More Calculus!!?

Question

A rectangular storage container with an open top is to have a volume of 10m³. The lenth of its base is twce the with. Material for the base costs $10 m². Material for the sides costs $6 m². Gine the cost oft he materials for the cheapest such container.
I did:
w=h
V=lwh
2w²h=10m³
2w³=10m³
w³=5
w= 1.71
2w= 3.42

I'm not sure I'm doing this right.

Answer 1

This question is fairly complicated to explain, but here goes:

first of all, w does not equal to h, and remeber, cheapest means minimum, you have to differentiate something
to find h, we have V=lwh, l=2w, V=10, therefore, V=2w^2h,
h=5/w^2,
the problem is open top (the top surface area does not count, so the surface area=
l*w+2(l*h)+2(w*h)
substitute l and h in terms of w, you should get
2w^2+30/w, remember, 2w^2 is the base as state in the question (length of the base.....) and 30/w is the sides.

we have the cost of base =10, and cost of side is 6, so total cost is 10*(area f base)+6*(area of sides)=
20w^2+180/w

differentiate that function, we have 40w-180/w^2 and we solve for zero... so 40w=180/w^2, 180=40w^3, w^3=4.5, w=1.65
so:
l=3.3
w=1.65
h=1.83

total cos equals
$163.54

Answer 2

From question:
(1) V = 10 = w * l * h
(2) l = 2 * w
(3) C_{base} = 10 * A_{base}
(4) C_{sides} = 6 * A_{sides}

We can create a function of total cost:
(5) C = C_{base} + C_{sides}

Substituting (3) & (4) into (5):
(6) C = 10 * A_{base} + 6 * A_{sides}

We can find the areas:
(7) A_{base} = l * w
(8) A_{sides} = 2 * l * h + 2 * w * h

Substituting (7) & (8) into (6):
(9) C = 10 * l * w + 12 * h * (l + w)

Substituting (2) into (9):
(10) C = 20 * w^2 + 36 * h * w

Substituting (2) into (1):
(11) 10 = 2 * h * w^2
(11) h = 5 / w^2

Substituting (11) into (10):
(12) C = 20 * w^2 + 180 / w

Optimize (12) by finding relative extrema:
(13) dC/dw = 0 = 40 * w - 180 / w^2
(13) 0 = (4 * w^3 - 18) / w^2
(13) 0 = w^3 - 4.5
(13) w = 4.5^(1/3)
(13) w ~ 1.6509636244473133419373049762045

Find whether this is a max or min by taking second derivative:
(14) d^2C/dw^2 = 40 + 360 / w^3
(14) d^2C/dw^2|(w^3 = 4.5) = 40 + 360 / 4.5 > 0

Since the second derivative is positive, we know that the extrema at this point in a minimum. Therefore we know that our calculated point in the minimum cost.

Solving other variables;
Substituting (13) into (2):
(15) l ~ 2 * 1.6509636244473133419373049762045
(15) l ~ 3.3019272488946266838746099524091

Substituting (13) into (11):
(15) h ~ 5 / (1.6509636244473133419373049762045* 1.6509636244473133419373049762045)
(15) h ~ 1.8344040271636814910414499735612

Therefore, the dimensions of the cheapest box are:
Length ~ 3.30 units
Width ~ 1.65 units
Height ~ 1.83 units

Answer 3

Your assumption w = h is not stated and it is wrong.
Therefore solution will be like this:

Since l = 2w the volume is 2*h*w^2 = 10 or

h = 5/w^2. (1)

The cost function is:

y = 10*l*w + 6*(2*l*h + 2*w*h)
= 20*w^2 + 12*h*3*w, using (1)
= 20*w^2 + 180/w

To find the cost minimum y' = 40*w - 180/w^2 = 0, which gives w^3 = 9/2 or w = 1.651. Then l = 3.302 and h = 1.834.

Answer 4

I haven't seen you derive to get the answer:

The volume 10m^3 = 2w^2*h
Thus, h = 10m^3 / 2w^2 = 5m^3 / w^2

Now, the area of the base is 2w^2, and its cost is 20w^2
There are two measure for sides wh = 5w m^3 / w^2 = 5m^3 / w
2wh = 10m^3/w
Their cost is 6*2*(5/w + 10/w) = 180/w

The overall cost is 20w^2 + 180/w

Let's derive it per w:

0=40w - 180/w^2

40w^3 - 180 = 0

w^3 = 180/40 = 4.5

w=cbrt(4.5) cbrt means cubic root

Let us derive one more time to find if this is a minimum point:
40 + 360/w^3 = 40 + 360/4.5 = 40 + 80 = 120 > 0. That's a minimum point.

And the cost is:
20*cbrt(4.5)^2 + 180/cbrt(4.5) = $163.54...

Answer 5

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Answer 6

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