An open box has a square base and a surface area of 240 square
inches. What dimensions (width × length × height) will produce a
box with maximum volume?

Question

Samantha Adkison · Accepted Answer

# 4sqrt5# inches is the length of one edge of the square base and#2 sqrt 5 # , inches the height of the box.

Let # x# be the length of one edge of the square base and #y# , the height of the box . Surface area of the box is #s= x^2+4 x y =240 or 4 x y = 240-x ^2 or y= (240-x ^2)/(4x)# or #y=60/x-x/4# . Volume of the box is #v= x^2*y# or #v= x^2(60/x-x/4) = 60 x-x^3/4# differentiating w.r.t #x# we get, #v^' = 60 - (3 x^2)/4# For maximizing #v^'=0# #60 - (3 x^2)/4= 0 or 240 - 3 x^2 =0 or 80 - x^2=0# or # ((4sqrt5)^2-x^2 )=0 or (4 sqrt5 +x)(4 sqrt5 -x)=0# #:. x = 4 sqrt 5 or x = - 4 sqrt5 # , but #x != - 4 sqrt5 # #:.x = 4 sqrt5 # When #x<4 sqrt5 , v^' >0 # , slope increasing When #x>4 sqrt5 , v^' <0 # , slope decreasing. #:.x =4 sqrt5 # gives the maximum volume. When #x= 4 sqrt 5# #y = (240-80)/(16sqrt5)= 10/sqrt5= 2 sqrt5 # # 4sqrt5# inches is the length of one edge of the square base and #2 sqrt 5 # inches , the height of the box. [Ans]

An open box has a square base and a surface area of 240 square inches. What dimensions (width × length × height) will produce a box with maximum volume?