Tasks 05-02 - The Derivative as Rate of Change

Section 05: Differential Calculus

Problem 1: Computing Derivatives from Definition (x)

Use the limit definition \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\) to find the derivative at the given point:

  1. \(f(x) = 5x - 3\) at \(a = 2\)

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

  3. \(f(x) = 4\) at \(a = 3\)

  4. Explain what the derivative value tells you about each function’s behavior at the given point.

  1. \(f(x) = 5x - 3\) at \(a = 2\):

\[\begin{align} f'(2) &= \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} \\ &= \lim_{h \to 0} \frac{[5(2+h) - 3] - [10 - 3]}{h} \\ &= \lim_{h \to 0} \frac{10 + 5h - 3 - 7}{h} \\ &= \lim_{h \to 0} \frac{5h}{h} \\ &= \lim_{h \to 0} 5 = 5 \end{align}\]

Result: \(f'(2) = 5\)

  1. \(f(x) = x^2 + 2x\) at \(a = 1\):

\[\begin{align} f'(1) &= \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \\ &= \lim_{h \to 0} \frac{[(1+h)^2 + 2(1+h)] - [1 + 2]}{h} \\ &= \lim_{h \to 0} \frac{[1 + 2h + h^2 + 2 + 2h] - 3}{h} \\ &= \lim_{h \to 0} \frac{4h + h^2}{h} \\ &= \lim_{h \to 0} \frac{h(4 + h)}{h} \\ &= \lim_{h \to 0} (4 + h) = 4 \end{align}\]

Result: \(f'(1) = 4\)

  1. \(f(x) = 4\) at \(a = 3\):

\[\begin{align} f'(3) &= \lim_{h \to 0} \frac{f(3+h) - f(3)}{h} \\ &= \lim_{h \to 0} \frac{4 - 4}{h} \\ &= \lim_{h \to 0} \frac{0}{h} = 0 \end{align}\]

Result: \(f'(3) = 0\)

  1. Interpretation:

    • Part a: \(f'(2) = 5\) means the function is increasing at a constant rate of 5 units. This makes sense because \(f(x) = 5x - 3\) is linear with slope 5.

    • Part b: \(f'(1) = 4\) means at \(x = 1\), the function is increasing at a rate of 4 units per unit change in x. The tangent line at \((1, 3)\) has slope 4.

    • Part c: \(f'(3) = 0\) means the function is not changing at all - it’s constant. The “tangent line” is horizontal. This makes sense because \(f(x) = 4\) is a horizontal line.

Problem 2: Average vs. Instantaneous Rate of Change (x)

A ball is thrown upward. Its height (in meters) after \(t\) seconds is given by: \[h(t) = -5t^2 + 20t + 2\]

  1. Compute the average rate of change of height from \(t = 1\) to \(t = 3\) seconds.

  2. Estimate the instantaneous rate of change at \(t = 2\) by calculating \(\frac{h(2.01) - h(2)}{0.01}\).

  3. Verify your answer from (b) by computing \(h'(2)\) using the limit definition.

  4. Interpret the meaning of \(h'(2)\) in the context of the ball’s motion.

  1. Average rate of change from t = 1 to t = 3:

    • \(h(1) = -5(1)^2 + 20(1) + 2 = 17\) meters
    • \(h(3) = -5(9) + 20(3) + 2 = -45 + 60 + 2 = 17\) meters
    • \(\text{ARC} = \frac{h(3) - h(1)}{3 - 1} = \frac{17 - 17}{2} = 0\) m/s

    Interpretation: On average, between t = 1 and t = 3, the ball returns to the same height, so the average velocity is zero.

