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

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 :)

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.

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}