# Thread: Help Related rates

1. ## Help Related rates

Hi people this is my last problem so sorry for asking too much because its our finals day tommorow...

1.) A conical tent with no floor is to have a capacity of 1000cubic meters. Find the dimensions that minimize the amount of canvas required:

2.) A box with a square base has an open top. The area of the material in this box is 100cm^2. What should be the dimensions in order to make the volume as large as possible?

Just give me formula Ill be differentiating this:

THanks SOOO much for all the big help!
My Next bunch of questions will be about integrals (i hope if i pass)

2. Originally Posted by ^_^Engineer_Adam^_^
Hi people this is my last problem so sorry for asking too much because its our finals day tommorow...

1.) A conical tent with no floor is to have a capacity of 1000cubic meters. Find the dimensions that minimize the amount of canvas required:
The curved surface area of a cone is slant height * perimiter of base/2,

so if the height is h, and the radius of the base is r, we have slant height:

s = sqrt(h^2+r^2)

So the area of canvas required is:

A = sqrt(h^2+r^2) (2 pi r)/2

The volume is:

V = (1/3) (pi r^2) h = 1000,

so:

h = 3000/(pi r^2),

and area of canvas:

A = sqrt([3000/(pi r^2)]^2+r^2) (2 pi r)/2

But if A is minimise then so is A^2, so you may as well minimise:

A^2 = ([3000/(pi r^2)]^2+r^2) pi^2 r^2

RonL

3. Originally Posted by ^_^Engineer_Adam^_^
2.) A box with a square base has an open top. The area of the material in this box is 100cm^2. What should be the dimensions in order to make the volume as large as possible?
Area of material is:

A= b^2 + 4*b*h = 100,

where h is the height of the box and b the side of the base in cm.

So:

h = [100 - b^2]/(4 b) ... (1)

The volume is:

V = b^2 h = b [100 - b^2]/4 ... (2)

So to find the b that maximises the volume we solve dV/db=0 (determining
which solutions correspond to maxima/minima if necessary), then solve for
h using (1)

RonL