SOLUTION: A box is constructed out of two different types of metal. The metal for the top and bottom, which are both square, costs $5 per square foot and the metal for the sides costs $2 per
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Question 1099943: A box is constructed out of two different types of metal. The metal for the top and bottom, which are both square, costs $5 per square foot and the metal for the sides costs $2 per square foot. Find the dimensions that minimize cost if the box has a volume of 15 cubic feet. Found 2 solutions by ankor@dixie-net.com, ikleyn:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A box is constructed out of two different types of metal.
The metal for the top and bottom, which are both square, costs $5 per square foot and the metal for the sides costs $2 per square foot.
Find the dimensions that minimize cost if the box has a volume of 15 cubic feet.
:
let s = one side of the square base
let h = the height of the box
then
s^2 * h = 15 cu/ft
Therefore
h =
:
surface area = top, bottom, 4 side areas
SA = 2s^2 + 4hs
replace h with
SA = 2s^2 + 4s*
cancel the s
SA = 2s^2 + (60/s)
Cost of the box
C(x) = 5(2x^2) + 2(60/s)
C(x) = 10x^2 +
:
Graphically, y axis = the cost
minimum cost when x=1.9 ft
find the height
h =
h = 4.155 ft
:
The dimensions for minimum cost: 1.9 by 1.9 by 4.155 ft Cost about $99.26 (green)
You can put this solution on YOUR website! A box is constructed out of two different types of metal.
The metal for the top and bottom, which are both square, costs $5 per square foot and the metal for the sides costs $2 per square foot.
Find the dimensions that minimize cost if the box has a volume of 15 cubic feet.
~~~~~~~~~~~~~~~~~~~~~
The solution by other tutor is not exactly correct.
So I place below a corrected version.
let s = one side of the square base
let h = the height of the box
then
s^2 * h = 15 cu/ft
Therefore
h =
surface area = top, bottom, 4 side areas
SA = 2s^2 + 4hs
replace h with
SA = 2s^2 + 4s*
cancel the s
SA = 2s^2 + (60/s)
Cost of the box
C(x) = 5(2x^2) + 2(60/x) <<<---=== One correction is HERE
C(x) = 10x^2 +
To find the minimum, take the derivative <<<---=== Second correction is here and in all following lines
C'(x) = 20x -
The minimum is at x = = = 1.817 (approximately).
Graphically, y axis = the cost
minimum cost when x = 1.817 ft
find the height
h =
h = 4.543 ft
:
The dimensions for minimum cost: 1.817 by 1.817 by 4.543 ft. Cost about $99.06 (green)
