please help... i need this find value math?

Question

see the pic it
please help i need this
http://img527.imageshack.us/my.php?image=maxareauf7.png

Area please help?
consider a symmetric cross inscrived in a circle of radius r

a)Write the area A of the cross as a function of x find the value of x that maximizes the area

b)Write the area A of the cross as a function of @ find the value of @ that maximizes the area

c)Show that the critical numbers of parts a and b yield the same maximum are. What is that area?

Answer 1

I found it easier to work with one eighth of the circle. Draw the sketch and draw the line y = x And see that the area of the shape in the first quadrant below the line y = x can be written as follows:
A = x√(r^2 - x^2) - (1/2)(r^2 - x^2)
A = x^2/2 - r^2/2 + x√(r^2 - x^2)
Take the derivative and set it equal to zero:
A ' = x + x(1/2)(r^2 - x^2)^(-1/2)(-2x) + √(r^2 - x^2)
A ' = x - x^2/√(r^2 - x^2) + √(r^2 - x^2) = 0
x√(r^2 - x^2) - x^2 + r^2 = 0
-2x^2 + r^2 + x√(r^2 - x^2)(r^2 - x^2) = 0
x√(r^2 - x^2) = 2x^2 - r^2
Square both sides:
x^2(r^2 - x^2) = 4x^4 - 4r^2x^2 + r^4
r^2x^2 - x^4 = 4x^4 - 4r^2x^2 + r^4
5x^4 - 5r^2x^2 + r^4 = 0
Let w = x^2:
5w^2 - 5r^2 + r^4 = 0
Use the quadratic formula:
w = [(5^r^2 +/- √(25r^4 - 20r^4)]/10
w = r^2/2 +/- (r^2/10)√5
x = +/- √[r^2/2 +/- (√5)/10]
x = r√[(1/2) +/- (√5)/10]
x = r√(1/2 +/- √5/10)
Θ = 2(arc cos(x/r))
Θ/2 = arc cos(x/r)
The 2 values for x yield values for Θ/2 that are compliments of eachother. Due to symmetry about the line y = x these 2 values denote the same solution.
Substituting the smaller value for x in the original equation gives us(after some messy simplification):
A = r^2[-√5/10 + (1/2)√(1/2 - √5/10)]
Since this is 1/8th of the shape it needs to be multiplied by 8:
A(total) = 8[-√5/10 + (1/2)√(1/2 - √5/10)]r^2
[Interestingly this calculates to 2.417r^2. Dividing that by π shows it's 76.9% of the circle]