AP Calculus BC 5.11 Solving Optimization Problems Study Notes - New Syllabus
AP Calculus BC 5.11 Solving Optimization Problems Study Notes- New syllabus
AP Calculus BC 5.11 Solving Optimization Problems Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- A function’s derivative can be used to understand some behaviors of the function.
Key Concepts:
- Solving Optimization Problems
Solving Optimization Problems
Solving Optimization Problems
To solve optimization problems effectively, you apply calculus concepts (especially derivatives) to find where a function reaches its maximum or minimum values under real-world constraints. These problems involve modeling a situation with an equation, differentiating, and interpreting results.
Key Terms:
- Objective function: The function to be optimized (maximized or minimized).
- Constraint: A condition or equation that restricts the variables in the problem.
General Strategy:
- Define variables based on the problem.
- Write an equation for the quantity to be optimized (objective function).
- Write the constraint equation that relates the variables.
- Simplify the objective function using the constraint (reduce to one variable).
- Differentiate the simplified function.
- Find critical points by solving when derivative = 0.
- Use the second derivative test or endpoints to confirm max or min.
Example:
A square piece of cardboard 24 cm on each side is made into an open box by cutting equal squares from the corners and folding up the sides. What size square should be cut to maximize the volume?
▶️Answer/Explanation
Let square cut size = \( x \). Then dimensions after folding are:
\( \text{Length} = 24 – 2x \), \( \text{Width} = 24 – 2x \), \( \text{Height} = x \)
Volume: \( V = x(24 – 2x)^2 = 4x^3 – 96x^2 + 576x \)
\( \dfrac{dV}{dx} = 12x^2 – 192x + 576 \)
Critical points: \( x = 4 \), \( x = 12 \) → check 2nd derivative.
At \( x = 4 \), max volume = \( 1024 \text{ cm}^3 \)
Final Answer: Cut squares of size 4 cm.
Example:
Find the point on the curve \( y = \sqrt{x} \) that is closest to the point (4, 0).
▶️Answer/Explanation
Let the point be \( (x, \sqrt{x}) \).
Distance squared:
\( D^2 = (x – 4)^2 + x = x^2 – 7x + 16 \)
Derivative: \( f'(x) = 2x – 7 \Rightarrow x = 3.5 \)
Closest point: \( (3.5, \sqrt{3.5}) \approx (3.5, 1.87) \)
Example:
A company sells pens for $\$20$ each and sells 100 pens daily. For every $\$1$ increase in price, 5 fewer pens are sold. What price maximizes revenue?
▶️Answer/Explanation
Let price increase = \( x \).
New price = \( 20 + x \), quantity sold = \( 100 – 5x \)
Revenue: \( R(x) = (20 + x)(100 – 5x) = -5x^2 + 0x + 2000 \)
Vertex of parabola: \( x = 0 \), max at original price.
Final Answer: Price should remain $\$20$ for max revenue.
Example:
A can must hold 500 cm³. Find the dimensions of the cylinder that minimize surface area.
▶️Answer/Explanation
Let radius = \( r \), height = \( h \).
Constraint: \( \pi r^2 h = 500 \Rightarrow h = \dfrac{500}{\pi r^2} \)
Surface area: \( A = 2\pi r^2 + 2\pi rh = 2\pi r^2 + \dfrac{1000}{r} \)
Derivative: \( A’ = 4\pi r – \dfrac{1000}{r^2} \Rightarrow r^3 = \dfrac{250}{\pi} \)
\( r \approx 4.3 \), \( h \approx 8.6 \)
Final Answer: Radius ≈ 4.3 cm, Height ≈ 8.6 cm.
Example:
A rectangle is to be formed using 100 m of fencing. What is the maximum area that can be enclosed?
▶️Answer/Explanation
Let width = \( x \), then length = \( \dfrac{100 – 2x}{2} \)
Area: \( A = x(50 – x) = 50x – x^2 \)
Derivative: \( A’ = 50 – 2x = 0 \Rightarrow x = 25 \)
Max area at square: \( 25 \times 25 = 625 \text{ m}^2 \)
Example:
A boat is 3 km from the shore and must reach a point 5 km down the beach. It can row at 4 km/h and run at 6 km/h. Where should it land to minimize time?
▶️Answer/Explanation
Let landing point be \( x \) km along the beach.
Rowing distance: \( \sqrt{x^2 + 9} \), running distance = \( 5 – x \)
Time = \( \dfrac{\sqrt{x^2 + 9}}{4} + \dfrac{5 – x}{6} \)
Derivative: Use chain rule, solve \( T'(x) = 0 \)
Approximate solution: \( x \approx 2.29 \text{ km} \)
Final Answer: Land about 2.29 km from closest point.
Example:
Given: Revenue function \( R(x) = 100x – x^2 \), Cost function \( C(x) = 20x + 10 \). What is the output that maximizes profit?
▶️Answer/Explanation
Profit: \( P(x) = R(x) – C(x) = (100x – x^2) – (20x + 10) = -x^2 + 80x – 10 \)
Derivative: \( P'(x) = -2x + 80 \Rightarrow x = 40 \)
Final Answer: Maximum profit at 40 units of output.