Posted by Alexis on Friday, April 25, 2008 at 4:55pm.
a)
x^2 + y ^2 = 4
area = 4 |x y| (so it works in each quadrant)
y = sqrt(4-x^2)
area = 4 x sqrt(4-x^2)
b)
the area can be 0 (at x = 0 and x = 2)and it has a maximum at x = y = sqrt 2
to show that take derivative and set to zero
0 = -4 x^2/sqrt(4-x^2) +4 sqrt(4-x^2)
0 = -4x^2 + 16 - 4 x^2
8 x^2 = 16
x = sqrt 2
so domain of area is 0 to x = sqrt 2 where the area is 8
c)
L = 2 x + 2 y
= 2 x + 2 sqrt(4-x^2)
= 2 ( x + sqrt (4-x^2))
d)
Q goes from x = 0 to x = 2 in Quadrant 1
What do you mean?
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