Question

If \( x, y \) are two positive integers such that \( x + y = 20 \) and the maximum value of \( x^3 y \) is \( k \) at \( x = a, y = \beta \), then \( \frac{k}{\alpha^2 \beta^2} = ? \) 

• A. \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \)
• B. \( \frac{\alpha}{\beta} - \frac{\beta}{\alpha} \)
• C. \( \frac{\alpha}{\beta} \)
• D. \( \frac{\alpha + \beta}{\alpha \beta} \)

Hint:

To maximize a function subject to a constraint, express one variable in terms of the other and take the derivative to find critical points. Verify the maximum using second derivative or substitution.

Answer 1

The Correct Option is C

Step 1: Express the function in terms of \( x \) and \( y \) 
We are given the equation: \[ x + y = 20 \] and we need to maximize the function: \[ f(x) = x^3 y. \] Using \( y = 20 - x \), we rewrite: \[ f(x) = x^3 (20 - x). \] 
Step 2: Differentiate \( f(x) \) and find the critical points 
To maximize \( f(x) \), we take the derivative: \[ \frac{d}{dx} (x^3 (20 - x)) = 3x^2 (20 - x) - x^3. \] Setting this to zero: \[ 3x^2 (20 - x) - x^3 = 0. \] \[ x^2 (3(20 - x) - x) = 0. \] \[ x^2 (60 - 4x) = 0. \] Ignoring \( x^2 = 0 \) (as \( x>0 \)), we solve: \[ 60 - 4x = 0 \Rightarrow x = 15. \] Thus, \( x = 15 \) and \( y = 20 - 15 = 5 \). 
Step 3: Compute the Maximum Value 
Substituting \( x = 15 \) and \( y = 5 \) into \( f(x) \): \[ k = 15^3 \times 5. \] 
Step 4: Compute \( \frac{k}{\alpha^2 \beta^2} \) 
\[ \frac{k}{\alpha^2 \beta^2} = \frac{(15^3 \times 5)}{(15^2 \times 5^2)}. \] \[ = \frac{15^3 \times 5}{15^2 \times 5^2} = \frac{15 \times 5}{5^2}. \] \[ = \frac{15}{5} = \frac{\alpha}{\beta}. \] 
Final Answer: \( \boxed{\frac{\alpha}{\beta}} \).