Answer
a. $V(x)=x(5-x)(15-2x)=2x^3-25x^2+75x$
b. Domain $0\lt x\lt 5$, see graph.
c. $V(1.962)=66.019in^3$ with $x=1.962in$
d. See explanations.
Work Step by Step
a. Use the figure given in the Exercise; the three dimensions of the box are: height=$x$, depth=$10-2x$, width=$(15-2x)/2$. Thus the volume is given as $V(x)=x(10-2x)(15-2x)/2=x(5-x)(15-2x)=2x^3-25x^2+75x$
b. The three sides of the box need to be positive and we have $x\gt0, x\lt5, x\lt15/2$, which leads to $0\lt x\lt 5$ for the domain. See graph for $V(x)$ in the domain.
c. With the help of the graph, the maximum volume can be found as $V(1.962)=66.019in^3$ with $x=1.962in$
d. Take the derivative of $V(x)$ to get $V'=6x^2-50x+75$. Letting $V'=0$, we get $x_1=\frac{25-5\sqrt 7}{6}\approx1.962cm$ and $x_2=\frac{25+5\sqrt 7}{6}\approx6.371cm$ where only $x_1$ is in the domain. As $V''(1.962)=12x-50=12(1.962)-50\lt0$, the function is concave down and has a maximum at $x_1=1.962cm$.
