... for r > 0, it must be a global minimum.The cup has a radius of 4 3 √πcm and a height of 4 3 √π.106ax xxxxx∆12 3 in-2n-1bf( )ξ1Figure 4. 2: Divide -and- Conquer Strategy for ... f(r),2πr −128r2= 0,2πr 3 − 128 = 0,r = 4 3 √π.The second derivative of the surface area is f(r) = 2π +256r 3 . Since f( 4 3 √π) = 6π, r = 4 3 √πis a local minimum off(r). ... derivative does not exist and thus x = 2 is acritical point. For x < 2, f(x) < 0 and for x > 2, f(x) > 0. x = 2 is a local minimum.Solution 3. 16Let r be the radius and h the height...