Finding max and min of a function of several variables( 1 more time )

[tnutty]

Homework Statement

Find min/max/saddle point of the function :

F(x,y) = y^2 - 2*y*cos(x) ; 1 <= x <= 7

F_x = 2y*sin(x)
F_xx = 2y*cos(x)
F_xy = 2cos(x)

F_y = 2y - 2cos(x)
F_yy = 2


Finding its critical points :

F_x = 2y*sin(x) = 0
F_y = 2y - 2cos(x) = 0

Solving for y in F_y

2y = -2cos(x)
y = -cos(x)

inputing this value into F_x

2( -cos(x) ) * (sin(x) ) = 0;

realizing trig identities, somehow.

2( - cos(x) ) * (sin(x) ) = -2sin(x)cos(x) = -2sin(2x)

so ,

-2sin(2x) = 0

That function equals 0 when x = n*pi

but we have boundaries so ,

1<= 2n*pi <= 7
or
1/2 <= n*pi <= 7/2

the only solutions is n = 1, so n*pi = pi. ( but the book has critical points as 0,pi,2pi )

I won't do on because I know I did something wrong. I don't think they accounted for the
extra 2 in 2n*pi, otherwise the critical points would be 0,pi,2pi.

 

[HallsofIvy]

Homework Statement

Find min/max/saddle point of the function :

F(x,y) = y^2 - 2*y*cos(x) ; 1 <= x <= 7

F_x = 2y*sin(x)
F_xx = 2y*cos(x)
F_xy = 2cos(x)

F_y = 2y - 2cos(x)
F_yy = 2


Finding its critical points :

F_x = 2y*sin(x) = 0

so either y= 0 or sin(x)= 0=> $x= n\pi$

F_y = 2y - 2cos(x) = 0
If y= 0, then -2cos(x)= 0 so x is an odd multiple of $\pi/2$: $x= (2n+1)\pi/2$.
If $x= 2n\pi$, cos(x)= 1 so 2y- 2= 0 or y= 1.
If $x= (2n+1)\pi$, cos(x)= -1 so 2y+ 2= 0 or y= -1.

Since $1\le x\le 7$, the critical points are $(\pi/2, 0)$, $(3\pi/2, 0)$, $(\pi, -1)$, and $(2\pi, 1)$.

Solving for y in F_y

2y = -2cos(x)
y = -cos(x)

inputing this value into F_x

2( -cos(x) ) * (sin(x) ) = 0;

realizing trig identities, somehow.

2( - cos(x) ) * (sin(x) ) = -2sin(x)cos(x) = -2sin(2x)

so ,

-2sin(2x) = 0

That function equals 0 when x = n*pi

but we have boundaries so ,

1<= 2n*pi <= 7
or
1/2 <= n*pi <= 7/2

the only solutions is n = 1, so n*pi = pi. ( but the book has critical points as 0,pi,2pi )

I won't do on because I know I did something wrong. I don't think they accounted for the
extra 2 in 2n*pi, otherwise the critical points would be 0,pi,2pi.