a rectangular sheet of paper with perimeter 57 cm. is rolled into a right cylinder. if (x) represents the…
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
Answer
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<x<28.5).
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)