Question

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=?`

Answer 1

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`

Answer 2

`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`