The local maximum value of the function \( f(x)=\displaystyle\frac{b^2}{a^4}(a^2x^2-x^4),a,b\gt 0 \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{b^2}{4} \)
\( \displaystyle f'(x)=\frac{b^2}{a^4}(2a^2x-4x^3) \)
\( \displaystyle =\frac{2b^2x}{a^4}(a^2-2x^2) \)
\( \displaystyle f'(x)=0\ \Rightarrow\ x=0 \text{ or } x^2=\frac{a^2}{2} \)
\( \displaystyle f''(x)=\frac{b^2}{a^4}(2a^2-12x^2) \)
\( \displaystyle \text{At } x^2=\frac{a^2}{2}:\ f''(x)=\frac{b^2}{a^4}(2a^2-6a^2) \)
\( \displaystyle =-\frac{4b^2}{a^2}\lt 0 \)
So \( x=\pm\frac{a}{\sqrt{2}} \) give local maxima (x = 0 gives \( f''(0)=\frac{2b^2}{a^2}\gt 0 \), a local minimum).
\( \displaystyle f\left(\pm\frac{a}{\sqrt{2}}\right)=\frac{b^2}{a^4}\left(a^2\cdot\frac{a^2}{2}-\frac{a^4}{4}\right) \)
\( \displaystyle =\frac{b^2}{a^4}\cdot\frac{a^4}{4} \)
\( \displaystyle =\frac{b^2}{4} \)
Shortcut: Put \( t=x^2\ge 0 \): \( a^2t-t^2 \) is a downward parabola with vertex at \( t=\frac{a^2}{2} \) and largest value \( \frac{a^4}{4} \). Multiplying by \( \frac{b^2}{a^4} \) gives \( \frac{b^2}{4} \).