Applications Of Derivatives Question 310
Question: Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are
[DCE 1999]
Options:
A) (10, 10)
B) $ (5,15) $
C) (13, 7)
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ x+y=20 $ and $ z=xy^{3} $
therefore $ z=y^{3}(20-y)=20y^{3}-y^{4} $
therefore $ \frac{dz}{dy}=60y^{2}-4y^{3}=0 $
therefore $ 4y^{2}(15-y)=0 $
So, either $ y=0,, $ or $ y=15 $
Now, $ \frac{d^{2}z}{dy^{2}}=120y-12y^{2} $ ; At $ y=0,\frac{d^{2}z}{dy^{2}}>0 $ $ y=0 $ is the point of minima and at $ y=15,\frac{d^{2}z}{dy^{2}}<0 $ $ y=15 $ is the point of maxima.
Hence the required parts is (5, 15).
BETA
