Capstone Practice: Mixed Application Sets and Full Solutions

How to Use This Capstone Chapter

This chapter is practice-heavy: you will integrate curve sketching, motion, related rates, and optimization in multi-step tasks. Each guided solution follows the same layout so you can copy it on homework/exams: (A) Variables → (B) Equations → (C) Derivative work → (D) Domain → (E) Conclusion (with interpretation).

In the mixed sets, problems are grouped by the workflow translate → compute → interpret. The goal is not just getting a derivative, but deciding what the derivative means and why your final answer makes sense.

Mixed Application Sets (Grouped by Skill)

Set 1: Translate → Compute → Interpret (Curve Sketching + Optimization)

• 1.1 A manufacturer models profit (in dollars) from producing x units by P(x)= -0.02x^3 + 3x^2 + 120x - 500 for 0 ≤ x ≤ 120. (a) Find the production level that maximizes profit. (b) Justify why your choice is a maximum (not minimum). (c) Interpret the sign of P'(x) near your answer.
• 1.2 A function is defined by f(x)= (x-1)^2(x+2). (a) Identify critical points and intervals of increase/decrease. (b) Determine where the graph is concave up/down. (c) Sketch a labeled graph including intercepts, critical points, and inflection point(s).
• 1.3 A rectangular pen is built against a straight wall, using 80 m of fencing for the other three sides. (a) Express area as a function of one variable. (b) Maximize area. (c) Explain why endpoints matter here.

Set 2: Translate → Compute → Interpret (Motion + Curve Features)

• 2.1 A particle moves on a line with position s(t)= t^3 - 6t^2 + 9t (meters), for 0 ≤ t ≤ 6. (a) Find when the particle is at rest. (b) Determine intervals where it moves right/left. (c) Find total distance traveled. (d) Identify when speed is increasing.
• 2.2 A car’s velocity is v(t)= 12t - 3t^2 (m/s) for 0 ≤ t ≤ 6. (a) When is the car speeding up? (b) When does it reach maximum speed? (c) Compute displacement and distance traveled.
• 2.3 A particle’s acceleration is a(t)= 6 - 2t and v(0)=1 (m/s). (a) Find v(t). (b) Find when velocity is zero. (c) Interpret the sign of acceleration at that time.

Set 3: Translate → Compute → Interpret (Related Rates + Optimization)

• 3.1 A 10 ft ladder leans against a wall. The bottom slides away at 2 ft/s. (a) Find how fast the top slides down when the bottom is 6 ft from the wall. (b) State the domain restriction that prevents impossible configurations.
• 3.2 A spherical balloon is inflated so that its radius increases at 0.4 cm/s. (a) Find dV/dt when r=10 cm. (b) Interpret units.
• 3.3 A right circular cylinder must have volume 500 cm³. Minimize surface area (including top and bottom). (a) Write surface area in one variable using the constraint. (b) Find the minimizing dimensions. (c) Explain why the critical point gives a minimum.

Guided Solutions (Modeled Layout)

Guided Solution 1 (Optimization with Endpoints): Fencing Against a Wall (Problem 1.3)

(A) Variables. Let x be the side lengths perpendicular to the wall (two sides), and y be the side parallel to the wall (one side). Units: meters.

(B) Equations. Fencing used: 2x + y = 80. Area: A = xy.

(C) Derivative work. Use the constraint to write y = 80 - 2x. Then A(x)= x(80 - 2x)= 80x - 2x^2. Differentiate: A'(x)= 80 - 4x. Set A'(x)=0: 80 - 4x=0 gives x=20. Then y=80 - 2(20)=40.

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(D) Domain. Physical lengths require x ≥ 0, y ≥ 0. From y=80-2x ≥ 0, we get x ≤ 40. So 0 ≤ x ≤ 40.

(E) Conclusion (with interpretation). Compute area at the critical point and endpoints: A(0)=0, A(40)=0, A(20)= 20·40=800. The maximum area is 800 m^2 when x=20 m and y=40 m. Why endpoints matter: the domain is closed and bounded; the “best” configuration could occur at an endpoint (here it does not, but you must check).

