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 March 13th, 2017, 04:40 PM #1 Member   Joined: Jan 2017 From: Toronto Posts: 82 Thanks: 1 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, 05:37 PM   #2
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Quote:
 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, 06:28 PM   #3
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Quote:
 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, 07:59 PM   #4
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Quote:
 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, 03:02 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,524 Thanks: 641 I would have done this slightly differently, writing the Lagrange multiplier equation as $\nabla (3lw+ 2lh+ 2wh)= \lambda \nabla (lwh)$. That gives the three equations $3w+ 2h= \lambda wh$, $3l+ 2h= \lambda lh$ and $2l+ 2w= \lambda lw$. Now, since a specific value for $\lambda$ is not necessary for the solution, eliminate [tex]\lambda[/b] buy dividing one equation by another. Dividing the first equation by the second, $\frac{3w+ 2h}{3l+ 2h}= \frac{w}{l}$. Dividing the first equation by the third, $\frac{3w+ 2h}{2l+ 2w}= \frac{h}{l}$. We also have the condition $lwh= V$. That gives three equations to solve for l, w, and h.
 March 14th, 2017, 04:21 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

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