Calc optimization: "An ice cube that is 15 inches on each side is melting at the..."

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kamurphy

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Calc optimization: "An ice cube that is 15 inches on each side is melting at the..."

  1. An ice cube that is 15 inches on each side is melting at the rate of 1.5in3 per minute. How fast isthe length of its side decreasing?
 
kamurphy said:
  2. An ice cube that is 15 inches on each side is melting at the rate of 1.5in3 per minute. How fast isthe length of its side decreasing?
Click to expand...
What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33
 
stuck on optimization

I am completely stuck and have no idea where to begin...
 
kamurphy said:
  2. An ice cube that is 15 inches on each side is melting at the rate of 1.5in3 per minute. How fast isthe length of its side decreasing?
Click to expand...
Then decide what is meant by each side is melting at the rate of 1.5in3 per minute. Get back to us with the work you have done so far.
 
answer?

They are each decreasing at 1.5 inch^3/min, which is a total 9 inch^3/min (1.5*6 sides)?
 
kamurphy said:
  1. An ice cube that is 15 inches on each side is melting at the rate of 1.5in3 per minute. How fast isthe length of its side decreasing?
Click to expand...
That means \(\displaystyle \dfrac{dV}{dt} = -1.5\)

continue....
 
kamurphy said:
  1. An ice cube that is 15 inches on each side is melting at the rate of 1.5in3 per minute. How fast is the length of its side decreasing?
Click to expand...
You seem to be completely stuck so I will help you out this time.

I will use s for the length of the side of the cube

Each side is melting at the rate of 1.5in3 per minute. This means the V is changing with respect to time at the rate of 1.5in3 per minute. That means \(\displaystyle \dfrac{dV}{dt}= -1.5in^3\). The reason you know that V is changing (and not s or something else) is because the change is in CUBIC units (in your case cubic inches). Cubic is always Volume

You want to know how fast is the length of its side decreasing. That is you want to know \(\displaystyle \dfrac{ds}{dt}\)equals.

So you have \(\displaystyle \dfrac{dV}{dt}\) and \(\displaystyle \dfrac{ds}{dt}\). V and s!! Can you think of a formula for a cube the involves V and s. How about V=s3

Now we take the derivative of V=s3 wrt t.

This gives us \(\displaystyle \dfrac{dV}{dt}= 3s^2 \dfrac{ds}{dt}\)

Now solve for \(\displaystyle \dfrac{ds}{dt}\)

So \(\displaystyle \dfrac{ds}{dt} = \dfrac{\dfrac{dV}{dt}}{3s^2} = \dfrac{-1.5}{3s^2}= \dfrac{-1}{2s^2}\) This result to the left is the answer to your problem.

Now usually you would be asked what is the rate of change of the side of the cube when s equals some specific amount, like when s=.1 inches. So you would simply plug in .1 for s and simplify.

You need to understand that \(\displaystyle \dfrac{ds}{dt}\) is not a constant. Picture a big cube. Now to remove 1.5 in3 from this big cube can be done by removing very little from the sides of each cube. So \(\displaystyle \dfrac{ds}{dt}\) is a small negative number at this time.

Now if you wait a bit the cube will be much smaller (after all it is losing 1.5 in3 each minute). Now for volume of the cube to decrease by 1.5in3 each side will have to decrease more than with the initial big cube. So \(\displaystyle \dfrac{ds}{dt}\) is a larger negative number than before. I hope that I motivated that \(\displaystyle \dfrac{ds}{dt}\) is not constant and that it is getting more and more negative as time goes on (at least until the cube vanishes).
 
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