Applications Of Derivatives Question 311
Question: The maximum and minimum values of $ x^{3}-18x^{2}+96x $ in interval (0, 9) are
[RPET 1999]
Options:
A) 160, 0
B) 60, 0
C) 160, 128
D) 120, 28
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ y=x^{3}-18x^{2}+96x $
therefore $ \frac{dy}{dx}=3x^{2}-36x+96=0 $
$ x^{2}-12x+32=0\Rightarrow (x-4),(x-8)=0 $ , $ x=4,,8 $
Now, $ \frac{d^{2}y}{dx^{2}}=6x-36 $
At $ x=4,\frac{d^{2}y}{dx^{2}}=24-36=-12<0 $
At $ x=4 $ function will be maximum
and $ {{[f(4)]} _{max\text{.}}}=64-288+384=160 $
At $ x=8 $ , $ \frac{d^{2}y}{dx^{2}}=48-36=12>0 $ At $ x=8 $ ,
function will be minimum and $ {{[f(8)]} _{min\text{.}}}=128. $
BETA
