# How to set this volume problem?

• April 6th 2009, 11:39 AM
Kaitosan
How to set this volume problem?
Let V be the volume of the region bounded between two expanding concentric spheres. at time t = 0 the spheres have radii of 6 cm and 40 cm. The radius of the smaller sphere is increasing at a rate of 8 cm/sec, while the radius of the larger sphere is increasing at a rate of 2 cm/sec. At what value of t (t>0) will the volume V be a maximum?

I had no idea to set this problem up because I don't know how to install the variable "t". The book set up the equation as v = (4/3)pi(40+2t)^3 - (4/3)pi(6+8t)^3.

I'm just hoping that someone can make sense out of the equation, how the radii were replaced by expressions with "t".
• April 6th 2009, 01:01 PM
sinewave85
Quote:

Originally Posted by Kaitosan
Let V be the volume of the region bounded between two expanding concentric spheres. at time t = 0 the spheres have radii of 6 cm and 40 cm. The radius of the smaller sphere is increasing at a rate of 8 cm/sec, while the radius of the larger sphere is increasing at a rate of 2 cm/sec. At what value of t (t>0) will the volume V be a maximum?

I had no idea to set this problem up because I don't know how to install the variable "t". The book set up the equation as v = (4/3)pi(40+2t)^3 - (4/3)pi(6+8t)^3.

I'm just hoping that someone can make sense out of the equation, how the radii were replaced by expressions with "t".

Hi, Kaitosan.

First, lets look at how they introduced t into the equation. We are told, "At time t = 0 the spheres have radii of 6 cm and 40 cm. The radius of the smaller sphere is increasing at a rate of 8 cm/sec, while the radius of the larger sphere is increasing at a rate of 2 cm/sec." That means that you can describe the radii of the circles in terms of t since the radius is a function of time. Let R be the radius of the eternal circle and r be the radius of the internal circle.

R = 40 cm(the starting radius) + 2 cm/sec (the rate of expansion) *t sec(the time) = 40 + 2t

r = 6 cm(the starting radius) + 8cm/sec (the rate of expansion) *t sec(the time) = 6 + 8t

Those values for R and r are substituted into the standard equations for the volume of a circle.

$V = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi(40 + 2t)^3 - \frac{4}{3}\pi(6 + 8t) ^3$

To find the maximum value for V, you need to differentiate that equation in order to find and test the critical numbers.

$V^{\prime} = \frac{4}{3}\pi(3)(40 + 2t)^{(3 -1)}(2) - \frac{4}{3}\pi(3)(6 + 8t)^{(3 - 1)}(8)$

$V^{\prime} = 8\pi(40 +2t)^{2} - 32\pi(6 + 8t)^{2}$

$0 = 8\pi(40 +2t)^{2} - 32\pi(6 + 8t)^{2}$

$0 = -63t^2 -56t + 364$

$t = -2.8889, 2 \mbox{ by the quadratic formula}$

The only critical value of t > 0 is 2, and testing will confirm that V is maximized at t = 2 seconds. Hope that helps.
• April 6th 2009, 04:56 PM
Kaitosan
Thanks, that helps. I'm just not used to that kind of problem.