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 November 3rd, 2010, 06:05 PM #1 Newbie   Joined: Nov 2010 From: CA Posts: 25 Thanks: 0 Optimization problem a Donor is willing to build a new hangar for the program with the following stipulations - the hangar must be in the shape of a half cylinder - the hangar to have an exact volume of 225000 cubit feet We would like to minimize the cost of the building. 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. what should the dimensions of the building be to minimize the total cost ? How can i solve for this !! help Thanks November 3rd, 2010, 06:20 PM #2 Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Optimization problem To simplify your question, I downloaded a pic [attachment=0:uaud2u5l]???.GIF[/attachment:uaud2u5l] Do you mean this? Attached Images  ???.GIF (23.2 KB, 1407 views)  November 3rd, 2010, 07:40 PM #3 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Optimization problem First, we should define the cost function. Let W be the width of the building, which will be the diameter of the half-cylinder, and let L be the length of the building, which will be the height of the half-cylinder. The area F of the floor will be F = L?W, the area R of the roof will be $R=\pi\cdot\frac{W}{2}\cdot L$ and the area S of the sides will be $S=\pi\cdot\left(\frac{W}{2}\right)^2$. The volume V of the building is $V=\frac{1}{2}\pi\cdot\left(\frac{W}{2}\right)^2\cd ot L=225000\text{ ft.^3}$. Let: $C_f$ be the cost per square foot for the foundation. $C_s$ be the cost per square foot for the sides. $C_r$ be the cost per square foot for the roof. So the cost C, in dollars will be: $C=C_fF+C_sS+C_rR = C_fLW+C_s\pi\cdot\left(\frac{W}{2}\right)^2+C_r\pi \cdot\frac{W}{2}\cdot L$ From the formula for volume, we see that $L=\frac{8V}{\pi W^2}$, giving $C=C_f\left(\frac{8V}{\pi W^2}\right)W+C_s\pi\cdot\left(\frac{W}{2}\right)^2 +C_r\pi\cdot\frac{W}{2}\cdot\left(\frac{8V}{\pi W^2}\right)$ Simplification yields: $C=\frac{8V(2C_f+\pi C_r)}{2\pi}\cdot W^{-1}+\frac{\pi C_s}{4}\cdot W^2$ Now, differentiating C with respect to W and equating to zero gives: $-\frac{8V(2C_f+\pi C_r)}{2\pi}\cdot W^{-2}+\frac{\pi C_s}{2}\cdot W=0$ Multiply through by $2\cdot W^2$ $-\frac{8V(2C_f+\pi C_r)}{\pi}+\pi C_s\cdot W^3=0$ Solve for W: $W=2\cdot\left(\frac{V(2C_f+\pi C_r)}{\pi^2C_s}\right)^{\frac{1}{3}}$ $L=\frac{8V}{\pi\left(\frac{8V(2C_f+\pi C_r)}{\pi^2C_s}\right)^{\frac{2}{3}}}=\frac{2(\pi VC_s^2)^{\frac{1}{3}}}{\left(2C_f+\pi C_r\right)^{\frac{2}{3}}}$ All that's left to do now is plug in the values given. November 3rd, 2010, 08:10 PM #4 Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Optimization problem Quote:  Originally Posted by genie a Donor is willing to build a new hangar for the program with the following stipulations - the hangar must be in the shape of a half cylinder - the hangar to have an exact volume of 225000 cubit feet We would like to minimize the cost of the building. 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. what should the dimensions of the building be to minimize the total cost ? How can i solve for this !! help Thanks
So, let $x$be the width of the foundation, and$y$ be the length. If the hangar in the shape of half a cylinder, the max height of the vault should be$\frac{x}{2}$.

The areas:

vault: $\frac{\pi}{2}xy$ ;

sidewall(both two): $\frac{\pi}{4}x^2$ ;

foundation: $xy$ .

The volume:

$\frac{\pi}{8}x^2y=V=\text{225000 c.f. (i)}$
total cost: (let $a$be the vaultcost,$b$ be the sidewallcost, $c$be the foundationcost, where $a=\text{\15}$, $b=\text{\20}$, $c=\text{\30}$)

$\text{total cost=vaultcost+sidewallcost+foundationcost}=\frac{ a\pi}{2}xy+\frac{b\pi}{4}x^2+cxy\text{ (ii)}$

Construction an auxiliary function using (i)and (ii) as below:

$F(x,y)=\frac{a\pi}{2}xy+\frac{b\pi}{4}x^2+cxy+\lam bda{(\frac{\pi}{8}x^2y-V)}\text{ (iii)}$ .

Get the first order partial derivative of $F(x,y)$ with $x ,y$.

$\frac{\partial{F}}{\partial{x}}=\frac{a\pi}{2}y+\f rac{b\pi}{2}x+cy+\frac{\lambda\pi}{4}xy\text{ (iv)}$ ;

$\frac{\partial{F}}{\partial{y}}=\frac{a\pi}{2}x+cx +\frac{\lambda\pi}{8}x^2\text{ (v)}$ ;

Let (iv)and (v) equal to zeros, and with$\frac{\pi}{8}x^2y-V=0$, we get a group of equations:

$\frac{a\pi}{2}y+\frac{b\pi}{2}x+cy+\frac{\lambda\p i}{4}xy=0$
$\frac{a\pi}{2}x+cx+\frac{\lambda\pi}{8}x^2=0\text{ (vi)}$
$\frac{\pi}{8}x^2y-V=0$

The solution of $x$,$y$and $\lambda$

$x=2\sqrt[3]{\frac{(a\pi+2c)V}{b\pi^2}}$

$y=2\sqrt[3]{\frac{b^2\pi{V}}{(a\pi+2c)^2}}$

$\lambda=-2\sqrt[3]{\frac{b(a\pi+2c)^2}{\pi{V}}}$

where $a=\text{\15}$, $b=\text{\20}$, $c=\text{\30}$ , and $V=\text{225000 c.f.}$

 November 3rd, 2010, 08:34 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Optimization problem I considered using optimization with constraint (Lagrange multipliers), but in the end decided to go with what I knew better. Glad to see our answers agree.
November 3rd, 2010, 08:38 PM   #6
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Joined: Oct 2010
From: Changchun, China

Posts: 492
Thanks: 14

Re: Optimization problem

Quote:
 Originally Posted by MarkFL I considered using optimization with constraint (Lagrange multipliers), but in the end decided to go with what I knew better. Glad to see our answers agree.

I checked my answer for a few times, and found blunders out. Fortunately, it follows what you got

 November 3rd, 2010, 09:13 PM #7 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Optimization problem So the width should be 99.22 feet, the length should be 58.20 feet, the top height should be 49.61 feet,which will lead to the min cost... And $\frac{\pi}{8}(99.22)^2(58.20)=225000$ Maybe the door should be in the roof not in the sidewall
 November 3rd, 2010, 09:33 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Optimization problem This results in a total cost (to the nearest penny) of $463,938.62 November 3rd, 2010, 09:38 PM #9 Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Optimization problem Quote:  Originally Posted by MarkFL This results in a total cost (to the nearest penny) of$463,938.62
Well well well, this's really a big money.

November 3rd, 2010, 10:25 PM   #10
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Joined: Nov 2010
From: CA

Posts: 25
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Re: Optimization problem

Quote:
 Originally Posted by stainburg To simplify your question, I downloaded a pic [attachment=0:rx0c59eq]???.GIF[/attachment:rx0c59eq] Do you mean this?
yes, the picture you gave is such a hangar ( did u get it from a game or something ? lol ^^!

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