Gloria would like to construct a box with volume of exactly 45ft3 using only metal and wood. The metal costs $14/ft2 and the wood costs $5/ft2. If the wood is to go on the sides, the metal is to go on the top and bottom, and if the length of the base is to be 3 times the width of the base, find the dimensions of the box that will minimize the cost of construction. Round your answer to the nearest two decimal places.
If the width is w, and the height is h, then we have
w*3w*h = 45
So, h = 15/w^2
The cost is
c = 5*2(wh + 3wh) + 14*2*w*3w = 84w^2 + 600/w
dc/dw = 168w - 600/w^2
minimum cost is when dc/dw = 0
Now just finish it off.
Im still very confused on what to do with those values
let the width be x ft
then the length is 3x ft
let the height be y ft
we know x^2 y = 45 or y = 45/x^2
surface of sides = 2(3xy) + 2(xy) = 8xy
surface of top and bottom = 2(3x^2) = 6x^2
cost = 5(8xy) + 14(6x^2)
= 40x(45/x^2) + 84x^2
= 1800/x + 84x^2
d(cost)/dx = -1800/x^2 + 168x
= 0 for a min cost
168x = 1800/x^2
x^3 = 1800/168 = 10.714...
x = appr 2.20 ft
find the dimensions
huh? just keep going.
dc/dw = (168w^3-600)/w^2 = 24(7w^3-25)/w^2
So, dc/dw = 0 when x = ∛(25/7)
Thus, the dimensions of the box are
width = ∛(25/7)
length = 3∛(25/7)
height = 15/(25/7)^(2/3) = 3∛(49/5)
check: ∛(25/7) * 3∛(25/7) * 3∛(49/5) = 45
okay, yeah, I got really confused with the derivative and how to solve to get w
Don't forget your Algebra I now that you're doing calculus ...
Go with oobleck's numbers, I messed up in my volume expression, duhh!
I had x^2 y = 45, should have been 3x^2 y = 45 like oobleck had