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March 13th, 2017, 04:40 PM   #1
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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.
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March 13th, 2017, 05:37 PM   #2
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Quote:
Originally Posted by zollen View Post
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)
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March 13th, 2017, 06:28 PM   #3
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Quote:
Originally Posted by zollen View Post
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?
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March 13th, 2017, 07:59 PM   #4
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Quote:
Originally Posted by zollen View Post
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

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

$h = \left(\dfrac {3V}{ 2} \right)^{2/3}$
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March 14th, 2017, 03:02 AM   #5
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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.
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March 14th, 2017, 04:21 AM   #6
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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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