  2. Estimate using small interval:

    • \(h(2) = -5(4) + 20(2) + 2 = -20 + 40 + 2 = 22\) meters
    • \(h(2.01) = -5(2.01)^2 + 20(2.01) + 2 = -5(4.0401) + 40.2 + 2 = -20.2005 + 40.2 + 2 = 21.9995\) meters
    • \(h'(2) \approx \frac{21.9995 - 22}{0.01} = \frac{-0.0005}{0.01} = -0.05\) m/s

  3. Computing \(h'(2)\) using the limit definition:

\[\begin{align} h'(2) &= \lim_{h \to 0} \frac{h(2+h) - h(2)}{h} \\ &= \lim_{h \to 0} \frac{[-5(2+h)^2 + 20(2+h) + 2] - 22}{h} \\ &= \lim_{h \to 0} \frac{[-5(4 + 4h + h^2) + 40 + 20h + 2] - 22}{h} \\ &= \lim_{h \to 0} \frac{[-20 - 20h - 5h^2 + 40 + 20h + 2] - 22}{h} \\ &= \lim_{h \to 0} \frac{22 - 5h^2 - 22}{h} \\ &= \lim_{h \to 0} \frac{-5h^2}{h} \\ &= \lim_{h \to 0} (-5h) = 0 \end{align}\]

Result: \(h'(2) = 0\) m/s

  1. Interpretation:
    • At \(t = 2\) seconds, the instantaneous velocity is 0 m/s
    • This means the ball has reached its maximum height and momentarily stops before falling back down
    • This is the turning point of the parabolic trajectory
    • After this instant, the ball will begin descending (velocity becomes negative)

The graph shows the parabolic trajectory of the ball, with height on the y-axis and time on the x-axis. The curve reaches its maximum height of 22 meters at t = 2 seconds, marked with a point and a horizontal dashed line. The derivative h’(2) = 0 is annotated at the peak, confirming the momentary zero velocity at maximum height.

Problem 3: Marginal Cost Analysis (xx)

A manufacturer’s total cost function (in euros) for producing \(x\) units is: \[C(x) = 500 + 10x + 0.05x^2\]

  1. Determine the average cost per unit when producing 100 units.

  2. Compute \(C'(100)\) by calculating \(C(101) - C(100)\).

  3. Investigate the behavior of marginal cost: Is it increasing, decreasing, or constant? Calculate \(C'(50)\) and \(C'(150)\) using the approximation method.

  4. Argue whether the company should expand from 100 to 101 units if they can sell each unit for €25.

  1. Average cost per unit at x = 100:

    • Total cost: \(C(100) = 500 + 10(100) + 0.05(100)^2 = 500 + 1000 + 500 = 2000\) euros
    • Average cost: \(AC(100) = \frac{C(100)}{100} = \frac{2000}{100} = 20\) euros/unit

  2. Marginal cost at x = 100:

    • \(C(100) = 2000\) euros
    • \(C(101) = 500 + 10(101) + 0.05(101)^2 = 500 + 1010 + 510.05 = 2020.05\) euros
    • \(C'(100) \approx C(101) - C(100) = 2020.05 - 2000 = 20.05\) euros/unit

    Interpretation: The 101st unit costs approximately €20.05 to produce.

  3. Marginal cost at different production levels:

    At x = 50:

    • \(C(50) = 500 + 500 + 125 = 1125\) euros
    • \(C(51) = 500 + 510 + 130.05 = 1140.05\) euros
    • \(C'(50) \approx 1140.05 - 1125 = 15.05\) euros/unit

    At x = 150:

    • \(C(150) = 500 + 1500 + 1125 = 3125\) euros
    • \(C(151) = 500 + 1510 + 1140.05 = 3150.05\) euros
    • \(C'(150) \approx 3150.05 - 3125 = 25.05\) euros/unit

    Comparison: | Production Level | Marginal Cost | |—————–|—————| | 50 units | €15.05/unit | | 100 units | €20.05/unit | | 150 units | €25.05/unit |

    Conclusion: Marginal cost is increasing as production increases. This reflects diseconomies of scale - each additional unit becomes more expensive to produce at higher production levels.

