How do you find the max and min values of function fxy6x220x

Solution

The critical values on x^2+y^2<16 are when
df/dx=df/dy=0
df/dx=-12x-20y=0
df/dy=-12y-20x=0
y=12x/20 from first
-12*12x/20-20x=0
-144x=-400x
so x=0
y=0
(0,0) is the only critical value on the inside

On the boundary we have x^2+y^2=16
since f(x,y)=-6(x^2+y^2)-20xy-4
the minimum or maximum occur when they occur for g(x,y)=-xy
a polar coordinate will be handy
we can parameterize x=4cos(t), y=4sin(t), t from 0 to 2

g=-16sin(t)cos(t)=-8sin(2t)

which is maximum when sin(2t)=-1, 2t=3/2, t=3/4

x=4*(-2/2)=-22, y=22

the minimum occurs when sin(2t)=1, 2t=/2, t=/4

x=y=22

So we check at the end these critical values

x=y=0, f(0,0)=-4

f(-22,22)=-6*16+20-4

f(22,22)=-6*16-20-4

So the absolute maximum occurs for x=y=0

and the absolute minimum when x=y=2