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 November 9th, 2011, 03:53 PM #1 Newbie   Joined: Nov 2010 From: CA Posts: 25 Thanks: 0 Find the dimension of the rectangular box Hello everyone, please help me with these problems Thanks in advance. 1/Find the dimension of the rectangular box of maximum volume that has three of its faces in the coordinate planes, one vertex at the origin, and another vertex in the first octant on the plane 2x+3y+5z=36 2/ Use polar coordinate to evaluate $\int_{R}\int_\ f(x,y) dA$ where $f(x,y)= e^{-((x^2)+(y^2))} and R ={(x,y), (x^2)+ (y^2) <=4, x>=0, y >=0}$
November 9th, 2011, 08:33 PM   #2
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Re: Find the dimension of the rectangular box

Hello, genie!

Quote:
 1. Find the dimension of the rectangular box of maximum volume that has three of its faces in the coordinate planes, one vertex at the origin, and another vertex in the first octant on the plane $2x\,+\,3y\,+\,5z\:=\:36$

I would use Lagrange multipliers.

$\text{Maximize: }\,V \:=\:xyz,\,\text{ with the constraint: }\:2x\,+\,3y\,+\,5z\:=\:36$

$\text{W\!e have: }\:f(x,\,y,\,z,\,\lambda) \;=\;xyz\,+\,\lambda(2x\,+\,3y\,+\,5z\,-\,36)$

$\text{Set the partial derivatives equal to zero, and solve.}$

[color=beige]. . [/color]$\begin{Bmatrix}\frac{\partial V}{\partial x}=&yz\,+\,2\lambda=&[1] \\ \\ \\\frac{\partial V}{\partial y}=&xz\,+\,3\lambda=&[2] \\ \\ \\ \frac{\partial V}{\partial z}=&xy\,+\,5\lambda=&[3] \\ \\ \\ \frac{\partial V}{\partial \lambda}=$

[color=beige]. . [/color]$\begin{Bmatrix} \text{From [1]:}=&\lambda=&-\frac{yz}{2}=&[5] \\ \\ \\ \text{From [2]:}=&\lambda=&-\frac{xz}{3}=&[6] \\ \\ \\ \text{From [3]:}=&\lambda=&-\frac{xy}{5}=&[7] \end{Bmatrix}=$

[color=beige]. . [/color]$\begin{Bmatrix} \text{Equate [5] and [6]:}=&-\frac{yz}{2} \:=\:-\frac{xz}{3} &\Rightarrow=&3y \:=\:2x &[8] \\ \\ \\ \text{Equate [6] and [7]:}=&-\frac{xz}{3} \:=\:-\frac{xy}{5} &\Rightarrow=&5z \:=\:3y &[9] \end{Bmatrix}=$

$\text{Hence: }\:2x\,=\,3y\,=\,5z$

$\text{Substitute into [4]: }\:2x\,+\,2x\,+\,2x\,-\,36 \:=\:0 \;\;\;\Rightarrow\;\;\;6x \:=\:36 \;\;\;\Rightarrow\;\;\;x \:=\:6$

$\text{Substitute into [8]: }\:3y\:=\:2(6) \;\;\;\Rightarrow\;\;\;y \:=\:4$

$\text{Substitute into [9]: }\:5z \:=\:3(4) \;\;\;\Rightarrow\;\;\;z \:=\:\frac{12}{5}$

$\text{Therefore, the dimensions are: }\:(x,\,y,\,z) \;=\;\left(6,\,4,\,\frac{12}{5}\right)$

November 10th, 2011, 04:16 AM   #3
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Re: Find the dimension of the rectangular box

[color=#000000]2.

$x=r\cos(t)$ $y=r\sin(t)$ and $dA=r\;dr\;dt$

$\iint_{R}f(x,y)\;dA=\int_{0}^{\frac{\pi}{2}}\int_{ 0}^{2}r\cdot e^{-r^2}\;dr\;dt=\frac{\left(e^4-1\right) \pi }{4 e^4}$

A plot of the given region.

[attachment=0:2barqvmz]graph.gif[/attachment:2barqvmz]
.[/color]
Attached Images
 graph.gif (4.9 KB, 404 views)

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# find the dimension of the box that has one vertex on the plane

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