Guided Solution 2 (Motion + Distance vs Displacement): Position s(t)= t^3 - 6t^2 + 9t (Problem 2.1)

(A) Variables. t in seconds, s(t) in meters. Velocity v(t)=s'(t), acceleration a(t)=v'(t).

(B) Equations. s(t)= t^3 - 6t^2 + 9t. Then v(t)= 3t^2 - 12t + 9 and a(t)= 6t - 12.

(C) Derivative work. At rest means v(t)=0: 3t^2 - 12t + 9=0 → divide by 3: t^2 - 4t + 3=0 → (t-1)(t-3)=0. So rest at t=1 and t=3. Determine direction by sign of v(t): test intervals (0,1), (1,3), (3,6). For t=0.5, v>0 (right). For t=2, v<0 (left). For t=4, v>0 (right).

Displacement on [0,6] is s(6)-s(0). Compute: s(0)=0, s(6)=216 - 216 + 54=54. Displacement = 54 m.

Total distance requires adding absolute changes over monotonic intervals: evaluate s(0), s(1), s(3), s(6). s(1)=1 - 6 + 9=4. s(3)=27 - 54 + 27=0. s(6)=54. Distance = |s(1)-s(0)| + |s(3)-s(1)| + |s(6)-s(3)| = |4-0| + |0-4| + |54-0| = 4 + 4 + 54 = 62 m.

Speed increasing occurs when velocity and acceleration have the same sign. Here a(t)=6(t-2) is negative for t<2 and positive for t>2. Combine with velocity signs: on (0,1), v>0 and a<0 → speed decreasing; on (1,2), v<0 and a<0 → speed increasing; on (2,3), v<0 and a>0 → speed decreasing; on (3,6), v>0 and a>0 → speed increasing.

(D) Domain. Given 0 ≤ t ≤ 6.

(E) Conclusion (with interpretation). The particle rests at t=1 and t=3, moves right on (0,1) and (3,6), left on (1,3). Displacement is 54 m while total distance is 62 m, showing that reversing direction increases distance without necessarily increasing net change.

Guided Solution 3 (Related Rates with Domain Check): Ladder Sliding (Problem 3.1)

(A) Variables. Let x be the distance of the ladder’s bottom from the wall (ft), y the height of the top on the wall (ft). Ladder length is constant 10 ft. Given dx/dt = 2 ft/s. Find dy/dt when x=6.

(B) Equations. Right triangle constraint: x^2 + y^2 = 10^2 = 100.

(C) Derivative work. Differentiate implicitly with respect to t: 2x dx/dt + 2y dy/dt = 0. Solve: dy/dt = -(x/y) dx/dt. When x=6, compute y from the constraint: y= sqrt(100-36)=8 (take positive height). Then dy/dt = -(6/8)(2)= -12/8 = -1.5 ft/s.

(D) Domain. The triangle requires 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10. Also note the formula dy/dt = -(x/y)dx/dt is undefined at y=0 (when the ladder is flat), so computations must avoid that endpoint.

(E) Conclusion (with interpretation). When the bottom is 6 ft from the wall, the top slides down at 1.5 ft/s (negative sign indicates downward motion).

Guided Solution 4 (Constraint Optimization): Cylinder with Fixed Volume (Problem 3.3)

(A) Variables. Let radius r (cm) and height h (cm). Volume fixed at 500 cm³.

(B) Equations. Volume: V= πr^2 h = 500. Surface area (top + bottom + side): S = 2πr^2 + 2πrh.

(C) Derivative work. Use the constraint to eliminate h: h = 500/(πr^2). Then S(r)= 2πr^2 + 2πr(500/(πr^2)) = 2πr^2 + 1000/r. Differentiate: S'(r)= 4πr - 1000/r^2. Set to zero: 4πr = 1000/r^2 → 4πr^3 = 1000 → r^3 = 250/π → r = (250/π)^{1/3}. Then h = 500/(πr^2). Using r^3=250/π, we get r^2 = (250/π)^{2/3}, so h = 500/(π(250/π)^{2/3}) = 2(250/π)^{1/3}. Thus h = 2r.

(D) Domain. r>0, h>0. The surface area expression S(r)=2πr^2 + 1000/r requires r>0.

