Calculus Minimization Problem... ?

Question

A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

Walk through please!

The answer is 163.44
But I don't know how to get it.

Answer 1

x = width
2x = length
10/2x^2 = 5/x^2 = height
Area of base = 2x^2
Area of sides = A = 2(x)(5/x^2) +2(2x)(5/x^2)
A = 10/x + 20/x =30/x
Cost = C = 10(2x^2) +6(30/x) = 20x^2 + 180/x
dC/dx =40x -180/x^2 = 0
40x^3 - 180 =0
x^3 = 180/40 = 9/2
x = 1.65 m
Cost = 20(1.65)^2 +180/1.65 = $163.54

Answer 2

First lets assign some variables
Let x = width
2x = length

and to figure out a variable expression for the height use the volume formula V = lwh
height = V/(lw)

In our case V = 10 so:

10/(2x*x) = height which simplifies to
5/x^2 = height.

Next we write an equation F that we want to optimize.

There are five sides that we need to find the cost of producing. Multiply cost/sq foot by area of each side.

Base = 10(2x*x)=20x^2
Side 1 = 6(x*5/x^2)=30/x = Side 2
Side 3 = 6(2x*5/x^2)=60/x = Side 4

F = 20x^2 +30/x + 30/x + 60/x + 60/x

Combine like terms

F = 20x^2 + 180/x = 12x^2 + 180x^(-1)

Now differentiate

F' = 40x - 180x^(-2)
F' = 40x - 180/x^2

Set the derivative = 0

0 = 40x - 180/x^2

180/x^2 = 40x/1

Cross multiply and solve

180 = 40x^3
45 = x^3
3.577 = x

plug x into F

F = 20(4^2) + 180/4
F = 20*16 + 45
F = 320 +45
F = $365

Answer 3

Let L be the length, W be the width, and H be the height.
L = 2W
Volume = LWH = 10 => H = 10/(LW) = 5/W^2
Objective: cost = f(W) = 10LW + 6(2L+2W)H = 20W^2 + 180/W, where LW is the area of the base, and (2L+2W)H is the area of 4 sides.
Solve f'(W) = 0 for W,
40W - 180/W^2 = 0, edited here
W^3 = 180/40 = 4.5
W = 4.5^(1/3), m
f(4.5^(1/3)) = 20*4.5^(2/3) + 180/4.5^(1/3) = $163.54, the cheapest cost
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The answer you gave is little bit off. Check it!