A candy company needs a custom box for their truffles. The box they've chosen is in the shape of a cylinder with a hemisphere of the same radius on top. The total volume of the box is v= 1/2 ((4pir^3)/(3))+ (pi)(r^2)(y-r). y is height and r is radius.Originally, the candy box was designed to have a height of 6 inches and a radius of 2 inches, but the box needs to be slightly shorter. You need to adjust the box so that the height is 5.75 inches but v remains constant. find the value of dr/dy at the point r=2 and y=6. use the value to approximate the new radius
applications of derivative
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sure thing!
my first step:
I took the derivative
0= 2 (pi) (r^2) (dr/dy) + (pi) [2r(y-r) dr/dy + (r^2) (1- dr/dy) ] = 0
then i plugged in y=6 and r =2
8 pi (dr/dy) + pi ( 16(dr/dy) + 4(1- (dr/dy)) = 0
then I solved for dr/dy to get dr/dy = -1/5
now I'm stuck, I have dr/dy , but i don't know how to get this to use the new radius. Also, I'm not sure if my work is right, but that's my best attempt..... Help would be greatly appreciated
my first step:
I took the derivative
0= 2 (pi) (r^2) (dr/dy) + (pi) [2r(y-r) dr/dy + (r^2) (1- dr/dy) ] = 0
then i plugged in y=6 and r =2
8 pi (dr/dy) + pi ( 16(dr/dy) + 4(1- (dr/dy)) = 0
then I solved for dr/dy to get dr/dy = -1/5
now I'm stuck, I have dr/dy , but i don't know how to get this to use the new radius. Also, I'm not sure if my work is right, but that's my best attempt..... Help would be greatly appreciated
This problem is still really bothering me!
So I plugged in (-1/5) to the original derivative and also y=5.75 to solve for r.
And I ended up getting 1.2R^2-2.3R=0 (which makes no sense), and to be sure I graphed it and the minimum was x=.95 but the radius must be greater then two if the height decreases. mind boggling....
I really hate not understanding a problem, so if anyone could help me out I would really appreciate it.
So I plugged in (-1/5) to the original derivative and also y=5.75 to solve for r.
And I ended up getting 1.2R^2-2.3R=0 (which makes no sense), and to be sure I graphed it and the minimum was x=.95 but the radius must be greater then two if the height decreases. mind boggling....
I really hate not understanding a problem, so if anyone could help me out I would really appreciate it.
\(\displaystyle V=\frac{2\pi}{3}r^{3}+{\pi}r^{2}(y-r)\)
With y=6 and r=2, the volume ends up being \(\displaystyle \frac{64\pi}{3}\)
Now, using the same volume, but letting the height be 23/4:
\(\displaystyle \frac{64\pi}{3}=\frac{2}{3}{\pi}r^{3}+{\pi}r^{2}(\frac{23}{4}-r)\)
Solving for r, we find the radius increases to r=2.052 when the height decreases to 5.75=23/4.
But, this cubic can be difficult to solve by hand. Let's try this:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But, solving V for y, we get:
\(\displaystyle y=\frac{{\pi}r^{3}+3V}{3{\pi}r^{2}}\)
\(\displaystyle \frac{dy}{dr}=\frac{{\pi}r^{3}-6V}{3{\pi}r^{3}}\)
\(\displaystyle \frac{dr}{dy}=\frac{3{\pi}r^{3}}{{\pi}r^{3}-6V}\)
Letting \(\displaystyle V=\frac{64\pi}{3}\), it becomes:
\(\displaystyle \frac{dr}{dy}=\frac{3r^{3}}{r^{3}-128}\)
If r=2, then \(\displaystyle \frac{dr}{dy}=\frac{-1}{5}\)
This may be too much info, but I thought it may be helpful.
Now, we know that\(\displaystyle dy=\frac{r^{3}-128}{3r^{3}}dr\)
We are told that y decreases 1/4 units.
\(\displaystyle \frac{-1}{4}=\frac{r^{3}-128}{3r^{3}}dr\)
Let r=2 and solve for dr, the change in radius.
Doing so results in \(\displaystyle dr=\frac{1}{20}=.05\)
The radius increases .05 to 2.05 when the height decreases to 5.75.
Just as dr/dy shows, the radius increases 1 unit for every 5 the height decreases.
So, if the height goes down .25 (from 6 down to 5.75), then the radius will increase \(\displaystyle \frac{1}{5}\cdot \frac{1}{4}=\frac{1}{20}=.05\).
Remember, this is an approximation using differentials. You may not get the original volume right on the money...but very close.
With y=6 and r=2, the volume ends up being \(\displaystyle \frac{64\pi}{3}\)
Now, using the same volume, but letting the height be 23/4:
\(\displaystyle \frac{64\pi}{3}=\frac{2}{3}{\pi}r^{3}+{\pi}r^{2}(\frac{23}{4}-r)\)
Solving for r, we find the radius increases to r=2.052 when the height decreases to 5.75=23/4.
But, this cubic can be difficult to solve by hand. Let's try this:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But, solving V for y, we get:
\(\displaystyle y=\frac{{\pi}r^{3}+3V}{3{\pi}r^{2}}\)
\(\displaystyle \frac{dy}{dr}=\frac{{\pi}r^{3}-6V}{3{\pi}r^{3}}\)
\(\displaystyle \frac{dr}{dy}=\frac{3{\pi}r^{3}}{{\pi}r^{3}-6V}\)
Letting \(\displaystyle V=\frac{64\pi}{3}\), it becomes:
\(\displaystyle \frac{dr}{dy}=\frac{3r^{3}}{r^{3}-128}\)
If r=2, then \(\displaystyle \frac{dr}{dy}=\frac{-1}{5}\)
This may be too much info, but I thought it may be helpful.
Now, we know that\(\displaystyle dy=\frac{r^{3}-128}{3r^{3}}dr\)
We are told that y decreases 1/4 units.
\(\displaystyle \frac{-1}{4}=\frac{r^{3}-128}{3r^{3}}dr\)
Let r=2 and solve for dr, the change in radius.
Doing so results in \(\displaystyle dr=\frac{1}{20}=.05\)
The radius increases .05 to 2.05 when the height decreases to 5.75.
Just as dr/dy shows, the radius increases 1 unit for every 5 the height decreases.
So, if the height goes down .25 (from 6 down to 5.75), then the radius will increase \(\displaystyle \frac{1}{5}\cdot \frac{1}{4}=\frac{1}{20}=.05\).
Remember, this is an approximation using differentials. You may not get the original volume right on the money...but very close.