Question

If $x^3 + y^3 = 3axy$, then at $\left(\frac{3a}{2}, \frac{3a}{2}\right)$ the value of $3a y'' + 40$ is:

• A. $-5$
• B. $0$
• C. $8$
• D. $1$

Hint:

Implicit Second Derivative:

• Use chain/product rule when differentiating again.
• Plug known values only at the end.
• Be careful with signs and coefficients.

Answer 1

The Correct Option is C

Differentiate implicitly: \[ 3x^2 + 3y^2 y' = 3a(y + x y') \Rightarrow x^2 + y^2 y' = a(y + x y') \Rightarrow y' = \frac{ay - x^2}{y^2 - ax} \] At $x = y = \frac{3a}{2}$: \[ y' = \frac{a \cdot \frac{3a}{2} - \left(\frac{3a}{2}\right)^2}{\left(\frac{3a}{2}\right)^2 - a \cdot \frac{3a}{2}} = -1 \] Differentiate again to find $y''$ and evaluate to get $y'' = -\frac{32}{3a} \Rightarrow 3a y'' + 40 = -32 + 40 = \boxed{8}$