Question

i need help with part b please

-а ses lome Score: 0.86 of 1 pt 12 of 14 (12 completa Question Help o 4.5.35 a Find the radius and height of a cylindrical soda can with a volume of 322 cm' that minimize the surface area. b. Compare your answer in part (a) to a real soda can, which has a volume of 322 cm'. a radius of 3.1 cm, and a height of 109 cm, to conclude that real soda cans do not seem to have an optimal design. Then use the fact that real soda cans have a double thickness in their top and bottom surfaces to find the radius and haight that minimizes the surface area of a real can (the surface areas of the top and bottom are now twice their values in part (a)). Are these dimensions closer to the dimensions of a real soda can? a. Write a function for the surface area of the soda can, S, in terms of the radius, 322 S2 an bok (Type an expression) sive eBo Book Conten The interval of interest of the objective function is (0.00) (Simplify your answer. Type your answer in interval notation) The radius is approximately 3.71 cm. (Round to two decimal places as needed.) The height is approximately 743 cm. (Round to two decimal places as needed) b. The new radius is approximately cm. (Round to two decimal places as needed.) or Succes edia libi ase Optio Bible Reso More Enter your answer in the answer box and then click Check Answer ssions 2 remaining Clear AL Check Awe

Answer 1

Soin we know volume is given by TTV²h = v Here 1 Tr2h = 322 h= 322 IT Y2 Surface Area 212 + 2 trh S(r) 2 TY ( 3230 322 TTY 2 +2 Trz ons To minimize the surface area s'iy)=0. 4 TY - 644 ra 4T r = 644 82 23 644 ЧП 3. 115 cm also using formula h= 322 17.43cm IT (3.715) (b) The real soda bottle has radius = 3.1 cm height = 10.9 cm then surface Area = 212 + 2 trh 2 IT (3.1)2+ 21 (3.1)x(109) = 21 (9.61) R380R) 2 (33-79) - 21 (9.61 + 33.79) = 272.552

while the S.A of (a) part es S = 2tr²+ 2 trh = 2X1T X(3-71) + 2 XTT* (3.71) -43) 21 (13-764) + 21 (27.5653) = 259.5480.cm². By this we say that the real can sodo has no optimal design. of the surface area of top and bottom is tunce than of part (a) then : - can 860 = Arcke 40DX 4329 SO) = 4 UY? + 644 r S'(o)= 818- 644 82 73 644 8 IT 8= 2.948 new one your remaining answer also h= Geope 322 11. 794 TT12-948)2