  4. Decision analysis for expanding to 101 units:

    • Revenue from 101st unit: €25
    • Cost of 101st unit: €20.05 (marginal cost)
    • Marginal profit: \(25 - 20.05 = €4.95\)

    Recommendation: YES, expand to 101 units

    • The company makes a profit of €4.95 on the 101st unit
    • Since \(MR (€25) > MC (€20.05)\), producing more increases total profit
    • Continue expanding until marginal cost equals marginal revenue (€25)

Two side-by-side graphs are shown. Left: The total cost function is a convex upward-curving parabola starting at 500, with a marked point at x = 100. Right: The marginal cost is a straight line MC(x) = 10 + 0.1x (increasing linearly), intersecting a horizontal dashed line at price = 25 euros. The intersection point at x = 150 units is labeled as the optimal production level.

Note: The optimal production is 150 units, where MC = Price = €25.

Problem 4: Revenue and Marginal Revenue (xx)

A company sells widgets. The demand function is \(p = 80 - 0.4x\), where \(p\) is the price per widget (in euros) and \(x\) is the quantity demanded.

  1. Show that the revenue function is \(R(x) = 80x - 0.4x^2\).

  2. Compute \(R'(75)\) using the approximation \(R(76) - R(75)\).

  3. Assess whether the company should increase or decrease production if they are currently producing 75 widgets. Substantiate your answer.

  4. Decide: At what production level is revenue maximized? (Hint: Find where \(R'(x) = 0\) by using the fact that \(R'(x) = 80 - 0.8x\).)

  1. Revenue function derivation:

    • Revenue = Price × Quantity
    • \(R(x) = p \cdot x = (80 - 0.4x) \cdot x\)
    • \(R(x) = 80x - 0.4x^2\) ✓

  2. Marginal revenue at x = 75:

    • \(R(75) = 80(75) - 0.4(75)^2 = 6000 - 0.4(5625) = 6000 - 2250 = 3750\) euros
    • \(R(76) = 80(76) - 0.4(76)^2 = 6080 - 0.4(5776) = 6080 - 2310.4 = 3769.6\) euros
    • \(R'(75) \approx R(76) - R(75) = 3769.6 - 3750 = 19.6\) euros/unit

    Interpretation: The 76th widget will generate approximately €19.60 in additional revenue.

  3. Production decision at x = 75:

    • Marginal revenue: \(R'(75) = 19.6\) euros/unit (positive)
    • Since \(R'(75) > 0\), revenue is still increasing

    Recommendation: Increase production beyond 75 units

    • Each additional widget (near 75) adds approximately €19.60 to total revenue
    • Continue increasing production until marginal revenue becomes zero
    • At that point, revenue will be maximized

  4. Revenue maximization:

    • Given: \(R'(x) = 80 - 0.8x\)
    • Set \(R'(x) = 0\): \[80 - 0.8x = 0\] \[0.8x = 80\] \[x = 100 \text{ widgets}\]

    Verification:

    • \(R(100) = 80(100) - 0.4(100)^2 = 8000 - 4000 = 4000\) euros
    • This is indeed the maximum revenue

    Check with nearby values:

    • \(R(99) = 80(99) - 0.4(9801) = 7920 - 3920.4 = 3999.6\) euros
    • \(R(101) = 80(101) - 0.4(10201) = 8080 - 4080.4 = 3999.6\) euros
    • Both are less than \(R(100) = 4000\), confirming the maximum

Two side-by-side graphs are shown. Left: The total revenue function is a downward-opening parabola with maximum at x = 100 widgets (revenue = 4000 euros), marked with a point and dashed vertical line. Right: The marginal revenue is a decreasing straight line MR(x) = 80 - 0.8x that crosses zero at x = 100, confirming that revenue is maximized where MR = 0.

Summary: Produce 100 widgets to maximize revenue at €4,000.

Problem 5: Profit Optimization (xx)

A small business has the following functions:

  • Cost: \(C(x) = 200 + 8x + 0.02x^2\)
  • Revenue: \(R(x) = 50x - 0.3x^2\)

where \(x\) is the number of items produced and sold.

