There is one critical point at (2, 4), but this point happens to fall on one of the boundaries of the region. We'll get to that point in a moment.
Along the boundary 
, we have
which attains a maximum value of
Along 
, we have
which attains a maximum of
Along 
, we have
which attains a maximum of
So over the given region, the absolute maximum of 
is 1578 at (2, 44).
move te dotty thing to the midle them sumbtract 5 1/2
u are very smart
You can tell based off the translation of graphs.
Vertical:
y=f(x) + k : graph shifts k units up
y=f(x) - k : graph shifts k units down
Horizontal:
y=f(x - k) : graph shifts k units right
y= f(x+k) : graph shifts k units left
X=6
Z=8 so X:Z
=. 6/8
So 2 goes into both 6 and 8 so divide numerator and denominator by 2 which = 3/4
One question at a time, please. I will focus on #16 and ignore #18.
slope: 3/4 line passes thru (-8,2)
Write out y = mx + b. Subst. 2 for y. Subst. -8 for x Subst. (3/4) for m:
2 = (3/4)(-8) + b. Find b. 2 = -6 + b => b = 8
So your equation is y = (3/4)x + 8. Please, use ( ) around those fractions!
