Question

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must have a volume of 4640 cubic feet. The hemispherical ends cost twice as much per square foot of surface area as the sides. Find the dimensions that will minimize the cost. (Round your answers to three decimal places.)

Answer 1

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere  $=\frac{2}{3}\pi r^3$

Volume of a Cylinder $=\pi r^2 h$

Therefore:

The Volume of the Solid formed

$=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Area of the Hemisphere =$2\pi r^2$

Curved Surface Area of the Cylinder =$2\pi rh$

Total Surface Area=

$2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh$

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

$=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh$

Recall: $h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Therefore:

$C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}$

The minimum cost occurs at the point where the derivative equals zero.

$C^{'}=\frac{-27840+32\pi r^3}{3r^2}$

$When \:C^{'}=0$

$-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518$

Recall:

$h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet$

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet