# Quadratic Problem/Frustrum of a cone

• Nov 20th 2009, 04:49 AM
manich44
Hi Everyone,

Can someone have a look at the attachment and let me know where I am going wrong?

Thanks....Mark
• Nov 20th 2009, 08:38 AM
Hello Mark
Quote:

Originally Posted by manich44
Hi Everyone,

Can someone have a look at the attachment and let me know where I am going wrong?

Thanks....Mark

I can't see anything wrong at all.

Obviously, you haven't shown us the complete question, but if you want the value of $r$ that gives the volume of the frustum to be $3.46 \text{ m}^3$, it looks OK to me. The positive root of your quadratic $\approx 0.856$, and this value of $r$ gives the volume to be $3.48$ - the slight error being due to the approximation $\frac{3V}{\pi h} \approx 3.2$, where $3.178$ would have been better.

Is this not the correct answer, then?

• Nov 20th 2009, 09:11 AM
manich44
RE Complete question
Sorry the complete question is....change the size of the upper radius "r" (without any changes to "h" height, and "R" lower Radius) to make the weight of this concrete pile to be at least 8300kg (density of used concrete = 2400kg/m3)
Calculate the new upper Radius r

r=55cm R=1.2m h=104cm

Thanks....
• Nov 20th 2009, 09:20 AM
OK, I got that much. And that gives the volume $3.46 \text{ m} ^3$ as you said. So where is the problem? Do you have a different answer for $r$? If so, I suspect the answer you have may be wrong!