Find local minima/maxima #y=(sinx-cosx+sqrt(2))/(sinx+cosx+2)# ?

Question

Quinn Anderson · Accepted Answer

#y=(sinx-cosx+sqrt(2))/(sinx+cosx+2)#

#=>y=(sqrt2(1/sqrt2sinx-1/sqrt2cosx+1))/(sqrt2(1/sqrt2sinx+1/sqrt2cosx+sqrt2)#

#=>y=(cos(pi/4)sinx-sin(pi/4)cosx+1)/(cos(pi/4)sinx+sin(pi/4)cosx+sqrt2)#

#=>y=(sin(x-pi/4)+1)/(sin(x+pi/4)+sqrt2)#

We know #-1 <=sin(x-pi/4)<=1 #
and #-1 <=sin(x+pi/4)<=1 #

when #sin(x-pi/4)=-1#

The minimum value of numerator is =0,

Denominator is always #>0#

So #y_"min"=0#

When #sin(x-pi/4)=1=sin(2npi+pi/2)#

#=>x=2npi+(3pi)/4#
Then

#sin(x+pi/4)=sin(2npi+(3pi)/4+pi/4)=sin((2n+1)pi)=0#

So #y=(1+1)/sqrt2=2/sqrt2=sqrt2#

When

#sin(x+pi/4)=-1#

The value of denominator #sqrt2-1#

Then #sin(x+pi/4)=-1=sin((2n+1)pi+pi/2)#

#=>x=(2n+1)pi+pi/2-pi/4#

so #sin(x-pi/4)=sin((2n+1)pi+pi/2-pi/4-pi/4)=sin((2n+1)pi)=0#

then numerator #=1#

And #y=1/(sqrt2-1)=sqrt2+1#

Hence #y=sqrt2+1#

Graph shows that #y_"min"=0# satisfies the graphical solution. But #y_"max"# obtained from graph is #~~2sqrt2# cannot be obtained by the method adopted here.

Emery Adkins · Answer

See below.

Another approach.

Making

#u = cos x#
#v=sin x#
#u^2+v^2 = 1#

now with #f(u,v) = (v - u + sqrt[2])/(v + u + 2)#

the minima/maxima problem can be stated as

#min//max f(u,v)#

subjected to

#u^2+v^2=1#

This problem can be successfully handled with the so called Lagrange multipliers

#L(u,v,lambda) = (v - u + sqrt[2])/(v + u + 2) + lambda (u^2 + v^2 - 1)#

The stationary conditions are

#L_u =2 lambda u (2 + u + v)^2 -2 - sqrt[2] - 2 v =0#
#L_v = 2 lambda v (2 + u + v)^2+2 - sqrt[2] + 2 u =0#
#L_lambda = u^2+v^2-1=0#

Solving for #u,v,lambda# gives

#((u,v,f(u,v),"type"),(1/sqrt2,-1/sqrt2,0,min),(-1/6 (4 +sqrt[2]),-1/6 (4 + sqrt[2]),2 sqrt[2],max))#

Attached a plot showing the maximum in red and the minimum in green.

NOTE

Equating the #lambda# values at #L_u# and #L_v# we obtain the system of equations

#{(2(u^2+v^2)+(2-sqrt2)u+(2+sqrt2)v = 0),(u^2+v^2 = 1):}#

that can be reduced to

#{(2+(2-sqrt2)u+(2+sqrt2)v = 0),(u^2+v^2 = 1):}#

Grace Ashcraft · Answer

undefined To find the local minima/maxima of the function $y = \frac{\sin(x) - \cos(x) + \sqrt{2}}{\sin(x) + \cos(x) + 2}$, we'll first find the critical points by taking the derivative of the function with respect to $x$, setting it equal to zero, and solving for $x$. Then, we'll use the second derivative test to determine whether these critical points correspond to local minima, maxima, or points of inflection.

1. Take the derivative of $y$ with respect to $x$:
$$ \frac{dy}{dx} = \frac{(\cos(x) + \sin(x))(\sin(x) + \cos(x) + 2) - (\sin(x) - \cos(x) + \sqrt{2})(\cos(x) - \sin(x))}{(\sin(x) + \cos(x) + 2)^2} $$

2. Set $\frac{dy}{dx} = 0$ and solve for $x$.

3. Once you find the critical points, evaluate the second derivative at these points. If the second derivative is positive, it indicates a local minimum; if negative, it indicates a local maximum. If the second derivative is zero, the test is inconclusive.

4. Plug these critical points back into the original function to find the corresponding $y$ values.

These $x$ and $y$ values represent the local minima/maxima of the function.