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
so its this picture of a rectangle inscribed in a circle. the circle has a radius of 2. the vertexes of the circle are present in every quadrant, but you cant tell exactly whgat point they represent. point Q (x,y) is the vertex of the rectangle in Quadrant I and is on the circle. the equation of the circle is : x^2 + y^2 = 4
a.) express the area of the rectangle as a function of x.
b.) give the domain of the area
c.) epxress the perimeter as a function of x.
d.) give the domain of Q
PLEASE HELP IM SO LOST. and friday afternoon is my scheduled math homework day. (dont ask im psycho) please this is bother ing me like crazyy
2 answers
What do you mean?
1 answer