81.9k views
4 votes
F the height remains fixed and the side of the base is decreasing by 0.002 meter/yr, what rate is the volume decreasing when the height is 180 meters and the width is 200 meters?

1 Answer

6 votes

The original volume equation looks like this: V = 1/3 * h * (x^2)
After the side is reduced by 0.002, the new volume would look like V1 = 1/3 * h * (x-0.002) ^ 2

Then we have:
V-V1 = 1/3*h*(x^2) - 1/3*h*(x – 0.002) ^2


= 1/3 * h *(x^2 - (x – 0.002) ^2)


= 1/3 * h * (0.004x - 0.00004)


The rate of decreasing is computed by:


(V-V1)/V * 100% = [1/3 * h *(0.004x - 0.00004)] / [1/3 * h * (x ^ 2)] * 100% this would be equal to (0.004x - 0.00004) / (x^2) * 100%

So replace x by 200, you’ll get:

(0.004(200) - 0.00004) / (200^2) * 100%

= 0.001999% is the rate of decreasing.

User Sandcar
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.