You can
put this solution on YOUR website!An open-top box is to be constructed
from a 6 foot 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. Find the function V that represents the
volume of the box in terms of x. Graph this
function and show the graph over the valid
range of the variable x...
Using the graph, what is the value of x that
will produce the maximum volume?
We start out with this:
____________________________
| 8' |
| |
| |
|6' |6'
| |
| |
| |
| |
|____________________________|
8'
Then we mark an x by x square on each
corner like this, which makes the distance
between the squares on the longer side become
8-2x, and on the shorter side 6-2x:
_____________________________
| x | 8-2x | x |x
|____|x x|____|
| |
| |
| |6-2x
| |
|____ ____|
| x | | x |x
|____|__________________|____|
Notice that to have a square at
all, x must be greater than 0,
and x must be less than half
of the shorter side, or 3. So
the valid range (actually called
the "domain") of x is (0, 3)
Now when we cut the squares
out, we have this:
____________________
| 8-2x |x
____|x |____
| x x |
| |
| |6-2x
| |
|____ ____|
x |x | x
|__________________|x
Now we mark some dotted line to fold
the sides up
____________________
| 8-2x |x
____|x_ _ _ _ _ _ _ _ _|____
| x | 8-2x | x |
| | | |
| | 6-2x| |6-2x
| | | |
|____| _ _ _ _ _ _ _ _ _|____|
x |x | x
|__________________|x
When we fold the sides up, the
dimensions of the box will be
8-2x feet wide, 6-2x units high
and x units high.
V = length × width × height
V = (8-2x)(6-2x)x
V = x(8-2x)(6-2x)
V = (8x-2x²)(6-2x)
V = 48x-16x²-12x²+4x³
V = 4x³-28x²+48x
We get some points between 0 and 3:
x | y
-----------
.2 | 8.512
.4 | 14.976
.6 | 19.584
.8 | 22.528
1.0 | 24.0
1.1 | 24.244
1.2 | 24.192
1.3 | 23.868
1.4 | 23.296
1.6 | 21.504
1.8 | 19.008
2.0 | 16.0
2.5 | 7.5
2.9 | 1.276
and plot then and draw
a smooth curve through
then we get
However the graph is meaningless except when x is
between 0 and 3, so we'll chop the graph off there:
We see that as we cut out
larger and larger squares
the volume V grows to
24.244 ft² when we cut a
1.2 ft square, but if we
cut out a larger square
the volume begins to
shrink:
If you are taking algebra
and not calculus, then all
you can do is estimate from
the graph that the highest
point on the graph is where
x is about 1.1 feet, as is
shown by the vertical green
line below:
So, if you are taking algebra
and not calculus, you can only
estimate that the value of x
that will produce the most
volume is about 1.1 feet.
Now if you are taking
calculus, you will be
able to calculate this
value of x precisely. Ignore
the following if you are only
taking algebra:
V = 4x³-28x²+48x
Taking the derivative of V
with respect to x, we get
V' = 12x²-56x+48
Setting this derivative = 0,
we have
12x² - 56x + 48 = 0
Dividing through by 4
3x² - 14x + 12 = 0
Use the quadratic formula:
______
-b ± Öb²-4ac
x = —————————————
2a
where a = 3; b = -14; c = 12
_______________
-(-14) ± Ö(-14)²-4(3)(12)
x = ——————————————————————————
2(3)
_______
14 ± Ö196-144
x = ——————————————
6
__
14 ± Ö52
x = ——————————
6
_____
14 ± Ö4·13
x = —————————————
6
__
14 ± 2Ö13
x = ——————————
6
__
14 2Ö13
x = ———— ± —————
6 6
__
x = 7/3 ± Ö13/3
__
Using the +, x = 7/3 + Ö13/3, which
is one solution and equals about 3.535183758 which is outside
the domain for x., and so must be ignored.
__
Using the -, x = 7/3 - Ö13/3, which
is one solution and equals about 1.131482908 which is the
value we are looking for. So the exact answer is
__
x = 7/3 - Ö13/3
__
x = (7 - Ö13)/3 or about
x = 1.131482908 feet.
Edwin
