Hello,

The following problem indicates that the answer is -1/5, but I am inclined to think that it is -1/3 based on my derivative calculations. Is this an error of mine or theirs?Note:solving only part A of the problem is necessary (not the whole thing), and that is all I provided.

A candy company needs a custom box for their truffles. The total volume of the box is $\displaystyle V = \frac{1}{2}\left(\frac{4\pi r^3}{3}\right) = \pi r^2(y - r)$, where $\displaystyle y$ is the height of the box and $\displaystyle r$ is the radius of the box. Originally, the candy box was designed to have a height of 6 inches and a radius of 2 inches, but the shipper suggests that the boxes be made slightly shorter. You now need to adjust the radius so that the height is reduced to 5.75 inches but the volume remains constant.

Problem

A.Find the value of $\displaystyle dr/dy$ at the point $\displaystyle r = 2, y = 6$.

$\displaystyle V = 16\pi$

My Work:

$\displaystyle V = \pi r^2(y - r)$

$\displaystyle 16\pi = \pi r^2(y - r)$

$\displaystyle 16 = r^2(y - r)$

Implicit Differentiation:

$\displaystyle \frac{dr}{dy} (16 = r^2(y - r))$

$\displaystyle 0 = (r^2)'(y - r) + (r^2)(y - r)'$

$\displaystyle 0 = \left(2r\frac{dr}{dy}\right)(y - r) + (r^2)\left(1 - \frac{dr}{dy}\right)$

$\displaystyle \frac{dr}{dy}r^2 + \frac{dr}{dy}2r^2 - \frac{dr}{dy}2ry = r^2$

$\displaystyle \frac{dr}{dy} (r^2 + 2r^2 - 2ry) = r^2$

$\displaystyle \frac{dr}{dy} = \frac{r^2}{3r^2 - 2ry}$

Slope:

$\displaystyle \frac{dr}{dy} = \frac{(2)^2}{3(2)^2 - 2(2)(6)}= -\frac{1}{3}$