Respuesta :
Answer:
The height and the radius of the cylinder are 3.67 centimeters and 5.19 centimeters, respectively.
Step-by-step explanation:
The volume ([tex]V[/tex]) and the surface area ([tex]A_{s}[/tex]) of the cone, measured in cubic centimeters and square centimeters, respectively, are modelled after these formulas:
Volume
[tex]V = \frac{h\cdot r^{2}}{3}[/tex]
Surface area
[tex]A_{s} = \pi\cdot r \cdot \sqrt{r^{2}+h^{2}}[/tex]
Where:
[tex]h[/tex] - Height of the cylinder, measured in centimeters.
[tex]r[/tex] - Radius of the base of the cylinder, measured in centimeters.
The volume of the paper drinking cup is known and first and second derivatives of the surface area functions must be found to determine the critical values such that surface area is an absolute minimum. The height as a function of volume and radius of the cylinder is:
[tex]r = \sqrt{\frac{3\cdot V}{h} }[/tex]
Now, the surface area function is expanded and simplified:
[tex]A_{s} = \pi\cdot \sqrt{\frac{3\cdot V}{h} }\cdot \sqrt{\frac{3\cdot V}{h}+ h^{2}}[/tex]
[tex]A_{s} = \pi\cdot \sqrt{\frac{9\cdot V^{2}}{h^{2}} + 3\cdot V\cdot h }[/tex]
[tex]A_{s} = \pi\cdot \sqrt{3\cdot V} \cdot\sqrt{\frac{3\cdot V+ h^{3}}{h^{2}} }[/tex]
[tex]A_{s} = \pi\cdot \sqrt{3\cdot V}\cdot \left(\frac{\sqrt{3\cdot V + h^{3}}}{h}\right)[/tex]
If [tex]V = 33\,cm^{3}[/tex], then:
[tex]A_{s} = 31.258\cdot \left(\frac{\sqrt{99+h^{3}}}{h} \right)[/tex]
The first and second derivatives of this function are require to determine the critical values that follow to a minimum amount of paper:
First derivative
[tex]A'_{s} = 31.258\cdot \left[\frac{\left(\frac{3\cdot h^{2}}{\sqrt{99+h^{2}}}\right)\cdot h - \sqrt{99+h^{3}} }{h^{2}}\right][/tex]
[tex]A'_{s} = 31.258\cdot \left(\frac{3\cdot h^{3}-99-h^{3}}{h^{2}\cdot \sqrt{99+h^{2}}} \right)[/tex]
[tex]A'_{s} = 31.258\cdot \left(\frac{2\cdot h^{3}-99}{h^{2}\cdot \sqrt{99+h^{2}}} \right)[/tex]
[tex]A'_{s} = 31.258\cdot \left[2\cdot h\cdot (99+h^{2}})^{-0.5} -99\cdot h^{-2}\cdot (99+h^{2})^{-0.5}\right][/tex]
[tex]A'_{s} = 31.258\cdot (2\cdot h - 99\cdot h^{-2})\cdot (99+h)^{-0.5}[/tex]
Second derivative
[tex]A''_{s} = 31.258\cdot \left[(2+198\cdot h^{-3})\cdot (99+h)^{-0.5}-0.5\cdot (2\cdot h - 99\cdot h^{-2})\cdot (99+h)^{-1.5}\right][/tex]
Let equalize the first derivative to zero and solve the resultant expression:
[tex]31.258\cdot (2\cdot h - 99\cdot h^{-2})\cdot (99+h)^{-0.5} = 0[/tex]
[tex]2\cdot h - 99 \cdot h^{-2} = 0[/tex]
[tex]2\cdot h^{3} - 99 = 0[/tex]
[tex]h= \sqrt[3]{\frac{99}{2} }[/tex]
[tex]h \approx 3.672\,cm[/tex]
Now, the second derivative is evaluated at the critical point:
[tex]A''_{s} = 31.258\cdot \{[2+198\cdot (3.672)^{-3}]\cdot (99+3.672)^{-0.5}-0.5\cdot [2\cdot (3.672) - 99\cdot (3.672)^{-2}]\cdot (99+3.672)^{-1.5}\}[/tex]
[tex]A''_{s} = 18.506[/tex]
According to the Second Derivative Test, this critical value leads to an absolute since its second derivative is positive.
The radius of the cylinder is: ([tex]V = 33\,cm^{3}[/tex] and [tex]h \approx 3.672\,cm[/tex])
[tex]r = \sqrt{\frac{3\cdot V}{h} }[/tex]
[tex]r = \sqrt{\frac{3\cdot (33\,cm^{3})}{3.672\,cm} }[/tex]
[tex]r \approx 5.192\,cm[/tex]
The height and the radius of the cylinder are 3.672 centimeters and 5.192 centimeters, respectively.
