5) = 40.5\), \(R'(10) = 42\), \(R'(24) = -33.6\), each in thousand EUR per month.

  • Month 5: \(R'(5) > 0\), the revenue increases by about 40,500 EUR per month.
  • Month 10: \(R'(10) > 0\), the revenue increases by about 42,000 EUR per month.
  • Month 24: \(R'(24) < 0\), the revenue decreases by about 33,600 EUR per month.
  1. The template gives \(R'(20) = 0\) ✓: in month 20, the tangent line of \(R\) is horizontal, and the revenue neither grows nor falls at that moment.

By hand: \(R'(t) = -0.3t^2 + 4.8t + 24\). Then

\[R'(t) = 0 \iff -0.3t^2 + 4.8t + 24 = 0 \;\big|\; : (-0.3) \iff t^2 - 16t - 80 = 0\]

Quadratic formula: \(\Delta = 256 + 320 = 576\), \(\sqrt{576} = 24\), so \(t = \frac{16 \pm 24}{2}\), i.e. \(t = 20\) or \(t = -4\). Only \(t = 20\) lies in \([0; 30]\).

Problem 13: Calculator: Checking Tangent Lines (xx)

Let \(f(x) = x^3 + \frac{4}{x}\). A colleague claims:

  • The tangent line of \(f\) at \(x = 2\) is \(y = 11x - 10\).
  • The tangent line of \(f\) at \(x = 1\) is \(y = -x + 6\).

  1. Compute \(f'(2)\) and \(f'(1)\) with the d/dx template.

  2. Check both claims: a tangent line must have the slope \(f'(a)\) and pass through the point \((a, f(a))\). Correct the wrong claim.

  1. \(f'(2) = 11\) and \(f'(1) = -1\). (By hand: \(f'(x) = 3x^2 - \frac{4}{x^2}\), so \(f'(2) = 12 - 1 = 11\) and \(f'(1) = 3 - 4 = -1\).)

  2. At \(x = 2\): the slope 11 is right. But \(f(2) = 8 + 2 = 10\), and the line gives \(11 \cdot 2 - 10 = 12 \neq 10\). The line misses the point \((2, 10)\), so the claim is wrong. Correct tangent line:

\[y - 10 = 11(x - 2) \iff y = 11x - 12\]

At \(x = 1\): the slope \(-1\) is right, and \(f(1) = 1 + 4 = 5\), while the line gives \(-1 + 6 = 5\) ✓. This claim is right.