Question

Consider the function V=g(x), where g(x) =x(6-2x)(8-2x), with x being the length of a cutout in cm and V being the volume of an open box in cm³, where x must between 0 and 3 cm. Determine the maximum volume of the open box in cm³. Round your answer to two decimal places. Also, indicate or show what process you used to obtain this answer. (Hint: graphing)

Answer 1

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as $V=g(x)$, where $g(x)=x(6-2x)(8-2x)$ and $0\le x\le3$.

Expand the function to obtain:

$g(x)=4x^3-28x^2+48x$

Differentiate  wrt  x to obtain:

$g'(x)=12x^2-56x+48$

To find the point where the maximum value occurs, we solve

$g'(x)=0$

$\implies 12x^2-56x+48=0$

$\implies x=1.13,x=3.54$

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

$g''(x)=24x-56$, $g''(1.13)=24(1.13)-56=-28.88\:$

This means the maximum volume occurs at $x=1.13$.

Substitute $x=1.13$ into $g(x)=x(6-2x)(8-2x)$ to get the maximum volume.

$g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26$

The maximum volume of the open box is 24.26 cm³

See attachment for graph.