# Thread: max. percentage error

1. ## max. percentage error

if r=5cm and h=12cm to the nearest mm, what should we expect the max. percentage error in calculating V=(pi)(r^2)h to be?

Thanks!

2. Originally Posted by alwaizmini
if r=5cm and h=12cm to the nearest mm, what should we expect the max. percentage error in calculating V=(pi)(r^2)h to be?

Thanks!
Not enough info.

3. that's all the info that was given...

4. Originally Posted by alwaizmini
that's all the info that was given...
Well, then my advice is to move on to the next problem and quit sweating this one. The numbers that you were given will help to calculate the volume, but the problem gives no type of information regarding error.

5. Actually there is enough info Using the differential

Note dv/V is the relative error and dV/V*100 is max %error

dr = .1cm dh = .1cm

V=(pi)(r^2)h to be?

dV = 2pi rh dr +pir^2dh

dV/V = [2pi rh dr +pir^2dh]/(pi)(r^2)h

dV/V = 2dr/r + dh/h

dV/V = 2*.1/5 + .1/12= .0483

so max % eror is 4.83%

6. Originally Posted by alwaizmini
if r=5cm and h=12cm to the nearest mm, what should we expect the max. percentage error in calculating V=(pi)(r^2)h to be?

Thanks!
Calculus26 provided the most appropriate.
This is another way.

minimum $r = 5 -0.0499999 = 4.950001$
maximum $R = 5 +0.049999 = 5.049999$

minimum $h = 12 - 0.049999 = 11.950001$
maximum $H = 12 + 0.049999 = 12.049999$

minimum $V = \pi 4.950001^2 \cdot 11.950001 = 919.8741$

maximum $V = \pi 5.049999^2 \cdot 12.049999 = 965.4271$

average Volume = $\dfrac{919.8741 + 965.4271}{2} = 942.6506$

difference: $965.4271 - 919.8741 = 45.553$

ratio: $\dfrac{45.553}{942.6506} = 0.048324$

4.83 %

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### volume maximum percentage error given r and h

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