Applications Of Derivatives Question 36
Question: Find the minimum value of the function $ \frac{40}{3x^{4}+8x^{3}-18x^{2}+60} $ .
Options:
A) $ \frac{1}{3} $
B) $ \frac{2}{3} $
C) $ \frac{4}{3} $
D) $ \frac{5}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let $ y=\frac{1}{40}(3x^{4}+8x^{3}-18x^{2}+60) $
$ \Rightarrow \frac{dy}{dx}=\frac{1}{40}(12x^{3}+24x^{2}-36x) $ and $ \frac{d^{2}y}{dx^{2}}=\frac{1}{40}(36x^{2}+48x-36) $ Now $ \frac{dy}{dx}=0\Rightarrow x^{3}+2x^{2}-3x=0 $ or $ x(x-1)(x+3)=0 $ or $ x=0,1,-3 $ At $ x=0,,\frac{d^{2}y}{dx^{2}}=-36<0\therefore y $ is maximum at $ x=0 $
$ \Rightarrow $ The given function i.e., $ \frac{1}{y} $ is minimum at $ x=0 $
$ \therefore $ minimum value of the function $ =\frac{40}{60}=\frac{2}{3} $
BETA
