Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 39930 by Raj Singh last updated on 13/Jul/18

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

one is x and another is 6−x y=x^3 +(6−x)^3 (dy/dx)=3x^2 +3(6−x)^2 ×−1 =3x^2 −3(36−12x+x^2 ) =36x−108 for max/min(dy/dx)=0 36x−108=0 x=3 (d^2 y/dx^2 )=36>0 so at x=3 x^3 +(6−x)^3 is minimum min value =3^3 +(6−3)^3 =54 two numbers are 3,(6−x) =(3,3)

oneisxandanotheris6−xy=x3+(6−x)3dydx=3x2+3(6−x)2×−1=3x2−3(36−12x+x2)=36x−108formax/mindydx=036x−108=0x=3d2ydx2=36>0soatx=3x3+(6−x)3isminimumminvalue=33+(6−3)3=54twonumbersare3,(6−x)=(3,3)

Terms of Service

Privacy Policy

Contact: info@tinkutara.com