Minimum, Maximum Value of a Two-variable Function

Date: 6/27/96 at 21:43:5
From: Anonymous
Subject: Minimum, Maximum Value of Two-variable Function

Find f min and f max where x and y are real numbers and
f(x,y) = 2(sin x)(cos y)+3(sin x)(siny)+6(cos x).

      Thanks,
       Michael

Date: 6/28/96 at 14:6:1
From: Doctor Jerry
Subject: Re: Minimum, Maximum Value of Two-variable Function

Because of periodicity, it is sufficient to look at the square 
[0,2pi]x[0,2pi].  It is easy to check that on the boundaries, 
-2sqrt(10)<=f(x,y)<=2sqrt(10).  For example, on the y=0 and y=2pi 
boundaries, f[x,0] = f[x,2pi] = 2sin x +6 cos x = 2sqrt(10)sin(x+a).  
On the x = 0 and x = 2pi boundaries, f[0,y] = f[2pi,y] = 6.

Setting the two partial derivatives equal to zero gives
2cos x cos y-6 sin x +3 cos x sin y = 0
3cos y sin x-2sin x sin y = 0.

To solve the second equation, we want to assume that sin x is not 
zero.  Note that on the lines where sin x = 0, f = 6.  

If sin x is not zero, then from second equation, 
y = arctan(3/2) or pi+arctan(3/2).  
From first equation, x = arctan(sqrt(13)/6) or 
x = pi+arctan(sqrt(13)/6).

Evaluating f at these four points and recalling the boundary values, 
it follows that max f is 7 and min f is -7.  Moreover,

f(arctan(sqrt(13)/6),arctan(3/2)) = 7

f[pi+arctan(sqrt(13)/6),arctan(3/2)) = -7

-Doctor Jerry,  The Math Forum
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