  1. Determine the profit function \(P(x) = R(x) - C(x)\).

  2. Compute the marginal profit at \(x = 40\) using \(P(41) - P(40)\).

  3. Give the production level that maximizes profit using the formula \(P'(x) = R'(x) - C'(x)\) and the fact that:

    • \(R'(x) = 50 - 0.6x\)
    • \(C'(x) = 8 + 0.04x\)

  4. Verify that this production level gives \(MR = MC\).

  5. Graph the profit function and mark the optimal production point.

  1. Profit function: \[\begin{align} P(x) &= R(x) - C(x) \\ &= (50x - 0.3x^2) - (200 + 8x + 0.02x^2) \\ &= 50x - 0.3x^2 - 200 - 8x - 0.02x^2 \\ &= 42x - 0.32x^2 - 200 \end{align}\]

  2. Marginal profit at x = 40:

    • \(P(40) = 42(40) - 0.32(1600) - 200 = 1680 - 512 - 200 = 968\) euros
    • \(P(41) = 42(41) - 0.32(1681) - 200 = 1722 - 537.92 - 200 = 984.08\) euros
    • \(P'(40) \approx P(41) - P(40) = 984.08 - 968 = 16.08\) euros/unit

    Interpretation: The 41st item adds approximately €16.08 to profit.

  3. Optimal production level:

    • Marginal profit: \(P'(x) = R'(x) - C'(x) = (50 - 0.6x) - (8 + 0.04x)\)
    • \(P'(x) = 50 - 0.6x - 8 - 0.04x = 42 - 0.64x\)

    Set \(P'(x) = 0\): \[42 - 0.64x = 0\] \[0.64x = 42\] \[x = \frac{42}{0.64} = 65.625\]

    Optimal production: Approximately 66 items (rounding to nearest whole unit)

  4. Verification that MR = MC at x ≈ 66:

    • \(MR(66) = 50 - 0.6(66) = 50 - 39.6 = 10.4\) euros/unit
    • \(MC(66) = 8 + 0.04(66) = 8 + 2.64 = 10.64\) euros/unit

    These are approximately equal (small difference due to rounding). Let’s check at the exact value x = 65.625:

    • \(MR(65.625) = 50 - 0.6(65.625) = 50 - 39.375 = 10.625\) euros/unit
    • \(MC(65.625) = 8 + 0.04(65.625) = 8 + 2.625 = 10.625\) euros/unit

    Verification: \(MR = MC = €10.625/\text{unit}\) ✓

  5. Graph of profit function:

The graph shows the profit function as a downward-opening parabola. The maximum profit of approximately 878.53 euros is marked at x = 65.6 units with a dashed vertical line. Two break-even points (where the curve crosses the x-axis) are marked at approximately 5.2 and 126.2 units, showing the profitable production range.

Summary:

  • Maximum profit: €878.53 at 65.625 units
  • Break-even at approximately 5.2 and 126.2 units
  • Profitable range: between 5.2 and 126.2 units
  • Optimal: Produce 66 units for maximum profit

Problem 6: Advanced Derivative Computation (xxx)

  1. Use the limit definition to find \(f'(x)\) as a function of \(x\) (not just at a specific point) for \(f(x) = x^2 + 3x\).

  2. Show that your answer from (a) satisfies the property that \(f'(2) = 7\) by direct evaluation.

  3. Investigate: For what value of \(x\) is the tangent line to \(f\) horizontal? Give the point on the curve where this occurs.

  4. Verify by computing \(f'(1)\) and finding the equation of the tangent line at \((1, f(1))\).

  1. Finding \(f'(x)\) as a function:

\[\begin{align} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{[(x+h)^2 + 3(x+h)] - [x^2 + 3x]}{h} \\ &= \lim_{h \to 0} \frac{[x^2 + 2xh + h^2 + 3x + 3h] - [x^2 + 3x]}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2 + 3h}{h} \\ &= \lim_{h \to 0} \frac{h(2x + h + 3)}{h} \\ &= \lim_{h \to 0} (2x + h + 3) \\ &= 2x + 3 \end{align}\]

Result: \(f'(x) = 2x + 3\)

  1. Verification at x = 2:

    • \(f'(2) = 2(2) + 3 = 4 + 3 = 7\) ✓

    This matches the claim that \(f'(2) = 7\).

