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March 13th, 2017, 04:40 PM  #1 
Member Joined: Jan 2017 From: Toronto Posts: 64 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  
Member Joined: Jan 2017 From: Toronto Posts: 64 Thanks: 1  Quote:
(2V/3)^(1/3)* (2V/3)^(1/3) * (9V/4)^(1/3)  
March 13th, 2017, 06:28 PM  #3 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  
March 13th, 2017, 07:59 PM  #4  
Senior Member Joined: Sep 2015 From: CA Posts: 1,207 Thanks: 615  Quote:
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 $\nabla (C  \lambda (lwhV)) = \bf{0}$ 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: 2,424 Thanks: 614 
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, 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


Tags 
application, derivative, dimension, optimizating, optimization, partial 
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