Question

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.

Answer 1

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.

Answer 2

Im still very confused on what to do with those values

Answer 3

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

Answer 4

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

Answer 5

okay, yeah, I got really confused with the derivative and how to solve to get w

Answer 6

Don't forget your Algebra I now that you're doing calculus ...

Answer 7

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