Volume and Inequalities

Question

An open-top box is to be constructed from a 4 in by 8 in rectangular sheet of tin by cutting out squares of equal size at each corner, then folding up the resulting flaps. Let x denote the length of the side of each cut-out square. Assume negligible thickness.
(a) Find a formula for the volume, V, of the box as a function of x. V(x)= 

(b) For what values of xx does the formula from part (a) make sense in the context of the problem?
(inequalities)

(c) On a separate piece of paper, sketch a graph of the volume function.

(d) What, approximately, is the maximum volume of the box? Give answer to at least 4 decimal places.

Answer 1

I recommend you draw this out.

 

You start out with a flat piece of tin that is 4in by 8in.  When the corners are cut off, you end up with these dimensions:

 

length = 4 - 2x

width = 8 - 2x

 

The flaps that are folded upward (which will serve as the sides of the box) have dimensions of

 

x by (4 - 2x)        and               x by (8 - 2x)

 

 

There are four flaps, so you have two equal pairs.

 

After the fold, you will have a box with a base of dimensions (4 - 2x) and (8 - 2x), which we have established earlier.

 

Knowing this, your volume is length*width*height.

 

V(x) = x(4 - 2x)(8 - 2x)

V(x) = x(32 - 24x + 4x²)

V(x) = 32x - 24x² + 4x³

 

 

b)

 

Set the volume equation equal to zero and solve for x.

 

x(4 - 2x)(8 - 2x) = 0

 

x = 0     ,   x = 2      ,       and        x = 4

 

 

Based on the shape of the positive cubic function, the acceptable values of x are in the interval

 

0 < x < 2                or          x > 4

 

Since the volume must be positive in order for the box to maintain form.

 

 

c)

 

Use the values and information in part b to sketch the graph of V(x)

 

When you graph this function you have a cubic function the increases as x-increases as an end behavior.  Since you know the shape of the graph, the maximum value occurs somewhere in the interval

 

0 < x < 2

 

The volume must have a finite maximum, since the dimensions have restrictions.  That is why we reject  x > 4. 

 

Just use test points in the accepted interval to evaluate the volume and find the maximum volume.