Largest container box

The volume of the box is $v=x(18-2x)^2=x(18^2-72x+4x^2)=324x-72x^2+4x^3$ . Then take the first derivative with respect to $x$ .

$\dfrac{dv}{dx}=324-144x+12x^2$

The first derivative must be equal to zero. We have

$\dfrac{dv}{dx}=0$

$12x^2-144x+324=0$

$x^2-12x+27=0$

Solve for $x$ by using the quadratic formula. We have

$x=\dfrac{12\pm\sqrt{144-4(27)}}{2}$

$x=6 \pm 3$

The values of $x$ are:

$x=6+3=9$

and

$x=6-3=3$

$x=9$ can not be used because $18-2x=0$ , so we used $x=3$ .

So the desired volume is $v=3[18-2(3)]^2=432~cubic~inches$

Second Derivative Test:

$\dfrac{d^2v}{dx^2}=-144+24x$

when $x=3$

$\dfrac{d^2v}{dx^2}=-144+24(3)=-72$

$-72 <0$ , therefore, $v=432~cubic~inches$ is a maximum.