applications of derivative

[helen8315]

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

 

[mmm4444bot]

How may we help you? Can you tell us where you are stuck? Do you have a specific question about this exercise? What have you thought or done, thus far? Please don't make us guess.

 

[helen8315]

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

 

[helen8315]

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.

 

[galactus]

$\displaystyle V=\frac{2\pi}{3}r^{3}+{\pi}r^{2}(y-r)$

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.