• Mar 9th 2013, 10:28 AM
michellederz
An observatory is to be in the form of a right circular cylinder surmounted by a hemispherical dome. If the hemispherical dome costs 6 times as much per square foot as the cylindrical wall, what are the most economic dimensions for a volume of 16000 cubic feet?

The radius of the cylindrical base (and of the hemisphere) is _____ ft. (Round to the nearest tenth).
• Mar 9th 2013, 10:42 AM
majamin
Re: Derivative Word Problem Help Please!~
Here are some questions to help guide you. What are the equations of surface area for the lateral surface area of a cylinder, and that of a hemisphere? What would be the cost of the cylindrical part if the cost was C per ft^2 (if you need more help: you want to buy 2 kg of apples and the price is $3 per kg: what do you do to these two numbers to figure out the total cost?) Post your answers and further questions below and we'll go from there. • Mar 9th 2013, 12:17 PM michellederz Re: Derivative Word Problem Help Please!~ Quote: Originally Posted by majamin Here are some questions to help guide you. What are the equations of surface area for the lateral surface area of a cylinder, and that of a hemisphere? What would be the cost of the cylindrical part if the cost was C per ft^2 (if you need more help: you want to buy 2 kg of apples and the price is$3 per kg: what do you do to these two numbers to figure out the total cost?) Post your answers and further questions below and we'll go from there.

π = pi

Area: A = 2πr2+2πrh
C = (6)2πr2+2πrh

Cost: C = (32000/r) + (72πr2/3)

This could be wrong.. I'm not very good with these word problems. Are these equations correct?
• Mar 9th 2013, 05:18 PM
majamin
Re: Derivative Word Problem Help Please!~
Quote:

Originally Posted by michellederz
π = pi

Area: A = 2πr2+2πrh
C = (6)2πr2+2πrh

Good start. Because we are just calculating the wall of a cylinder the equation has to change to $\displaystyle A=2\pi rh$ (it doesn't include $\displaystyle 2\pi r^2$). The area of a hemisphere is half of $\displaystyle 4\pi r^2$ or $\displaystyle 2\pi r^2$. Let the cost of the cylindrical portion be C, so the hemispherical portion is 6C.

The cost will be $\displaystyle Cost = (CylindricalPortion) \times C + (HemisphericalPortion) \times 6C$

Quote:

Cost: C = (32000/r) + (72πr2/3)

This could be wrong.. I'm not very good with these word problems. Are these equations correct?
. I get $\displaystyle Cost = C\left(\frac{32000}{r}+\frac{32}{3}\pi r^2\right)$ where C is the cost as above. We need to minimize cost, so this is an opportune time for the derivative (I'll let you try to fill in the details). As a final answer, I get $\displaystyle r=7.8 ft$.
• Mar 10th 2013, 10:25 AM
michellederz
Re: Derivative Word Problem Help Please!~
Okay I think I got it now! Thank you so very much!