Sajang Yang: Dorm Room: Minimum Cost and Dimensions using Calculus

Question
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Dear Calculus Students:

After months of diligent work, I finally earned a promotion to Vice President for Development here at Dust-Mite U, which shocked quite a few because, as you may know, I am not more than two. I’m very proud of my fund raising accomplishments, but sometimes the gifts come with very strict limitations on how they can be used. We just received such a donation, and when I went looking for help, your enterprising and resourceful professor naturally referred me to you.

We have a somewhat eccentric alum who has made a major contribution in memory of his favorite Chia Pet Airplane that recently passed away in a bizarre gardening accident (it’s best we not discuss the details). As a fitting tribute to the dearly departed, the donor has designated that the funds be used to build a dorm in the shape of an airplane hanger, as shown below

There is an additional stipulation on the gift: the volume of the dorm must be exactly 225,000 cubic feet, which is one cubic foot for each sprout on the Chia Plane. If we do not need the entire amount of the donation for the dorm, then we may keep the additional money for any purpose we choose.

We are in the planning stages with the architects now, and we would obviously like to minimize the cost of the building. This is where I need your help. Currently, the construction costs for the foundation are $30 per square foot, the sides cost $20 per square foot to construct, and the roofing costs $15 per square foot. I need your expert advice on what the dimensions of the building should be to minimize the cost.

While the cost of the flooring and siding has been fairly stable, a further complicating factor is that the cost of the roofing material has been fluctuating dramatically for as long as I can remember (at least two months). Please give me your recommendations for the dimensions of the dorm given a price of $15 per sq. ft. That is; find r – the radius of the opening and l – the length of the building, and what the cost will be. Thank you.

It’s nap time now,

Frieda Jo

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Answer
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The area of foundation: $2\cdot r\cdot l$
The area of (two) sides: $2\cdot \left (\frac{\pi \cdot r^2}{2} \right )=\pi \cdot r^2$
The area of roof: $\frac{2\cdot \pi \cdot r}{2} \cdot l = \pi r l$
$The\ volume = 225,000 =\frac{\pi r^2}{2}\cdot l$
$\therefore l=\frac{2 \cdot 225,000}{\pi \cdot r^2}=\frac{450,000}{\pi \cdot r^2}$
$C = 30 \cdot 2r \cdot \frac{450,000}{\pi r^2}+ 20 \cdot \pi r^2+ 15 \cdot \pi r \cdot \frac{450,000}{\pi r^2} \\ \\ = \frac{60 \cdot 450,000}{\pi}r^{-1} + 20 \pi r^2 + 15 \cdot 450,000 r^{-1} \\ \\ = \left ( \frac{675 \cdot 10^4(4+\pi)}{\pi}\right )r^{-1} +20 \pi r^2 \\ \\$
$\frac{dC}{dr}= 40 \pi r + \left ( \frac{675 \cdot 10^4(4+\pi)}{\pi}\right )\left( \frac{-1}{r^2}\right ) \\ \\ \\ \\ =\frac{40 \pi^2 r^3 - 675 \times 10^4 \left( 4+ \pi \right) }{\pi r^2}$

To find critical number,

$r^3 = \frac{675 \times 10^4 \left(4+ \pi \right )}{40 \pi^2} \\ \\ r = \sqrt[3]{\frac{675 \times 10^4 \left(4+ \pi \right )}{40 \pi^2} }=49.61$

Therefore,
The minimum cost is 463,938.63 (dollars)
$r =49.61 \\ \\ l= \frac{450,000}{\pi r^2}=58.2$

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Sajang Yang

Monday, November 22, 2010

Dorm Room: Minimum Cost and Dimensions using Calculus

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