Question

a rectangular sheet of paper with perimeter 57 cm. is rolled into a right cylinder. if (x) represents the length of the paper, that then becomes the circumference of the open - base of the cylinder. the volume of the cylinder is given by: (v=pi(\frac{x}{2pi})^2(\frac{57 - 2x}{2})). using a graphing calculator, determine the following: a) what is the domain of the function? b) the length of the paper, (x), that maximizes the volume of the cylinder. (x=) c) the maximum volume of the cylinder. maximum volume = question help: video post to forum

Explanation:

Step1: Analyze domain constraints

The length \(x>0\) and the width \(\frac{57 - 2x}{2}>0\). Solving \(\frac{57 - 2x}{2}>0\) gives \(57-2x > 0\), then \(2x<57\), \(x < 28.5\). So the domain is \(0

Step2: Use graphing - calculator for maximum

Enter the volume function \(V=\pi(\frac{x}{2\pi})^2(\frac{57 - 2x}{2})\) into a graphing - calculator. Find the \(x\) - value of the maximum point on the graph in the domain \(0 < x<28.5\). After using the graphing - calculator's maximum - finding feature, we get \(x\approx19\).

Step3: Find maximum volume

Substitute \(x = 19\) into the volume function \(V=\pi(\frac{19}{2\pi})^2(\frac{57-2\times19}{2})\). First, \(\frac{19}{2\pi}\) is the radius \(r\) of the base, and \(\frac{57 - 2\times19}{2}=\frac{57 - 38}{2}=\frac{19}{2}\) is the height \(h\) of the cylinder. \(V=\pi\times\frac{19^{2}}{4\pi^{2}}\times\frac{19}{2}=\frac{19^{3}}{8\pi}\approx272.9\).

Answer:

a) \(0 < x<28.5\)
b) \(19\)
c) \(272.9\)