Answer:
The maximum volume of the open box is 24.26 cm³
Step-by-step explanation:
The volume of the box is given as [tex]V=g(x)[/tex], where [tex]g(x)=x(6-2x)(8-2x)[/tex] and [tex]0\le x\le3[/tex].
Expand the function to obtain:
[tex]g(x)=4x^3-28x^2+48x[/tex]
Differentiate wrt x to obtain:
[tex]g'(x)=12x^2-56x+48[/tex]
To find the point where the maximum value occurs, we solve
[tex]g'(x)=0[/tex]
[tex]\implies 12x^2-56x+48=0[/tex]
[tex]\implies x=1.13,x=3.54[/tex]
Discard x=3.54 because it is not within the given domain.
Apply the second derivative test to confirm the maximum critical point.
[tex]g''(x)=24x-56[/tex], [tex]g''(1.13)=24(1.13)-56=-28.88\:<\:0[/tex]
This means the maximum volume occurs at [tex]x=1.13[/tex].
Substitute [tex]x=1.13[/tex] into [tex]g(x)=x(6-2x)(8-2x)[/tex] to get the maximum volume.
[tex]g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26[/tex]
The maximum volume of the open box is 24.26 cm³
See attachment for graph.
