The maximum value of #f(x)=(3sinx-4cosx-10)(3sinx+4cosx-10)# is?

2 Answers
P dilip_k
Jul 25, 2018

#f(x)=(3sinx-4cosx-10)(3sinx+4cosx-10)#

#=((3sinx-10)-4cosx)((3sinx-10)+4cosx)#

#=(3sinx-10)^2-(4cosx)^2#

#=9sin^2x-60sinx+100-16cos^2x#

#=9sin^2x-60sinx+100-16+16sin^2x#

#=25sin^2x-60sinx+84#

#=(5sinx)^2-2*5sinx*6+6^2-6^2+84#

#=(5sinx-6)^2+48#

#f(x)# will be maximum when #(5sinx-6)^2# is maximum . It will be possible for #sinx=-1#

So

#[f(x)]_"max"=(5(-1)-6)^2+48=169#

A. S. Adikesavan
Jul 25, 2018

Maximum is 169. Minimum is 50 (perhaps, nearly). This is graphical illustration, for Dilip's answer.

Explanation:

Let #alpha = sin^(-1) ( 4/5 )#..Then

#f( x ) = 25 (sin ( x - alpha )-2)(sin (x + alpha) - 2)#
See graph.
graph{(y - 25 (sin (x- 0.9273)-2)(sin (x+ 0.9273)-2))(y-169)(y-50)=0[-20 20 20 230]}
graph{(y - 25 (sin (x- 0.9273)-2)(sin (x+ 0.9273)-2))(y-169)=0[-1.75 -1.5 167 171]}

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