Answer:
The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.
Explanation:
Volume of the Cylinder=400 cm³
Volume of a Cylinder=πr²h
Therefore: πr²h=400
[TeX]h=\frac{400}{\pi r^2}[/TeX]
Total Surface Area of a Cylinder=2πr²+2πrh
Cost of the materials for the Top and Bottom=0.06 cents per square centimeter
Cost of the materials for the sides=0.03 cents per square centimeter
Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)
C=0.12πr²+0.06πrh
Recall: [TeX]h=\frac{400}{\pi r^2}[/TeX]
Therefore:
[TeX]C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})[/TeX]
[TeX]C(r)=0.12\pi r^2+\frac{24}{r}[/TeX]
[TeX]C(r)=\frac{0.12\pi r^3+24}{r}[/TeX]
The minimum cost occurs when the derivative of the Cost =0.
[TeX]C^{'}(r)=\frac{6\pi r^3-600}{25r^2}[/TeX]
[TeX]6\pi r^3-600=0[/TeX]
[TeX]6\pi r^3=600[/TeX]
[TeX]\pi r^3=100[/TeX]
[TeX]r^3=\frac{100}{\pi}[/TeX]
[TeX]r^3=31.83[/TeX]
r=3.17 cm
Recall that:
[TeX]h=\frac{400}{\pi r^2}[/TeX]
[TeX]h=\frac{400}{\pi *3.17^2}[/TeX]
h=12.67cm
The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.