Differentiation Applications - IB Mathematics Applications and Interpretation

1. What is the maximum possible revenue from ticket sales?
   Answer: $200,000 Revenue $R$ is price times quantity: $R(q) = p \cdot q = (400 - 0.2q)q = 400q - 0.2q^2$. To maximize revenue, find the derivative: $R'(q) = 400 - 0.4q$. Set $R'(q)=0$ to find the critical point: $400 - 0.4q = 0 \implies q = 1000$. The maximum revenue is $R(1000) = 400(1000) - 0.2(1000)^2 = 400,000 - 200,000 = 200,000$.
2. What is the minimum length of fencing required, to the nearest meter?
   Answer: 200 m This is a classic optimization problem where you need to minimize perimeter given a fixed area constraint. When you see problems about minimizing fencing or materials, think calculus optimization or recognize standard geometric relationships. Let's set up the problem systematically. If the field has width $w$ (perpendicular to the river) and length $l$ (parallel to the river), then the area constraint gives us $l \times w = 5000$, so $l = \frac{5000}{w}$. Since the river provides one long side, you only need fencing for two widths and one length: $F = 2w + l = 2w + \frac{5000}{w}$. To minimize this, take the derivative and set it equal to zero: $\frac{dF}{dw} = 2 - \frac{5000}{w^2} = 0$. Solving: $2 = \frac{5000}{w^2}$, so $w^2 = 2500$ and $w = 50$ meters. This gives $l = \frac{5000}{50} = 100$ meters, and minimum fencing of $F = 2(50) + 100 = 200$ meters. Answer A (100 m) represents just the length parallel to the river, forgetting the two perpendicular sides. Answer B (141 m) might come from incorrectly assuming a square field ($\sqrt{5000} \approx 71$, then $2 \times 71 = 142$). Answer C (250 m) could result from using the wrong optimization setup or assuming all four sides need fencing. Strategy tip: In three-sided fencing problems, the optimal rectangle always has the unfenced side twice as long as each fenced side. Memorizing this 2:1 ratio can save time on similar problems.
3. What speed, in knots, will minimize the total cost per nautical mile?
   Answer: 20.0 Let $C_f$ be fuel cost and $v$ be speed. $C_f = kv^3$. Given $1350 = k(15^3)$, so $k = 1350/3375 = 0.4$. Total cost per hour is $C_h(v) = 0.4v^3 + 6400$. The cost per nautical mile, $C_{nm}$, is the cost per hour divided by the speed (miles per hour): $C_{nm}(v) = \frac{0.4v^3 + 6400}{v} = 0.4v^2 + \frac{6400}{v}$. To minimize this, find the derivative: $C'_{nm}(v) = 0.8v - \frac{6400}{v^2}$. Set to zero: $0.8v = \frac{6400}{v^2} \implies v^3 = \frac{6400}{0.8} = 8000$. Thus, $v = \sqrt[3]{8000} = 20$ knots.
4. At what time of day, approximately, is the temperature the lowest?
   Answer: 4:00 To find the minimum temperature, we need to find the derivative of $T(t)$ and find its roots. $T'(t) = 0.08t^3 - 2.7t^2 + 24t - 40$. Using a graphing calculator to solve $T'(t)=0$, we find critical points at approximately $t \approx 2.1$, $t \approx 4.0$, and $t \approx 27.6$. Since we're restricted to $0 \le t \le 24$, we evaluate $T(t)$ at the critical points and endpoints: $T(0) = 50$, $T(2.1) \approx 18.3$, $T(4.0) \approx 18.0$, $T(24) \approx 166$. The minimum temperature occurs at $t \approx 4.0$, which corresponds to 4:00.
