Answer:
The value of x that maximizes the volume enclosed by this box is 0.46 inches
The maximum volume is 3.02 cubic inches
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
The volume of the open-topped box is equal to
[tex]V=LWH[/tex]
where
[tex]L=(7-2x)\ in\\W=(2-2x)\ in\\H=x\ in[/tex]
substitute
[tex]V=(7-2x)(2-2x)x[/tex]
Convert to expanded form
[tex]V=(7-2x)(2-2x)x\\V=(14-14x-4x+4x^{2})x\\V=14x-14x^2-4x^2+4x^{3}\\V=4x^{3}-18x^{2} +14x[/tex]
using a graphing tool
Graph the cubic equation
Remember that
The domain for x is the interval -----> (0,1)
Because
If x>1
then
the width is negative (W=2-2x)
so
The maximum is the point (0.46,3.02)
see the attached figure
therefore
The value of x that maximizes the volume enclosed by this box is 0.46 inches
The maximum volume is 3.02 cubic inches