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 March 13th, 2017, 05:40 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Partial Derivative Application about optimization of dimension The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. Any tips would be grateful. March 13th, 2017, 06:37 PM   #2
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 Originally Posted by zollen The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. Any tips would be grateful.
Answer: It has a square base, and is one and one half times as tall as wide. If the volume is V the dimensions are
(2V/3)^(1/3)* (2V/3)^(1/3) * (9V/4)^(1/3) March 13th, 2017, 07:28 PM   #3
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 Originally Posted by zollen The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. Any tips would be grateful.
Are you told in the question that it has to have a square base? March 13th, 2017, 08:59 PM   #4
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 Originally Posted by zollen The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. Any tips would be grateful.
let the box have dimensions $l,~w,~h$

we want to minimize

$C = 2 l w + l w +2(lh + wh)$

subject to the constraint

$V = l w h$

to do this with the Lagrange Multiplier method we solve

$\nabla (C - \lambda (lwh-V)) = \bf{0}$

chugging through it we get the set of equations

$3w + 2h - w h \lambda = 0$

$3l + 2h - l h \lambda = 0$

$2(l+w) - l w \lambda = 0$

$l w h = V$

solving this is probably a bit of a mess but you end up with

$l = w =\left(\dfrac{2V}{3}\right)^{1/3}$

$h = \left(\dfrac {3V}{ 2} \right)^{2/3}$ March 14th, 2017, 04:02 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I would have done this slightly differently, writing the Lagrange multiplier equation as . That gives the three equations , and . Now, since a specific value for is not necessary for the solution, eliminate [tex]\lambda[/b] buy dividing one equation by another. Dividing the first equation by the second, . Dividing the first equation by the third, . We also have the condition . That gives three equations to solve for l, w, and h. March 14th, 2017, 05:21 AM #6 Newbie   Joined: Mar 2017 From: South Africa Posts: 1 Thanks: 0 Math Focus: CALCULUS AND LINEAR ALGEBRA You have to make use of the 1st derivative and second derivative to obtain your desired answers Tags application, derivative, dimension, optimizating, optimization, partial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post POISONXMITHIE Linear Algebra 1 October 20th, 2015 05:40 AM hyperbola Calculus 3 May 3rd, 2015 04:41 AM sachinrajsharma Calculus 1 May 18th, 2013 08:36 PM sachinrajsharma Calculus 3 May 8th, 2013 10:41 AM EXPLORE Calculus 2 January 28th, 2010 03:03 AM

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