Expert: Scotto - 1/9/2009

Question
QUESTION: 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) + pir^2(y - r) , where y is height of box and r is radius of box. This is 1/2 the volume of a sphere + voulme of cylinder. 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.
A. Find the value of dr/dy at the point r=2, y=6.
B. Use your value of dr/dy to approximate the new radius for these boxes.

I had dr/dy = -1/2 but this would decrease the rate when it should be increasing. How do I work it?


ANSWER: You might have got this part, but just for review:
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V=0.5(4πr^3)/3 + πr²(y-r).  Note that the volume V is a constant.
Note that 0.4*4=2.  Originally, the height y was 6 inches and the radius r was 2 inches.  Put y=6 and r=2 into the equation to find what the volume was.

Solve this equation for y in terms of r.
In doing so, you should get πr²(y-r) = (V - 2πr^3/3).

Further reduction gives y-r = (V - 2πr^3/3)/(πr²).

One last operation gives you the equation
[1] y = r + (V - 2πr^3/3)/(πr²).

Now that we know V and the new height y=5.75, we can find the new value for r.

Now about A and B.  
A.  The derivative dy/dr can be found by taking the derivative of y with respect to r in the equation [1].  dr/dy is 1/(dy/dr) with the variables converted.  Example: y=x², so x=√y.  Then dy = 2x dx, so dy/dx = 2x or dx/dy = 1/(2√y).

B. By Newton's law, the new value of a function can be approximated by V(x+h) = V(x) + hV'(x).

Note that [1] can be rewritten as
y = r + (K - 2r^3)/r² where K = V/π.

Now on to the question:
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I haven't checked it out to make sure it's correct, but the value of dr/dy should be negative.  This means as the radius is increasing, the height y would be decreasing to make the volume be the same.

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QUESTION: so I get dr/dy=.125
which makes my new r=1.91667
but this is smaller height and raduis so how would volume be the same?

ANSWER: from [1], after looking at it, I would divide it into terms.
We would get y = r + V/(πr²) - 2r/3.  Note that the r's can be combined.

That give y = r/3 + V/(πr²).  From here, we can say that
dy = dr(1/3 - 2V/(πr^3)).

Combining the fraction gives dy/dr = (πr^3 - 6V)/(3πr^3).

Our equation for V was V=0.5(4πr^3)/3 + πr²(y-r).
At r=2, y=6, note that the volume V is 16π/3 + 16π = 64π/3.

this gives dy/dr = (1/3 - (128π/3)/(8π))) = 1/3 - 16/3 = -5.

dr/dy would be 1 /(dy/dr), so dr/dy = -0.2.
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It would be more useful if you showed me how exactly you go dr/dy.
None that if y=x^3, then dy = 3x² dx.  To find dy/dx, divide both sides by dx.  To find dx/dy, divide both sides by 3x²dy and then note that x² can be replaced with y^(2/3), but if you know x, that works as well.  This shows that dy/dx really is a fraction.
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The first question stated that dr/dy was -1/2 and then it was given as 0.125.  I don't think either of these is write now that I worked through it.  It is negative, which means that as one of the two variables increases, the other decreases.





---------- FOLLOW-UP ----------

QUESTION: ok, I made a caluclation error and have dr/dy=-.2
when I put this back in the equation and use y=5.75
dr/dy=-r/(2y-r), -.2=-r/2(5.75-r), solving for r=1.91667.
This will not yield equal volume.

Answer
This is only an estimate to the change in volume assuiming the change is linear, but the change is merely almost linear for that small a change in r.

To find what the raidus should acutally be, put y = 5.75 in and keep the volume V the same, then solve for r.