Applications Of Derivatives Question 256
Question: 20 is divided into two parts so that product of cube of one quantity and square of the other quantity is maximum. The parts are
[RPET 1997]
Options:
A) 10, 10
B) 16, 4
C) 8, 12
D) 12, 8
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ x+y=20\Rightarrow y=20-x $ and $ x^{3}.y^{2}=z\Rightarrow z=x^{3}.y^{2} $
$ z=x^{3}{{(20-x)}^{2}} $
therefore $ z=400x^{3}+x^{5}-40x^{4} $
$ \frac{dz}{dx}=1200x^{2}+5x^{4}-160x^{3} $
Now $ \frac{dz}{dx}=0 $ , then $ x=12,\ 20 $
Now $ \frac{d^{2}z}{dx^{2}}=2400x+16x^{3}-480x^{2} $ ; $ {{( \frac{d^{2}z}{dx^{2}} )}_{x=12}}=-ive $
Hence $ x=12 $ is the point of maxima
$ \therefore x=12,\ \ y=8 $ .
BETA
