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AP Calculus AB 5.10 Introduction to Optimization Problems - MCQs - Exam Style Questions

AP Calculus AB-MCQs and Answers - All Topics

No-Calc Question

Consider a triangle in the \(xy\)-plane with base endpoints \((1,0)\) and \((5,0)\) and the third vertex on \(y=\ln(2x)-\dfrac{1}{2}x+5\) for \(\dfrac{1}{2}\le x\le 8\). What is the maximum area of such a triangle?

(A) \(\dfrac{19}{2}\)
(B) \(2\ln 2+9\)
(C) \(2\ln 4+8\)
(D) \(2\ln 16+2\)

▶️ Answer/Explanation

Base \(=4\). Height \(=y(x)=\ln(2x)-\tfrac{1}{2}x+5\).
\(A(x)=\tfrac{1}{2}\times 4\times y(x)=2\ln(2x)-x+10\).
\(A'(x)=\tfrac{2}{x}-1=0\Rightarrow x=2\).
\(A_{\max}=2\ln 4-2+10=2\ln 4+8\).
✅ Answer: (C) \(2\ln 4+8\)

Question

The position of an object attached to a spring is given by:

\[ y(t) = \frac{1}{6}\cos(5t) – \frac{1}{4}\sin(5t) \]

where t is time in seconds. In the first 4 seconds, how many times is the velocity of the object equal to 0?

A) Zero
B) Three
C) Five
D) Six
E) Seven

▶️ Answer/Explanation

Solution

To find when velocity equals zero:

1. First find the velocity function by differentiating position:

\[ v(t) = y'(t) = -\frac{5}{6}\sin(5t) – \frac{5}{4}\cos(5t) \]

2. Set velocity equal to zero and solve:

\[ -\frac{5}{6}\sin(5t) – \frac{5}{4}\cos(5t) = 0 \] \[ \Rightarrow \tan(5t) = -\frac{3}{2} \]

3. Find solutions in [0,4]:

  • The period of the trigonometric functions is \( \frac{2π}{5} \approx 1.26 \) seconds
  • In 4 seconds, there are \( \frac{4}{2π/5} \approx 3.18 \) periods
  • Each period has 2 zero-velocity points (max/min positions)
  • Total zero-velocity points: 3 full periods × 2 = 6 points

4. Graphical interpretation:

  • The position function \( y(t) \) is a sinusoidal wave
  • Velocity is zero at each peak and trough (max/min displacement)
  • 6 such points occur in the first 4 seconds

✅ Therefore, the correct answer is D) Six.

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