If x+y=2pi/4, find maximum and minimum of sinx+siny?

When
#x+y=(2pi)/4, x>=0, y>=0 :#
the maximum and the minimum value of # sinx+siny=?#

2 Answers

Answer:

Maximum #=\sqrt2#

Minimum #=-\sqrt2#

Explanation:

Given that

#x+y={2\pi}/4#

#x+y=\pi/2#

#y=\pi/2-x#

Now, setting #y=\pi/2-x# in given function as follows

#f(x, y)=\sin x+\siny#

#f(x)=\sin x+\sin(\pi/2-x)#

#f(x)=\sin x+\cosx#

#f'(x)=\cos x-\sin x#

#f''(x)=-sin x-cos x#

For minima & maxima, #f'(x)=0#

#\therefore \cos x-\sin x=0#

#\tan x=1#

#x=n\pi+\pi/4#

#\implies f''(\pi/4)= -\sin(\pi/4)-\cos(\pi/4)=-\sqrt2<0#

hence, function is maximum at #x=\pi/4# & the maximum value is given as

#f(\pi/4)=\sin(\pi/4)+\cos(\pi/4)=\sqrt2#

#\implies f''({5\pi}/4)= -\sin({5\pi}/4)-\cos({5\pi}/4)=\sqrt2>0#

hence, function is minimum at #x={5\pi}/4# & the minimum value is given as

#f({5\pi}/4)=\sin({5\pi}/4)+\cos({5\pi}/4)=-\sqrt2#

Jul 27, 2018

#x+y=(2pi)/4 implies y = pi/2 - x#

#:. sinx + siny = sinx + sin(pi/2 - x) #

#= sinx + cosx#

#= sqrt2( 1/sqrt2 sinx + 1/sqrt2 cosx)#

#= sqrt(2) sin(x + pi/4)#

So, maximum and minimum of "#sinx+siny#":

  • #sqrt2#

  • #- sqrt2#