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

1. Oct 19, 2009

### tnutty

1. The problem statement, all variables and given/known data

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 wont 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.

2. Oct 19, 2009

### HallsofIvy

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