SOLUTION: Identical squares are to be cut out of each corner of a piece of cardboard shaped like a rectangle with dimensions 5 ft. by 8 ft. The four squares are then discarded, and the sid

Algebra ->  Volume -> SOLUTION: Identical squares are to be cut out of each corner of a piece of cardboard shaped like a rectangle with dimensions 5 ft. by 8 ft. The four squares are then discarded, and the sid      Log On


   

Question 524300: Identical squares are to be cut out of each corner of a piece of cardboard shaped like a
rectangle with dimensions 5 ft. by 8 ft. The four squares are then discarded, and the sides
folded upwards to make a large box, with open top. What size square
should be cut out of each corner to maximize the volume of the box?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Starting with the formula for the volume of a rectangular prism ( a box):
V+=+L%2AW%2Ah
The dimensions of the initial piece of rectangular cardboard are given as 5ft. by 8ft. from which you will cut identical x by x squares from each corner.
So you first need to express the volume of the box in terms of the variable x.
The base of the new box can be expressed as:
%285-2x%29%288-2x%29 and the height as: x
The volume is then:
V+=+x%285-2x%29%288-2x%29 which ends up as:
V+=+4x%5E3-26x%5E2%2B40x
Now there are a couple of ways to determine the value of x required to make V a maximum:
1) Graph the cubic function above using your graphing calculator and find the relative maximum:
graph%28400%2C400%2C-5%2C5%2C-20%2C20%2C4x%5E3-26x%5E2%2B40x%29
The relative maximum of the volume function occurs at x = 1, so you can conclude the size of the squares to be cut from the corners is 1 by 1.
2) A second way is to use differential calculus by taking the first derivative of the volume function (this will give you the slope or the rate of change of the volume with respect to the x variable). Then you will set the result of this equal to zero which, when solved for x, gives you the x value at the relative maximum.
V+=+4x%5E3-26x%5E2%2B40x Take the first derivative.
dV%2Fdx+=+12x%5E2-52x%2B40 Set this equal to zero.
12x%5E2-52x%2B40+=+0 Factor a 4 to simplify.
4%283x%5E2-13x%2B10%29+=+0 so...
3x%5E2-13x%2B10+=+0 Solve the quadratic by factoring.
%283x-10%29%28x-1%29+=+0 Apply the zero product rule.
3x-10+=+0 or x-1+=+0 so...
x+=+10%2F3 or x+=+1
Discard the first solution x+=+10%2F3 as this will be too large for the initial size of the cardboard.
Solution: highlight%28x+=+1%29