On Paul's online math notes, in Applications of the derivative, Optimization, ex. 2, how does he go from C=10(2lw)+ 6(2wh+2lh)=60w^2+48wh??? I understand how he gets the equation, but not the answer. Thanks
can you provide mor einformation?
whats the setup to the problem, and maybe we can work thru it :)
Example 2 We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10/ft2 and the material used to build the sides cost $6/ft2. If the box must have a volume of 50ft3 determine the dimensions that will minimize the cost to build the box. this one right?
its the cost we want to minimize....
l = 3w; V = lwh = 50; h = V/lw h = 50/3ww = 50/3w^2 surface area = 2lw + 2wh + 2lh
cost = 2(10)lw + 2(6)wh + 2(6)lh cost = 2(10)(3w)w + 2(6)w(50/3w^2) + 2(6)(3w)(50/3w^2)
cost = 60w^2 +600w/3w^2 + 1800w/w^2 cost = 60w^2 +200/w + 600/w
cost' = 60(2w) -800/w^2 cost' = (120w^3 -800)/w^2 cost' = 40(3w^3 -20)/w^2 = 0 when 3w^3 -20 = 0
3w^3 = 20 w^3 = 20/3 w = cbrt(20/3) now where if anyplace did I go wrong lol
w=cbrt(20/3); l=20 ; h=50/3cbrt(400/9) maybe lol
apparently thats correct lol...its good to have an answer key to check :)
This is where I get lost... cost = 2(10)lw + 2(6)wh + 2(6)lh cost = 2(10)(3w)w + 2(6)w(50/3w^2) + 2(6)(3w)(50/3w^2) How do u got from 2(10)lw to 2(10)(3w)w?
length = 3 times the width in the information given; substitute 3w for l
I get the bottom you're replacing the h's
yeah; turning everything into w varaibale terms :)
i got cbrt(20/3) and was sure I messed it up lol....but its good
Oh right! But quickly, how do you solve for the height? 50/3ww?
V = 50 = lwh; solve for h and substitute you l value... h = 50/lw and l = 3w h = 50/3w^2
the volume of a box is (length) times (width) times (height) :)
ahhhh, thank you! the clouds are dispersing now! lol I'm going to try it again
good luck ;)
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