Asked by Amalia

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.
Answered by oobleck
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.
Answered by Amalia
Im still very confused on what to do with those values
Answered by Reiny
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
Answered by oobleck
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
Answered by Amalia
okay, yeah, I got really confused with the derivative and how to solve to get w
Answered by oobleck
Don't forget your Algebra I now that you're doing calculus ...
Answered by Reiny
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
There are no AI answers yet. The ability to request AI answers is coming soon!