Respuesta :
Answer:
The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.
Step-by-step explanation:
The volume ([tex]V[/tex]), in cubic centimeters, and surface area ([tex]A_{s}[/tex]), in square centimeters, formulas for the candle are described below:
[tex]V = \pi\cdot r^{2}\cdot h[/tex] (1)
[tex]A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h[/tex] (2)
Where:
[tex]r[/tex] - Radius, in centimeters.
[tex]h[/tex] - Height, in centimeters.
By (1) we have an expression of the height in terms of the volume and the radius of the candle:
[tex]h = \frac{V}{\pi\cdot r^{2}}[/tex]
By substitution in (2) we get the following formula:
[tex]A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)[/tex]
[tex]A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}[/tex]
Then, we derive the formulas for the First and Second Derivative Tests:
First Derivative Test
[tex]4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0[/tex]
[tex]4\pi\cdot r^{3} - 2\cdot V = 0[/tex]
[tex]2\pi\cdot r^{3} = V[/tex]
[tex]r = \sqrt[3]{\frac{V}{2\pi} }[/tex]
There is just one result, since volume is a positive variable.
Second Derivative Test
[tex]A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}[/tex]
If [tex]\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)[/tex]:
[tex]A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }[/tex]
[tex]A_{s} = 12\pi[/tex] (which means that the critical value leads to a minimum)
If we know that [tex]V = 3217\,cm^{3}[/tex], then the dimensions for the minimum amount of plastic are:
[tex]r = \sqrt[3]{\frac{V}{2\pi} }[/tex]
[tex]r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}[/tex]
[tex]r = 8\,cm[/tex]
[tex]h = \frac{V}{\pi\cdot r^{2}}[/tex]
[tex]h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}[/tex]
[tex]h = 16\,cm[/tex]
And the amount of plastic needed to cover the outside of the candle for packaging is:
[tex]A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h[/tex]
[tex]A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)[/tex]
[tex]A_{s} \approx 1206.372\,cm^{2}[/tex]
The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.