JAMES K. answered • 04/04/18

Tutor

5.0
(288)
Jim K - Math and Science

S=kh

^{2}w and from the circular log w^{2}+h^{2}= d^{2}.So, w = (d

^{2}-h^{2})^{.5},substitute this into the equation for S ...S = kh

^{2}(d^{2}-h^{2})^{.5}Now, differentiate this with respect to h...dS/dh = [kh

^{2 }(1/2 * (d^{2}-h^{2})^{-1/2}* (-2h)] + [(d^{2}-h^{2})^{1/2}* 2kh] = -2kh

^{3}/(d^{2}-h^{2})^{.5}^{ }+ 2kh(d^{2}-h^{2})^{.5}set this equal to zero to find the critical points, and solve ..

2kh

^{3}/(d^{2}-h^{2})^{.5}= 2kh(d-h^{2})^{.5}a little algebra gets you to ..

h

^{3}= 2hd^{2}-2h^{3 }then to ..3h

^{3 }- 2hd^{2}= 0 factor out an h ..h(3h

^{2}-2d^{2}) = 0 now set (3h^{2}-2d^{2}) = 0 , a little more algebra we get...h

^{2}= 2d^{2 }so h = d(2/3)^{.5}d = 29, so ..h = 29 (2/3)

^{.5 }= 23.68 inw = (29

^{2}- 23.68^{2})^{.5}= (280.257)^{.5 }= 16.74 in.