calculus optimization?

Question

We are going to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $ 10 / ft^2 and the material used to build the sides cost $ 6 / ft^2. If the box must have a volume of 50 ft^3 determine the dimensions that will minimize the cost to build the box.

Please step by step, thanks...

Answer 1

Please note that I posted my numbers before Som did.

l = 3w

V = lwh = 3w^2h = 50, so h = 50/3w^2

C = 20lw + 12lh + 12wh

Substitute for l and w:

C = 20(3w)w + 12(3w)(50/3w^2) + 12w(50/3w^2)
C = 60w^2 + 600/w + 200/w = 60w^2 + 800w^-1

Take derivative and set equal to zero:

C' = 120w - 800w^-2 = 0

120w^3 - 800 = 0
120w^3 = 800
w^3 = 6.67
w = 1.88

So l = 3w = 5.65 and h = 50/3w^2 = 4.71

Answer 2

width =x
length = 3x
height = h
Volume = V
V= 3x*x*h = 3x^2h = 50 ft^3
So, h= 50/ 3x^2
The cost of building the top and bottom will be: 2(3x^2)*10=60x^2
The cost of building the two ends will be 6)2(50/x^2)x =600/x
Cost of two sides =6* 2(50/x^2)3x= 1800/x
total cost C = 60x^2+ 600/x+1800/x
dC/dx = 120x -600/x^2-1800/x^2
120x^3 -600-1800= 0
120x^3 = 2400
x^3= 20
x = cuberoot(20) = width
3x = 3*cuberoot (20) = length
50/x^2 = 50/[cuberoot(20)]^2= height

Answer 3

Let base width be 'l' ft
Thus base length = 3*l ft
Since volume of box = length*breadth*height = 50ft^3
Thus height = 50/base length*base width = 50/(3*l^2)
Area of Top = Area of bottom = 3*l^2
Cost of top+bottom= $2*3*l^2*10 = $60*l^2
Total area of sides = 2*[length*height]+2*[breadth*height] = =2*height*[length+breadth]
=[2*50/(3*l^2)]*(3l+l)
=400*l/3*l^2
=400/(3*l)
Cost of sides= 6*400/(3*l)= $800/l

Total cost=800/l + 60*l^2
Differentiate with respect to l and equate to zero for minimum
thus
-800/l^2+120*l=0
l^3=20/3
l=1.882 ft
Length=5.646 ft
Breadth=1.882 ft
Height=4.705 ft