# What is the answer, correct to one decimal place, for the following problem?You are given a piece of sheet metal that is twice as long as it is wide and has an area of 800 m^2. Find the dimensions...

What is the answer, correct to one decimal place, for the following problem?

You are given a piece of sheet metal that is twice as long as it is wide and has an area of 800 m^2. Find the dimensions of the rectangular box that would contain a **maximum** volume if it were constructed from this piece of metal by cutting out squares of equal area at all four corners and folding up the sides. The box will not have a lid.

### 1 Answer | Add Yours

The area of the sheet metal is 800 m^2. The length is twice as long as the the width. Let the length be L. If the width is W, L = 2W.

800 = 2*W*W

=> W = sqrt(400)

=> W = 20

L = 40

Four squares are cut from the corners of the sheet of 40 x 20 cm and a box is constructed without a lid. Let the squares that are cut have a side of length s.

The volume of the box is : V = L*W*H, where L, W and H are the length, width and height of the the box.

The Length is (40 - 2s)

Width = 20 - 2s

HeightÂ = s

Volume = (40 - 2s)(20 - 2s)(s)

=> (40s - 2s^2)(20 - 2s)

=> 800s - 40s^2 - 80s^2 + 4s^3

=> 4s^3 - 120s^2 + 800s

To maximize volume find the derivative and solve for s.

12s^2 - 240s + 800 = 0

=> 3s^2 - 60s + 200 = 0

s1 = 15.77

s2 = 4.22

There are two roots of the equation, one giving a minimum point and the other the maximum point. The second derivative of the volume which is 24s - 240 is used to determine which of the values of s give the maximum. 24s - 240 has a negative value for the value of s that gives a maximum value.

For s = 15.7, 24s - 240 = 138.48

and for s = 4.2 , 24s - 240 = -138.72

This indicates that volume is minimum if s = 15.77 and maximum if s = 4.2.

For s = 4.2, the dimensions of the box are: 31.6, 11.6 and 4.2

**The box has a maximum volume when the dimensions are 31.6, 11.6 and 4.2 cm**