Answer
a. $V(x)=8x(12-x)(9-x)=8(x^3-21x^2+108x)$
b. $0\lt x\lt 9$. See graph.
c. $V(3.394)\approx1310in^3$ at $x=3.384in$
d. See explanations.
e. $x_1=2in$ and $x_2=5in$
f. See explanations.
Work Step by Step
a. Use the figure given in the Exercise; we can identify the three dimensions of the box as: height=$18-2x$, width=$24-2x$ and depth=$2x$. Thus the volume is given by $V(x)=2x(24-2x)(18-2x)=8x(12-x)(9-x)=8(x^3-21x^2+108x)$
b. The dimensions need to be positive; we have $x\gt0, x\lt12, x\lt9$ which lead to $0\lt x\lt 9$ for the domain. See graph of $V(x)$ over this domain.
c. With the help of the graph, the maximum volume can be found as $V(3.394)\approx1310in^3$ at $x=3.384in$
d. Take derivative of $V(x)$ to get $V'=8(3x^2-42x+108)=24(x^2-14x+36)$. Let $V'=0$ to get $x_1=\frac{14-\sqrt {52} }{2}=7-\sqrt {13}\approx3.394in$ and $x_2=7+\sqrt {13}\approx10.6in$ (discard this value as it is outside the domain). As $V''(x_1)=8(6x_1-42)=48(x_1-7)\lt0$, the function is concave down and has a maximum at $x_1=3.394in$.
e. Let $V(x)=1120$, we have $8(x^3-21x^2+108x)=1120$ or $x^3-21x^2+108x-140=0$ and we can factor the equation with the help of a graphing calculator as $(x-2)(x-5)(x-14)=0$. Within the domain, we have $x_1=2in$ and $x_2=5in$ which both yield a volume of $1120in^3$.
f. A possible issue that arises in part (b) is the restriction of dimensions where all sides need to be positive. With $0\lt x\lt 9$, the three dimensions will be limited in ranges of depth $(0,18)$, height $(0,18)$, width $(6,24)$ where the width can not be less than $6in$.
