Question

If an open box has a square base and a volume of 115 in.3 and is constructed from a tin sheet, find the dimensions of the box, assuming a minimum amount of material is used in its construction. (round your answers to two decimal places.)

Answer 1

Let h = height of the box,
x = side length of the base.

Volume of the box is
`V=x^(2) h = 115`.
So
`h = (115)/( x^(2) )`

Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is

`S = 2( x^(2) + hx + hx) \\ = 2 x^(2) + 4hx \\ = 2 x^(2) + 4( (115)/( x^(2) ) )x`

`= 2 x^(2) + (460)/(x)`
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.


`S' = 4x- (460)/( x^(2) ) = 0`

`4x = (460)/( x^(2) ) \\ 4x^(3) = 460 \\ x^(3) = 115 \\ x = \sqrt[3]{115} = 4.86`
Then
`h= (115)/(4.86^(2)) = 4.87`
So the box is 4.86 in. wide and 4.87 in. high.

Answer 2

length = 6.13 Width = 6.13 Height = 3.06 Let's use the variable L to represent the length of a bottom edge of the box. With that, the height of the box will be 115 / w^2. Since the box is open, the box will have a bottom, and 4 sides. No top. So the expression for the surface area of the box will be w^2 + 4w(115/w^2) = w^2 + 460/w Since we're looking for a local minimum, that will be at a point where the first derivative of the function is 0. So the first derivative is: 2w - 460/w^2 = 2w^3/w^2 - 460/w^2 = (2w^3 - 460)/w^2 = 2(w^3 - 230)/w^2 The above derivative has a real zero at the cube root of 230 which is approximately 6.126926 So the length of the sides of the box are 6.13 inches and the height of the box is 115/(6.13^2) = 3.06 inches.