Applications question!

[ruchika5]

A piece of wire 60cm long is cut into 2 pieces which are bent in order to create a square and a circle. If the square has side length 'x' cm

a) Find an expression for the radius of the circle in terms of x

b) Create an expression for the sum,S, of the two areas, in terms of x, giving the domain of S.

c) Not using calculus, find the length of the two pieces of wire in order to minimize the sum of the two areas.

Not quite sure if I'm on the right track with this question, I need help with part C.
I don't know how to use Latex at the moment but here goes..

for a) I wrote down     r=(30/2x)/π

b)S= x^2 +πr^2
   =x^2+ π((30/2x)/π)^2
   =x^2+ 1/π (30/2x)^2

not sure about the domain

Thanks!

[xZero]

correct me if im wrong but
a)

b) (we cant have negative length and the maximum for x is 4x=60, so x=15)

[luffy]

a) Perimeter of square + Circumference of circle = 60 cm
 Perimeter of square = x side length
                            = x
                            =
Therefore,  Circumference of circle perimeter of square  , where r = radius of circle
                Circumference of circle
              
               Radius of circle

b) A (of square)    
    A (of circle)      
                          
Expanding it, gives:
Simplifying further:
Thus, the sum, S,  
                          
                          
c) For part C, the latex would take me time. The way I would approach this question is to put the function, S(x) into turning point form (as it is a quadratic equation) using completing the square. Once that is done, you can easily find the minimum value. If you want more depth, let me know and I'll put up the working out.

Also, if I made an error, let me know and I will modify the post.