Square cardboard, side 20 inches, cut and folded into a box with lid. Shaded areas show pattern. Four \( x \)-sided squares and two rectangles discarded. Which is true? Volume \( V = lwh \).▶️ Answer/Explanation
Solution
Height = \( x \). Length = \( 20 – 2x \). Width = \( 10 – x \) (from \( 2x + 2y = 20 \), \( y = 10 – x \)).
Volume: \( V = x (20 – 2x)(10 – x) \).
Expand: \( V = 2x^3 – 40x^2 + 200x \).
Derivative: \( V’ = 6x^2 – 80x + 200 \).
Factor: \( 2(3x^2 – 40x + 100) = 2(3x – 10)(x – 10) = 0 \), so \( x = \frac{10}{3} \) or \( x = 10 \).
Check: \( x = \frac{10}{3} \approx 3.33 \) maximizes (second derivative \( 12x – 80 < 0 \)). \( x = 10 \) is invalid (negative dimensions).
Key:
- Max volume at \( x = \frac{10}{3} \).
- \( x = 10 \) not possible.
✅ Answer: D
▶️ Answer/Explanation
Solution
Wall on one side, fencing on three sides. Let length (parallel to wall) = \( l \), width = \( w \).
Fencing: \( l + 2w = 18 \), so \( l = 18 – 2w \).
Area: \( A = l \cdot w = (18 – 2w) \cdot w = 18w – 2w^2 \).
Maximize: Derivative \( A’ = 18 – 4w \). Set \( A’ = 0 \): \( 18 – 4w = 0 \), \( w = \frac{9}{2} \).
Then \( l = 18 – 2 \cdot \frac{9}{2} = 9 \).
Area: \( A = 9 \cdot \frac{9}{2} = \frac{81}{2} \, \text{m}^2 \).
✅ Answer: E
▶️ Answer/Explanation
Solution
Max speed at critical points or endpoints (0 to 3) due to Extreme Value Theorem.
Derivative: \( f'(m) = \frac{1}{10}(-6m^2 + 18m – 12) = -\frac{6}{10}(m-1)(m-2) \).
Critical points: \( m = 1, 2 \). Candidates: \( m = 0, 1, 2, 3 \).
Evaluate: \( f(0) = 7.0 \), \( f(1) = 6.5 \), \( f(2) = 6.6 \), \( f(3) = 6.1 \).
Max is 7.0 at \( m = 0 \).
✅ Answer: C
▶️ Answer/Explanation
Solution
Proportional growth: \( \frac{dw}{dt} = kw \). Solution: \( w(t) = 2e^{kt} \).
Given: \( t=0 \), \( w=2 \); \( t=2 \), \( w=3.5 \).
\( 3.5 = 2e^{2k} \), \( e^{2k} = 1.75 \), \( k = \frac{\ln(1.75)}{2} \).
Weight: \( w(t) = 2(1.75)^{\frac{t}{2}} \).
At \( t=3 \): \( w(3) = 2(1.75)^{1.5} \approx 4.63 \).
Rounded: \( 4.6 \) lbs.
✅ Answer: B