Question
If $a, b, c$ are positive numbers in A.P. such that their product is 64 , then the minimum value of $b$$(\mathrm{A})=2$(B) $=4$$(\mathrm{C})=1$(D) Does not exist
Step 1
This implies $b = \frac{a+c}{2}$. Show more…
Show all steps
Your feedback will help us improve your experience
Wendi Zhao and 60 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If three positive numbers $a, b$ and $c$ are in A.P. such that $a b c=8$, then the minimum possible value of $b$ is : (a) 2 (b) $4^{\frac{1}{3}}$ (c) $4^{\frac{2}{3}}$ (d) 4
Find two positive numbers whose product is 64 and whose sum is a minimum. Your answer: 8 and 8
If $n(A)=6$ and $n(B)=4$ then minimum value of $n(A-B)$ is (a) 2 (b) 7 (c) 6 (d) 4
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD