SOLUTION: An open-top box is to be constructed from a 6 by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of e

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: An open-top box is to be constructed from a 6 by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of e      Log On


   

Question 52505: An open-top box is to be constructed from a 6 by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of each side of the square to be cut out.
a) Find the function V that represents the volume of the box in terms of x.
Answer



b) Graph this function and show the graph over the valid range of the variable x..
Show Graph here



c) Using the graph, what is the value of x that will produce the maximum volume?
Answer

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
To find the volume of such a box, multiply the width by the length by the height.
The width, after cutting out the corner squares of x by x feet, will be (6-2x) feet and the length will be (8-2x) feet, and, of course, the height of the box will be x feet. So, the volume is expressed by:
V+=+%288-2x%29%286-2x%29x Expand this to get:
V+=+%2848-16x-12x%2B4x%5E2%29x Simplifying this:
V+=+4x%5E3-28x%5E2%2B48x This represents the volume of the box expressed in terms of x.
The graph of this cubic function looks like:
graph%28300%2C200%2C-5%2C6%2C-5%2C25%2C4x%5E3-28x%5E2%2B48x%29
The valid range of x is x = 0 to x = 3
Using the graph to find the value of x that will produce the maximum volume is a little more than x = 1.
The actual value can be found with a little elementary differential calculus, because it's a matter of finding the "relative" maximum of the cubic curve.
Take the first derivative of the function and set it equal to zero, then solve for x.
dV%2Fdx+=+12x%5E2-56x%2B48 Set this equal to zero.
12x%5E2-56x%2B48+=+0
The roots are:
x = 3.535... Ignore this solution as x is too large.
x = 1.131... This is the approximate value of x that will produce the largest volume.