Calculus question help

The Gateway Arch in St. Louis, Missouri was constructed using the hyperbolic cosne function. The equation used was y=693.8597-68.7672 cosh .0100333x, -299.2239 where x and y are measured in feet. Cross sections of the arch are equilateral trianges, and (x,y) traces the parth of the centers of the cross-sectional triangles. For each value of x, the area of the cross-sectional triangle is A=125.1406 cosh 0.0100333x.

A. How high above the ground is the center of the highest triangle?(at ground level, y=o)

B. What is the height of the arch? ( Hint: For an equilateral triangle, A= square root(3c^2), where c is one half the base of the triangle, ad the center of mass of the triangle is located at two-thirds the height of the triangle.)

C. How wide is the arch at ground level?( I know its 630ft can you show me how to get this?)

Let's simplify a little bit the things and say :

y = a - b cosh cx ;

A = d cosh cx ;

1.
The extreme values of the function (if any) are met where the derivative is null :

dy/dx = 0 ==> bc sinh cx = 0 ==> x = 0 ;
because ( d(cosh cx)/dx = c sinh cx ) ; ( sinh 0 = 0 ) ;

y(0) = a - b ; ( cosh 0 = 1 ) ;


2.
Area of a equilateral triangle is A = l² √3 / 4 ; where l is side of the triangle ;

l = √ ( 4A / √3 ) = 2√ ( A / √3 )

The height of the triangle is h = l√3 / 2 = 2√ ( A / √3 ) * √3 / 2 = √ ( A√3 ) ;

So h(x) = √ ( √3 d cosh cx )

3.
The maximum width must be the inner space between the bases of the triangles
at x = ± 299.2239 ;

So it is w = 2 * x + 2 * h(x) / 3 for x = 299.2239 ;
( the middle point is located one third of the height from the base ) ;

This verify :

w_max = 2*299.2239 + 2( √ ( √3 *125.1406 *cosh (0.0100333 *299.2239) ) ) / 3 =

= 629 . 6244 ft

You can do the other calculations .