Applications Of Derivatives Question 299
Question: If $ x+y=16 $ and $ x^{2}+y^{2} $ is minimum, then the values of x and y are
Options:
A) 3, 13
B) 4, 12
C) 6, 10
D) 8, 8
Show Answer
Answer:
Correct Answer: D
Solution:
$ x+y=16\Rightarrow y=16-x $
therefore $ x^{2}+y^{2}=x^{2}+{{(16-x)}^{2}} $
Let $ z=x^{2}+{{(16-x)}^{2}}\Rightarrow z’=4x-32 $
To be minimum of $ z,z’’>0 $ , and it is.
Therefore $ 4x-32=0\Rightarrow x=8\Rightarrow y=8 $
BETA
