Question
Bacteria in a certain culture increase at a rate proportional to the number present. If the number of bacteria doubles in three hours, in how many hours will the number of bacteria triple?
(A) \(\frac{3ln 3}{ln 2}\) (B) \(\frac{2ln 3}{ln 2}\) (C)\( \frac{ln 3}{ln 2}\) (D) \(ln\left ( \frac{27}{2} \right )\) (E)\(ln\left ( \frac{27}{2} \right ) \)
▶️ Answer/Explanation
Solution
1. Model the growth with exponential equation: \( N(t) = N_0 e^{kt} \)
2. Use doubling time to find growth rate \( k \):
\( 2N_0 = N_0 e^{3k} \) ⇒ \( 2 = e^{3k} \) ⇒ \( k = \frac{\ln 2}{3} \)
3. Find time \( t \) for population to triple:
\( 3N_0 = N_0 e^{kt} \) ⇒ \( 3 = e^{kt} \) ⇒ \( \ln 3 = kt \)
Substitute \( k \): \( t = \frac{\ln 3}{k} = \frac{\ln 3}{\frac{\ln 2}{3}} = \frac{3\ln 3}{\ln 2} \)
4. Verification:
Doubling time formula: \( t_{double} = \frac{\ln 2}{k} \)
Tripling time should be \( \frac{\ln 3}{k} = \frac{\ln 3}{\ln 2} \times t_{double} = \frac{\ln 3}{\ln 2} \times 3 \)
✅ Answer: A) \(\frac{3\ln 3}{\ln 2}\) hours
Question
What is the area of the largest rectangle that can be inscribed in the ellipse \(4x^{2}+9y^{2}=36\)
(A) \(\sqrt[6]{2}\) (B) 12 (C) 24 (D) \(\sqrt[24]{2}\) (E) 36
▶️ Answer/Explanation
Solution
1. Rewrite ellipse equation in standard form:
\(\frac{x^2}{9} + \frac{y^2}{4} = 1\) (a=3, b=2)
2. Parameterize the rectangle vertices as (x,y), (x,-y), (-x,y), (-x,-y)
3. Area function: \( A = (2x)(2y) = 4xy \)
4. Express y in terms of x from ellipse equation:
\( y = \frac{2}{3}\sqrt{9 – x^2} \)
5. Substitute into area: \( A(x) = 4x \cdot \frac{2}{3}\sqrt{9 – x^2} = \frac{8}{3}x\sqrt{9 – x^2} \)
6. Maximize A(x) by setting derivative to zero:
\( \frac{dA}{dx} = \frac{8}{3}\left(\sqrt{9-x^2} + x\cdot\frac{-x}{\sqrt{9-x^2}}\right) = 0 \)
Simplify: \( 9 – x^2 – x^2 = 0 \) ⇒ \( x = \frac{3\sqrt{2}}{2} \)
7. Find corresponding y: \( y = \frac{2}{3}\sqrt{9 – \frac{9\cdot2}{4}} = \sqrt{2} \)
8. Maximum area: \( A = 4 \cdot \frac{3\sqrt{2}}{2} \cdot \sqrt{2} = 12 \)
✅ Answer: B) 12