(E) Conclusion (with interpretation). The surface area is minimized when r = (250/π)^{1/3} cm and h = 2(250/π)^{1/3} cm (height equals diameter). Why this is a minimum: as r→0^+, 1000/r→∞; as r→∞, 2πr^2→∞. Since S goes to infinity on both ends and is smooth for r>0, the interior critical point must produce the global minimum.

Reflection Prompts (Reasonableness + Choice Explanations)

After any optimization problem

• State the domain in a complete sentence. What physical or contextual restriction created each inequality?
• Did you check endpoints? If you did not, explain precisely why (for example, an open domain or an unbounded objective).
• Explain why your critical point corresponds to a maximum or minimum: sign change of the derivative, second derivative test, or end behavior + single critical point.
• Units check: what are the units of the objective function, and do they match your final numerical value?

After any motion problem

• Write one sentence distinguishing displacement from total distance for your specific function and interval.
• When you said “speed is increasing,” did you compare signs of velocity and acceleration? Show the sign chart or a short justification.
• Does the magnitude of your answer fit the time scale? For example, if the time interval is small, would a huge distance be suspicious?

After any related rates problem

• List every quantity that changes with time and every quantity that is constant. Did you accidentally differentiate a constant as if it were variable?
• Where did you substitute numerical values: before or after differentiating? Explain why substitution should happen after differentiating.
• Domain check: are there configurations where your formula breaks (division by zero, impossible geometry)? Did your evaluation point avoid them?

Cumulative Review: Decide the Tool First

For each problem below, your first task is to write: Tool choice (curve sketching features, motion analysis, related rates, or optimization) and why. Then solve fully.

• R1 A function g(x)= x/(x^2+1) models a “response” level. Find where the response is increasing most rapidly and justify your choice using derivatives and domain reasoning.
• R2 A particle has velocity v(t)= (t-2)(t-5) on [0,6]. Find total distance traveled and the time(s) when the particle changes direction. (Be explicit about sign changes.)
• R3 A conical tank (vertex down) has height 12 m and radius 4 m. Water is poured in at 3 m³/min. How fast is the water level rising when the water is 6 m deep?
• R4 You want to design a closed-top box with square base that holds 256 in³. The material for the top and bottom costs twice as much per square inch as the material for the sides. Minimize total cost. (Your objective is cost, not area.)
• R5 A company models average cost per unit by C(x)= (0.01x^2 + 50x + 2000)/x for x>0. Find the production level that minimizes average cost and interpret what happens to C(x) as x becomes very large.

Final Mini-Project: Written Analysis + Labeled Graph

Project Prompt

A delivery drone travels along a straight line. Its position (meters) after t seconds is modeled by s(t)= -0.5t^3 + 6t^2 - 12t + 5 for 0 ≤ t ≤ 10. The drone’s battery temperature (in °C) is modeled as a function of time by T(t)= 20 + 3t - 0.2t^2 on the same interval.

Tasks (what to submit)

• 1) Motion analysis (derivative-based). Compute v(t) and a(t). Identify all times when the drone is at rest. Determine intervals where it moves forward/backward. Compute displacement and total distance traveled on [0,10].
• 2) Speed behavior. Determine when the drone is speeding up and slowing down. Your justification must reference the signs of v(t) and a(t) on intervals.
• 3) Temperature optimization. Find the maximum temperature on [0,10]. You must show domain/endpoints and justify why the maximum occurs where you claim.
• 4) Labeled graph (required). On a single set of axes, sketch a labeled graph of s(t) versus t on [0,10]. Mark and label: intercepts (if any in the interval), critical times (where v=0), and any inflection time(s) (where a=0). Use arrows to indicate direction of motion on each interval.
• 5) Interpretation paragraph (required). Write 8–12 sentences explaining what the math says about the drone’s trip: when it stops, when it reverses, how far it actually travels, and when temperature is most concerning. Include units in at least four places.

Required Solution Layout (use these headings in your submission)

• Variables (define t, s, v, a, T, and units)
• Equations (write the given models and derived formulas)
• Derivative Work (show derivatives, critical times, sign charts or interval tests)
• Domain (state and use 0 ≤ t ≤ 10)
• Conclusion (numerical answers + interpretation)

Grading Criteria (100 points total; setup and interpretation emphasized)