Question

Y 240 All boxes with a square base, an open top, and a volume of 60 ft have a surface area given by S(x)= x2 + where x is the length of the sides of the base. Find the absolute minimum of the surface area function on the interval (0,00). What are the dimensions of the box with minimum surface area? Determine the derivative of the given function S(x). 240 S'(x) = 2x- The absolute minimum value of the surface area function is 72.986 (Round to three decimal places as needed.) The dimensions of the box with minimum surface area are a base of length 4.932 ft and a height of 2.469 ft. (Round to three decimal places as needed.)

Answer 1

(120 3+3 240 x » s'(x) = 1 / 2 ( x² + 240 S(x) = x ²4 1 Aug s'cas = 2x - 240 x2 s'(x)=0 = 2x - 240 x² 2x²-240 & x²= 120 x²= 240 120 z x = (1203 S"(x) = d doc 1 - (2x - 240 = 2 +480 x² x= (120343 s" (3120) = 2+ 480 [(120)"}} 480 = 2 + = 2+ 4864 130 Thus sln) has a minimum at a= 1120 (12ots s(120)'s) ((1207²3, ²+ 240 (120) 93 = 72,986 ft

Ax² 60ff3 Go ft3 A x2 x=(120)t3 A = Go ft (120343 ft² = 2.4166 S (2120) = 72.986 ft² dimensions A=B=c=21466 ft