Basic Calculus - Optimization and Time Rates

Step 1: Draw an appropriate figure and label relevant quantities.
Step 2: Find a formula for the quantity to be maximized or minimized.
Step 3: Use conditions from the problem to eliminate variables, expressing the quantity as a function of one variable.
Step 4: Determine the interval of possible values for this variable based on physical restrictions in the problem.
Step 5: Use previous techniques to find the maximum or minimum value, if applicable.

Problem: Find the number in the interval [-2, 2] such that the difference from its square is maximized.
Function: 𝑓(𝑥) = 𝑥² - 𝑥, where 𝑥 ∈ [-2, 2].
Critical points: Derivative 𝑓′(𝑥) = 2𝑥 - 1; critical number at 𝑥 = 1/2.
Values at endpoints and critical number:
- 𝑓(-2) = 6
- 𝑓(2) = 2
- 𝑓(1/2) = -1/4
Conclusion: The maximum occurs at 𝑥 = -2.

Range Formula: 𝑅(𝜃) = 𝑣₀ g sin(2𝜃), where 𝜃 ∈ [0, 𝜋/2].
Find Critical Points: Derivative 𝑅′(𝜃) = 𝑣₀ g (2cos(2𝜃)) = 0, giving critical number 𝜃 = 𝜋/4 = 45°.
Comparison at endpoints and critical number:
- 𝑅(0) = 0,
- 𝑅(π/4) = 𝑣₀g,
- 𝑅(π/2) = 0.
Conclusion: Maximum range at 𝜃 = 45°.

Problem: Create a box from 24 cm long and 9 cm wide cardboard by cutting out squares of side 𝑠.
Volume Function: 𝑉(𝑠) = (24 - 2𝑠)(9 - 2𝑠)(𝑠) = 216𝑠 - 66𝑠² + 4𝑠³.
Critical Numbers: Determine by analyzing the derivative 𝑉′(𝑠) = 216 - 132𝑠 + 12𝑠².
Interval: 0 ≤ 𝑠 ≤ 4.5 (beyond these values, dimensions become zero).
Conclusion: Maximum volume occurs at 𝑠 = 2, therefore 𝑉(2)= 200 cm³.

Problem: Find dimensions of a cylinder with maximum volume inscribed in a cone of height 9 in and base radius 6 in.
Volume Function: 𝑉(𝑟) = π(𝑟²)(h), with h dependent on r due to similar triangles.
Critical Points: Upon calculating, the only critical number is 𝑟 = 4.
Maximum Volume: Occurs when 𝑟 = 4, height is found to be 3.
Conclusion: Cylinder dimensions: radius = 4 cm, height = 3 cm.

Rate of Change Definition: For a function x(t), the derivative dx/dt represents its rate of change with respect to time. A positive value indicates increase, negative indicates decrease.

Step 1: Provide a valid illustration of the problem.
Step 2: Identify quantities that change with respect to time, labeling constants distinctly.
Step 3: Write known numerical facts about the variables involved.
Step 4: Identify the specific rate of change to be found and under what conditions.
Step 5: Write an equation that shows the relationship between variables at any time.
Step 6: Differentiate this equation implicitly with respect to time.
Step 7: Substitute known values into the differentiated equation.
Step 8: Conclude with a statement that answers the problem and includes units.

Rate of Area Change: dA/dt = 2π cm²/s.
**Find dr/dt when r = 10 cm.
Area Relation: A = πr², thus dA/dt = 2πr(dr/dt), leading to dr/dt = 1/10 cm/s.
Conclusion: Radius increasing at this rate.

Problem Statement: Ladder length fixed at 10 m, bottom moves horizontally at -2 m/s, find dy/dt when x = 6.
Apply Pythagorean theorem: x² + y² = 100, with differentiation yielding an equation to solve for dy/dt.
Final Calculation: dy/dt = 3/2 m/s, indicating upward movement of the top of the ladder.

Calculating Rate Change: Determine dz/dt using Pythagoras and derivatives with respect to time, with given distances at t = 2 seconds.
Conclude: Distance between vehicles increasing at 20 m/s.

Given: Water pouring into a cone at rate of 8 cubic meters.
**Find dH/dt when the height is 4 m using relationships from volume formulas and differentiating them.
Final Rate: dH/dt = 2π meters/minute, showing how fast the water level rises.