  2. Finding where tangent is horizontal:

    • Horizontal tangent means slope = 0

    • Set \(f'(x) = 0\): \[2x + 3 = 0\] \[x = -\frac{3}{2}\]

    • Point on curve: \[f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) = \frac{9}{4} - \frac{9}{2} = \frac{9}{4} - \frac{18}{4} = -\frac{9}{4}\]

    Answer: The tangent line is horizontal at the point \(\left(-\frac{3}{2}, -\frac{9}{4}\right)\)

    This is the vertex of the parabola (minimum point).

  3. Tangent line at (1, f(1)):

    • Point: \((1, f(1)) = (1, 1 + 3) = (1, 4)\)
    • Slope: \(f'(1) = 2(1) + 3 = 5\)

    Tangent line equation using point-slope form: \[y - y_1 = m(x - x_1)\] \[y - 4 = 5(x - 1)\] \[y - 4 = 5x - 5\] \[y = 5x - 1\]

    Tangent line equation: \(y = 5x - 1\)

The graph shows the parabola f(x) = x^2 + 3x with two tangent lines. A horizontal tangent line (dashed) touches the curve at the vertex (-1.5, -2.25), which is the minimum point. A second tangent line y = 5x - 1 (dashed) touches the curve at the point (1, 4) with slope 5.

Problem 7: Comprehensive Business Optimization (xxx)

A tech startup develops mobile apps. Their analysis shows:

  • Development cost: \(C(x) = 5000 + 800x + 20x^2\) (in euros)
  • Revenue model: \(R(x) = 2000x - 15x^2\) (in euros)

where \(x\) is the number of apps developed per month.

Part A: Function Analysis

  1. Determine the profit function \(P(x)\).

  2. Compute the average cost per app when developing 20 apps per month.

  3. Compute the marginal cost and marginal revenue at \(x = 20\) using both the discrete difference and the derivative.

Part B: Optimization

  1. Decide whether the company should increase or decrease production from 20 apps. Substantiate mathematically.

  2. Investigate the optimal production level by finding where \(P'(x) = 0\), using:

    • \(R'(x) = 2000 - 30x\)
    • \(C'(x) = 800 + 40x\)

  3. Give the maximum profit and verify that \(MR = MC\) at this production level.

Part C: Strategic Analysis

  1. Argue whether it’s worth developing apps if the startup can only manage 10 apps per month.

  2. Assess the break-even points (where \(P(x) = 0\)) and interpret their business significance.

Part A: Function Analysis

  1. Profit function: \[\begin{align} P(x) &= R(x) - C(x) \\ &= (2000x - 15x^2) - (5000 + 800x + 20x^2) \\ &= 2000x - 15x^2 - 5000 - 800x - 20x^2 \\ &= 1200x - 35x^2 - 5000 \end{align}\]

  2. Average cost at x = 20:

    • Total cost: \(C(20) = 5000 + 800(20) + 20(400) = 5000 + 16000 + 8000 = 29000\) euros
    • Average cost: \(AC(20) = \frac{29000}{20} = 1450\) euros/app

  3. Marginal values at x = 20:

    Marginal Cost:

    • \(C(20) = 29000\) euros
    • \(C(21) = 5000 + 800(21) + 20(441) = 5000 + 16800 + 8820 = 30620\) euros
    • \(MC(20) = C(21) - C(20) = 30620 - 29000 = 1620\) euros/app

    Using the derivative: \(C'(20) = 800 + 40(20) = 800 + 800 = 1600\) euros/app

    Marginal Revenue:

    • \(R(20) = 2000(20) - 15(400) = 40000 - 6000 = 34000\) euros
    • \(R(21) = 2000(21) - 15(441) = 42000 - 6615 = 35385\) euros
    • \(MR(20) = R(21) - R(20) = 35385 - 34000 = 1385\) euros/app

    Using the derivative: \(R'(20) = 2000 - 30(20) = 2000 - 600 = 1400\) euros/app ✓

Part B: Optimization

  1. Production decision at x = 20:

    • \(MR(20) = 1400\) euros/app
    • \(MC(20) = 1600\) euros/app
    • \(MP(20) = MR - MC = 1400 - 1600 = -200\) euros/app