5. What is the displacement of the particle when it first comes to a momentary rest?
   Answer: 9 m The particle is at rest when its velocity is zero. Velocity is the derivative of displacement: $v(t) = s'(t) = 3t^2 - 18t + 15$. Set $v(t) = 0$: $3(t^2 - 6t + 5) = 0 \implies 3(t-1)(t-5) = 0$. The particle is at rest at $t=1$ and $t=5$. It first comes to rest at $t=1$ second. The displacement at this time is $s(1) = (1)^3 - 9(1)^2 + 15(1) + 2 = 1 - 9 + 15 + 2 = 9$ meters.
6. How many smartwatches should the company produce daily to maximize its profit?
   Answer: 650 Profit $P(x)$ is Revenue minus Cost: $P(x) = R(x) - C(x) = (-0.5x^2 + 800x) - (150x + 20000) = -0.5x^2 + 650x - 20000$. To find the maximum profit, we find the derivative of $P(x)$ and set it to zero. $P'(x) = -x + 650$. Setting $P'(x) = 0$ gives $x = 650$. The second derivative $P''(x) = -1 < 0$, confirming this is a maximum.
7. At what time $t$ is the population's rate of growth at a maximum?
   Answer: 15 days The rate of growth is the derivative of the population function, $R(t) = P'(t) = -3t^2 + 90t + 48$. To find when this rate is at a maximum, we need to find the derivative of $R(t)$ and set it to zero. $R'(t) = P''(t) = -6t + 90$. Setting $R'(t) = 0$ gives $-6t + 90 = 0$, which solves to $t=15$. This corresponds to the point of inflection of $P(t)$, where the gradient is steepest.
8. What should be the height of the can, in cm, to minimize the surface area, correct to one decimal place?
   Answer: 8.6 cm This is a classic optimization problem where you need to minimize surface area while maintaining a fixed volume constraint. When you see "minimize material" for containers, you're looking for the point where the derivative equals zero. Start by setting up your equations. For a cylinder with radius $r$ and height $h$: Volume = $\pi r^2 h = 500$, so $h = \frac{500}{\pi r^2}$. Surface area = $2\pi r^2 + 2\pi rh$ (top/bottom plus sides). Substitute the height expression into the surface area formula: $A = 2\pi r^2 + 2\pi r \cdot \frac{500}{\pi r^2} = 2\pi r^2 + \frac{1000}{r}$. To minimize, take the derivative and set it equal to zero: $\frac{dA}{dr} = 4\pi r - \frac{1000}{r^2} = 0$. Solving: $4\pi r = \frac{1000}{r^2}$, which gives $r^3 = \frac{250}{\pi}$, so $r = \sqrt[3]{\frac{250}{\pi}} ≈ 4.3$ cm. Now find the height: $h = \frac{500}{\pi(4.3)^2} ≈ 8.6$ cm. Looking at the wrong answers: B (5.4 cm) might come from incorrectly using the radius value as the height. C (4.3 cm) is actually the radius, not the height—a common mix-up. D (10.8 cm) could result from calculation errors in the optimization process. For optimization problems, always double-check which variable the question asks for. The minimum surface area occurs when height equals the diameter, but make sure you're reporting the right measurement.
9. What is the radius of the can, in cm, that minimizes the total cost of the material, correct to two decimal places?
   Answer: 4.57 Let radius be $r$ and height be $h$. Volume $V = \pi r^2 h = 1000$, so $h = \frac{1000}{\pi r^2}$. The cost function is $C(r) = 0.05(2\pi r^2) + 0.03(2\pi rh)$. Substituting for $h$, we get $C(r) = 0.1\pi r^2 + 0.06\pi r (\frac{1000}{\pi r^2}) = 0.1\pi r^2 + \frac{60}{r}$. To minimize cost, find the derivative: $C'(r) = 0.2\pi r - \frac{60}{r^2}$. Set $C'(r)=0$, so $0.2\pi r = \frac{60}{r^2}$, which gives $r^3 = \frac{60}{0.2\pi} = \frac{300}{\pi}$. Solving for $r$ gives $r = \sqrt[3]{\frac{300}{\pi}} \approx 4.57$ cm.
