Applications Of Derivatives Question 69
Question: What is the product of two parts of 20, such that the product of one part and the cube of the other is maximum ?
Options:
A) 75
B) 91
C) 84
D) 96
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let 20 be divided in two parts such that first part = x
$ \therefore $ Second part = 20-x Now, assume that $ P=x^{3}(20-x)=20x^{3}-x^{4} $ Now, $ \frac{dP}{dx}=60x^{2}-4x^{3}; $ and $ \frac{d^{2}P}{dx^{2}}=120x-12x^{2} $ Put $ \frac{dP}{dx}=0 $ for maxima or minima
$ \Rightarrow \frac{dP}{dx}=0 $
$ \Rightarrow 4x^{2}(15-x)=0\Rightarrow x=0,x=15 $
$ \therefore {{( \frac{d^{2}P}{dx^{2}} )}_{x=15}}=120\times 15-12\times (225) $
$ =1800-2700=-900<0 $
$ \therefore $ P is a maximum at $ x=15 $ .
$ \therefore $ First part = 15 and second part $ =20-15=15 $ Required product $ =15\times 5=75 $
BETA
