(a).
\Length of the box.
Height of the box.
Width of the box.
Volume of the box, .
.
.
.
The polynomial function is .
(b).
\Length of the box must be .
Height of the box must be .
Width of the box must be or
.
Therefore, the domain of is
.
Subustiute the values of in the obtained polynomial equation to get the values of
.
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
\
| \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
![]() | \
Graph :
\(1) Draw the coordinate plane.
\(2) Plot the point obtained in the above table.
\(3) Connect the points with the smooth curve.
\
(c).
\The company wants the box to have volume of .
Substiute the value of in the obtained polynomial equation
.
.
(d).
\.
when
.
.
Use synthetic substitution to verify .
The remainder is .
So is a factor of
.
Therefore, is a solution for
.
(a).
\The polynomial function is .
(b).
\Graph :
\(c).
\.
(d).
\ is a solution for
.