10. What production level, in thousands of units, maximizes the profit within the factory's capacity?
    Answer: 15.3 To find the maximum profit, we differentiate $P(x)$: $P'(x) = -3x^2 + 54x - 120$. We set $P'(x)=0$ and solve for $x$ using a GDC or the quadratic formula. The solutions are $x \approx 2.74$ and $x \approx 15.26$. We check the second derivative, $P''(x) = -6x + 54$. $P''(2.74) > 0$ (local minimum). $P''(15.26) < 0$ (local maximum). We must also check the endpoints of the domain [0, 20]. $P(0) = -50$, $P(15.26) \approx 485$, $P(20) = 430$. The local maximum at $x \approx 15.26$ is the global maximum on the interval. The closest answer is 15.3.
11. When is the value of the shares decreasing most rapidly?
    Answer: t = 6 months The rate of change of the value is $V'(t) = 0.3t^2 - 3.6t + 7.2$. The value is decreasing most rapidly when this rate is at its minimum. To find the minimum of $V'(t)$, we find its derivative, $V''(t)$, and set it to zero. $V''(t) = 0.6t - 3.6$. Setting $V''(t) = 0$ gives $0.6t = 3.6$, so $t = 6$. This is the vertex of the parabola $V'(t)$, which is its minimum point. Thus, the value decreases most rapidly at $t=6$ months.
12. Which of the following is the best interpretation of the statement $V'(10) = -250$?
    Answer: At the 10-week mark, the value of the cryptocurrency was decreasing at a rate of $250 per week. When you encounter derivative notation like $V'(10) = -250$, you're dealing with instantaneous rates of change. The derivative $V'(t)$ tells you how fast the value is changing at any specific moment in time, not cumulative changes over a period. Let's break down what $V'(10) = -250$ means: at exactly $t = 10$ weeks, the derivative (rate of change) of the cryptocurrency's value is $-250$ dollars per week. The negative sign indicates the value is decreasing, and the magnitude tells you the rate of that decrease. This matches option D perfectly. Now let's see why the other options miss the mark. Option A confuses the derivative with the original function—$V'(10)$ is not the same as $V(10)$. The derivative gives you the rate of change, not the actual value. Option B misinterprets the derivative as a total change over the entire 10th week, but derivatives measure instantaneous rates, not cumulative changes over time intervals. Option C suggests this represents the total accumulated loss after 10 weeks, which would require integrating the derivative function, not just evaluating it at one point. Remember this key distinction: $V(t)$ gives you the value at time $t$, while $V'(t)$ gives you how fast that value is changing at time $t$. On IB exams, questions often test whether you can distinguish between a function's value and its rate of change—always check whether you're looking at $f(x)$ or $f'(x)$.
13. Find the approximate number of units that should be produced to minimize the average cost per unit.
    Answer: 225 The average cost per unit is $A(x) = \frac{C(x)}{x} = 0.01x^2 - 4.5x + 800 + \frac{12000}{x}$. To minimize the average cost, we find its derivative and set it to zero: $A'(x) = 0.02x - 4.5 - \frac{12000}{x^2}$. Setting $A'(x) = 0$ gives $0.02x - 4.5 = \frac{12000}{x^2}$, which leads to the cubic equation $0.02x^3 - 4.5x^2 - 12000 = 0$. Using a graphing calculator's solver function, we find the positive real root is $x \approx 225.4$. The nearest whole number of units is 225.
14. For which time interval is the drone's altitude increasing?
    Answer: $0 < t < 10$ The altitude is increasing when its rate of change, the derivative $A'(t)$, is positive. $A'(t) = -0.3t^2 + 3t$. To find where $A'(t) > 0$, we first find the roots of $A'(t) = 0$. $-0.3t(t - 10) = 0$, so the roots are $t=0$ and $t=10$. Since the graph of $A'(t)$ is a downward-opening parabola, it is positive between its roots. Therefore, the altitude is increasing for the interval $0 < t < 10$.
