a tablet is 7m high is placed on a wall with its base9 m above the level of an observer's eye. how far from the wall should an observer stand in order that the angle subtended at the eye by the table be a maximum?

Question

a tablet is 7m high is placed on a wall with its base9 m above the level of an observer's eye. how far from the wall should an observer stand in order  that the angle subtended at the eye by the table be a maximum?

kaiz122 · Answer

help me please @saifoo.khan  @ParthKohli @KatrinaKaif

shubhamsrg · Answer

you attempting/preparing for  ISI ? coz this is a ques of ISI sample paper..

shubhamsrg · Answer

|dw:1340892427871:dw|
we have to find x for maximum y
now,
tan z = 9/x
tan(y+z) = 16/x
=>(tan y + tanz) /(1- tany tanz) = 16/x
=>tan y + 9/x = (16/x) (1- tany 9/x )
=> tan y +9/x = 16/x - 144/x^2  tany
tan y = (7/x) / (1+ 144/x^2) = 7x/(144 + x^2) = 7 / (144/x  +x)

now for y maximum,,tany will be maximum,,since y is acute.
now tany will be maximum for denominator of RHS be minimum
i.e. our problem is to minimize (144/x + x)
putting d/dx( 144/x +x) =0 ;
=> - 144/x^2 + 1 =0
=> x= 12 ,,which should be our ans

you can confirm this is the minimum value as d2 /dx^2 of this is +ve.
hope i was clear enough..