why discard the pie and the denominator 3 after finding the derivative?

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urimagic

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Hi friends,

I trust you are all well. Please could someone explain to me why is it that, after the step dV/dx, the pie and the denominator 3 is discarded?..The derivative of this expression had ALREADY been done in the step just above, so afterwards, whatever remains, remains....but the pie and the 3's are taken out...what rule allows for this?...any assistance will be greatly appreciated..thank you all..IMG20230504120232[1].jpg
 
zero divided by any number = zero. Also zero multiplied by any number = zero.

You have this:

\(\displaystyle 0 = \frac{64\pi}{3} - \frac{16\pi}{3}x - \frac{3\pi}{3}x^2\)

multiply every term by \(\displaystyle 3\) and also divide every term by \(\displaystyle \pi\)

\(\displaystyle \left(0 \times \frac{3}{\pi}\right) = \left(\frac{64\pi}{3} \times \frac{3}{\pi}\right) - \left(\frac{16\pi}{3}x \times \frac{3}{\pi}\right) - \left(\frac{3\pi}{3}x^2 \times \frac{3}{\pi}\right)\)

On the left side, you will have only \(\displaystyle 0\) and on the right side \(\displaystyle \pi\) and \(\displaystyle 3\) will be cancelled.

You will have now:

\(\displaystyle 0 = 64 - 16x - 3x^2\)
 
Presumably, you're trying to find a max/min point, so you're setting the derivative equal to zero. They just omitted the step, after the setting equal to zero, where they multiplied through by the constant [imath]\frac{3}{\pi}[/imath].

Eliz.
 
nasi112 said:
zero divided by any number = zero. Also zero multiplied by any number = zero.

You have this:

\(\displaystyle 0 = \frac{64\pi}{3} - \frac{16\pi}{3}x - \frac{3\pi}{3}x^2\)

multiply every term by \(\displaystyle 3\) and also divide every term by \(\displaystyle \pi\)

\(\displaystyle \left(0 \times \frac{3}{\pi}\right) = \left(\frac{64\pi}{3} \times \frac{3}{\pi}\right) - \left(\frac{16\pi}{3}x \times \frac{3}{\pi}\right) - \left(\frac{3\pi}{3}x^2 \times \frac{3}{\pi}\right)\)

On the left side, you will have only \(\displaystyle 0\) and on the right side \(\displaystyle \pi\) and \(\displaystyle 3\) will be cancelled.

You will have now:

\(\displaystyle 0 = 64 - 16x - 3x^2\)
Click to expand...
Hi nasi112,

So the reason for doing all that is basically to simplify that expression?
 
\(\displaystyle \frac{64\pi}{3} - \frac{16\pi}{3}x - \frac{3\pi}{3}x^2 = \frac{\pi}{3}[64 - 16x - 3x^2] = 0\)

So either \(\displaystyle \frac{\pi}{3}=0\ or\ 64 - 16x - 3x^2 = 0\) Well \(\displaystyle \frac{\pi}{3} \neq 0\) so it must be that \(\displaystyle 64 - 16x - 3x^2= 0\)
 
Steven G said:
\(\displaystyle \frac{64\pi}{3} - \frac{16\pi}{3}x - \frac{3\pi}{3}x^2 = \frac{\pi}{3}[64 - 16x - 3x^2] = 0\)

So either \(\displaystyle \frac{\pi}{3}=0\ or\ 64 - 16x - 3x^2 = 0\) Well \(\displaystyle \frac{\pi}{3} \neq 0\) so it must be that \(\displaystyle 64 - 16x - 3x^2= 0\)
Click to expand...
OOOHHHH beautiful.....I like this!!..Thanks Steven G..I do appreciate!
 
Steven G said:
\(\displaystyle \frac{64\pi}{3} - \frac{16\pi}{3}x - \frac{3\pi}{3}x^2 = \frac{\pi}{3}[64 - 16x - 3x^2] = 0\)

So either \(\displaystyle \frac{\pi}{3}=0\ or\ 64 - 16x - 3x^2 = 0\) Well \(\displaystyle \frac{\pi}{3} \neq 0\) so it must be that \(\displaystyle 64 - 16x - 3x^2= 0\)
Click to expand...
urimagic said:
OOOHHHH beautiful.....I like this!!..Thanks Steven G..I do appreciate!
Click to expand...
Yes, you do not need to divide by the constant to understand what is going on. IMO, there is no motivation to divide by pi/3, just realize that you have a product that equals 0. Then just set each factor to 0 and solve.
 
Steven G said:
Yes, you do not need to divide by the constant to understand what is going on. IMO, there is no motivation to divide by pi/3, just realize that you have a product that equals 0. Then just set each factor to 0 and solve.
Click to expand...
Got it, Thank you sir..
 
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