View Full Version : Diff. Calculus- application of derivatives
totoname
Sep 22, 2009, 02:50 AM
1. A closed cylindrical tank has a capacity of 16 pi cu.m. determine the radius of the tank that requires the minimum amount of material used.
4. with only 381.7m squared of materials, a closed cylindrical tank of max. volume is to be made. What should be the height of the tank in meters.
:D
Unknown008
Sep 22, 2009, 08:10 AM
1. The volume is 16\pi \,m^3
The volume of a cylinder is given by V = \pi r^2h
The area (material to be used) is given by A = 2\pi r^2 + 2\pi rh
Make 'h' in the volume formula the subject of the formula, so that you replace 'h' in the second formula:
V = \pi r^2h
\frac{V}{\pi r^2} = h
A = 2\pi r^2 + 2\pi rh
A = 2\pi r^2 + 2\pi r(\frac{V}{\pi r^2})
You know the value of V. Simplify A, then differentiate A with respect to r. Set the differentiation to zero and solve for r.
2. The volume of a cylinder is V = \pi r^2h
Its area A = 2\pi r^2 + 2\pi rh
You need to find the max of 'V', that means you have to differentiate V with respect to r. However, you don't have the value of 'h', so make 'h' the subject of formula in the area formula, then replace that in the volume formula.
A = 2\pi r^2 + 2\pi rh
\frac{A - 2\pi r^2}{2\pi r} = h
V = \pi r^2h
V = \pi r^2(\frac{A - 2\pi r^2}{2\pi r})
Differentiate with respect to r, then set it to zero and solve for r.
I hope it helped! :)
Post back your answers :)
totoname
Sep 26, 2009, 09:32 PM
1. The volume is 16\pi \,m^3
The volume of a cylinder is given by V = \pi r^2h
The area (material to be used) is given by A = 2\pi r^2 + 2\pi rh
Make 'h' in the volume formula the subject of the formula, so that you replace 'h' in the second formula:
V = \pi r^2h
\frac{V}{\pi r^2} = h
A = 2\pi r^2 + 2\pi rh
A = 2\pi r^2 + 2\pi r(\frac{V}{\pi r^2})
You know the value of V. Simplify A, then differentiate A with respect to r. Set the differentiation to zero and solve for r.
2. The volume of a cylinder is V = \pi r^2h
Its area A = 2\pi r^2 + 2\pi rh
You need to find the max of 'V', that means you have to differentiate V with respect to r. However, you don't have the value of 'h', so make 'h' the subject of formula in the area formula, then replace that in the volume formula.
A = 2\pi r^2 + 2\pi rh
\frac{A - 2\pi r^2}{2\pi r} = h
V = \pi r^2h
V = \pi r^2(\frac{A - 2\pi r^2}{2\pi r})
Differentiate with respect to r, then set it to zero and solve for r.
I hope it helped! :)
Post back your answers :)
thanks.
however I'm having problem in identifying the derivatives.
Unknown008
Sep 26, 2009, 09:52 PM
Ok, let me do the first one for you. Then, try the second one.
\begin{array}{rcl}A &=& 2\pi r^2 + 2\pi r (\frac{V}{\pi r^2}\\ &=& 2\pi r^2 + \frac{2(16\pi)}{r}\\ &=& 2\pi r^2 + \frac{32\pi}{r}\\ &=& 2\pi r^2 + 32\pi r^{-1} \end{array}
\frac{dA}{dr} = 4\pi r - 32 \pi r^{-2}\\
\begin{array}{rcl}0 &=& 4\pi r - 32 \pi r^{-2}\\
0 &=& 4\pi r^3 - 32 \pi \\
32\pi &=& 4\pi r^3\\
r^3 &=& \frac{32\pi}{4\pi} \\
r^3 &=& 8\\ r &=&2 \end{array}