15. At what rate is the machinery depreciating 5 years after purchase, in dollars per year?
    Answer: $10,580 per year The rate of change of value is given by the derivative $V'(t) = 150(-0.15)e^{-0.15t} = -22.5e^{-0.15t}$. At $t=5$, the rate is $V'(5) = -22.5e^{-0.75} \approx -10.58$ thousand dollars per year. Since this is negative, the machinery is losing value. The depreciation rate (positive value) is approximately $10,580 per year.
16. What is the maximum concentration of the drug in the bloodstream?
    Answer: 20.0 mg/L When you encounter a function modeling a real-world scenario and need to find a maximum value, you're looking for the highest point on the curve. This requires finding where the derivative equals zero and determining which critical point gives the maximum. To find the maximum concentration, you need to differentiate $C(t) = \frac{120t}{t^2 + 9}$ using the quotient rule. The quotient rule states that for $\frac{f(t)}{g(t)}$, the derivative is $\frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}$. Here, $f(t) = 120t$ so $f'(t) = 120$, and $g(t) = t^2 + 9$ so $g'(t) = 2t$. $C'(t) = \frac{120(t^2 + 9) - 120t(2t)}{(t^2 + 9)^2} = \frac{120t^2 + 1080 - 240t^2}{(t^2 + 9)^2} = \frac{1080 - 120t^2}{(t^2 + 9)^2}$ Setting $C'(t) = 0$: $1080 - 120t^2 = 0$, which gives $t^2 = 9$, so $t = 3$ (taking the positive value since time is positive). At $t = 3$: $C(3) = \frac{120(3)}{3^2 + 9} = \frac{360}{18} = 20$ mg/L. Option A (3.0 mg/L) is far too low and might result from calculation errors. Option B (13.3 mg/L) could come from incorrectly applying the quotient rule or arithmetic mistakes. Option C (40.0 mg/L) might result from substituting incorrectly or forgetting to square terms in the denominator. Remember: optimization problems in real-world contexts almost always require calculus. Practice the quotient rule thoroughly, as rational functions appear frequently in biological and pharmacological models.
17. What is the maximum possible volume of the box, in cm³?
    Answer: 2000 Let $x$ be the side length of the squares cut from the corners. The dimensions of the box will be length $(30-2x)$, width $(30-2x)$, and height $x$. The volume is $V(x) = x(30-2x)^2 = 4x^3 - 120x^2 + 900x$. To maximize the volume, find the derivative: $V'(x) = 12x^2 - 240x + 900$. Set $V'(x) = 0$ and solve for $x$, which gives $x=5$ or $x=15$. The domain is $0 < x < 15$, so $x=5$ is the only valid solution. The maximum volume is $V(5) = 5(30-2(5))^2 = 5(20)^2 = 2000$ cm³.
18. What is the width of the rectangular part of the window, in meters, that allows the maximum amount of light (i.e., maximizes the area)?
    Answer: 2.80 Let the width of the rectangle be $2r$ and the height be $h$. The radius of the semicircle is $r$. The perimeter is $P = 2r + 2h + \pi r = 10$. The area is $A = 2rh + \frac{1}{2}\pi r^2$. From the perimeter equation, $2h = 10 - 2r - \pi r$. Substitute this into the area equation: $A(r) = r(10 - 2r - \pi r) + \frac{1}{2}\pi r^2 = 10r - 2r^2 - \pi r^2 + \frac{1}{2}\pi r^2 = 10r - 2r^2 - \frac{1}{2}\pi r^2$. Differentiate with respect to $r$: $A'(r) = 10 - 4r - \pi r$. Set $A'(r)=0$: $10 = r(4+\pi)$. So $r = \frac{10}{4+\pi} \approx 1.40$. The question asks for the width, which is $2r$. Width = $2 \times 1.40 = 2.80$ meters.

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