    Decision: DECREASE production

    • The 21st app costs more to develop (€1600) than it generates in revenue (€1400)
    • Marginal profit is negative (−€200), so producing the 21st app reduces total profit
    • The company is past the optimal production level

  2. Optimal production level:

    • \(P'(x) = R'(x) - C'(x) = (2000 - 30x) - (800 + 40x)\)
    • \(P'(x) = 2000 - 30x - 800 - 40x = 1200 - 70x\)

    Set \(P'(x) = 0\): \[1200 - 70x = 0\] \[70x = 1200\] \[x = \frac{1200}{70} = \frac{120}{7} \approx 17.14 \text{ apps}\]

    Optimal: Develop 17 apps per month (rounding down to whole units)

  3. Maximum profit and verification:

    • Using \(x = 120/7 \approx 17.14\):
    • \(P(17.14) = 1200(17.14) - 35(17.14)^2 - 5000\)
    • \(= 20568 - 35(293.78) - 5000\)
    • \(= 20568 - 10282.3 - 5000 = 5285.7\) euros

    Verification that MR = MC:

    • \(MR(17.14) = 2000 - 30(17.14) = 2000 - 514.2 = 1485.8\) euros/app
    • \(MC(17.14) = 800 + 40(17.14) = 800 + 685.6 = 1485.6\) euros/app

    These are equal (small difference due to rounding), confirming optimal production. ✓

Part C: Strategic Analysis

  1. Profitability at 10 apps/month:

    • \(P(10) = 1200(10) - 35(100) - 5000 = 12000 - 3500 - 5000 = 3500\) euros

    Analysis:

    • At 10 apps, the company makes €3,500 profit
    • This is profitable, but not optimal
    • They should scale up to 17 apps to maximize profit
    • However, if capacity constraints limit them to 10, it’s still worth operating
    • Marginal profit at 10: \(P'(10) = 1200 - 700 = 500\) euros/app (still positive!)

  2. Break-even analysis:

    Solve \(P(x) = 0\): \[1200x - 35x^2 - 5000 = 0\] \[-35x^2 + 1200x - 5000 = 0\] \[35x^2 - 1200x + 5000 = 0\] \[7x^2 - 240x + 1000 = 0\]

    Using quadratic formula: \[x = \frac{240 \pm \sqrt{240^2 - 4(7)(1000)}}{2(7)} = \frac{240 \pm \sqrt{57600 - 28000}}{14}\] \[= \frac{240 \pm \sqrt{29600}}{14} = \frac{240 \pm 172.05}{14}\]

    • \(x_1 = \frac{240 + 172.05}{14} = \frac{412.05}{14} \approx 29.4\) apps
    • \(x_2 = \frac{240 - 172.05}{14} = \frac{67.95}{14} \approx 4.9\) apps

    Interpretation:

    • First break-even: 4.9 apps — Below this, the company loses money (fixed costs dominate)
    • Second break-even: 29.4 apps — Above this, the company loses money again (costs grow faster than revenue)
    • Profitable range: Between 5 and 29 apps per month
    • Optimal within range: 17 apps (maximum profit point)

Two side-by-side graphs are shown. Left: The profit function is a downward-opening parabola with maximum at approximately 17.1 apps (about 5286 euros), with break-even points at 4.9 and 29.4 apps. Right: Revenue and cost curves are plotted together, with the revenue curve (downward parabola) intersecting the cost curve (upward parabola) at the two break-even points. The gap between them represents profit.

Strategic Recommendations: 1. Optimal: Develop 17 apps/month for maximum profit (€5,286) 2. Viable range: 5-29 apps/month (profitable zone) 3. Current situation (20 apps): Still profitable but past optimal; reduce to 17 4. Minimum viable: Must develop at least 5 apps to cover fixed costs 5. Upper limit: Don’t exceed 29 apps - costs outweigh revenue beyond this point