Can you help me with this calculus problem please?

Question

A chocolate manufacturer uses an equilateral triangular prism package. If the volume of chocolate to be contained in the package is 400m^3, what dimensions of the package will use the monimum amount of materials?

Answer 1

As Jim observed, that's one mighty big piece of chocolate.

Let

V = volume of package
s = length of triangular side
w = width of rectangular side
K = surface area of package
b = surface area of triangular base

Given

V = 400 m³

Find minimum K for given V.

We have

b = (1/2)[(√3)/2]s² = [(√3)/4]s²
V = bw = [(√3)/4]s²w
w = V/{[(√3)/4]s²} = 4V/[(√3)s²]

K = 2b + 3sw = [(√3)/2]s² + 3sw
K = [(√3)/2]s² + 3sV/{[(√3)/4]s²}
K = [(√3)/2]s² + 12V/[(√3)s]
K = [(√3)/2]s² + 4(√3)V/s

Take the derivative of K and set equal to zero to minimize.

dK/ds = (√3)s - 4(√3)V/s² = 0
(√3)s = 4(√3)V/s²
s³ = 4V
s = (4V)^(1/3)

The second derivative of K determines maximum or minimum.

d²K/ds² = √3 + 8(√3)V/s³ > 0 since both terms are positve
Therefore it is a minimum.

w = 4V/[(√3)s²]
w = 4V/[(√3)(4V)^(2/3)]
w = (4V)^(1/3)/√3
w = s/√3

The dimension of the package for minimum surface area are:

s = (4V)^(1/3) = (4*400)^(1/3) = (1600)^(1/3)
s = (64*25)^(1/3) = 4(25)^(1/3) = 11.696071 m
w = s/√3 = 4(25)^(1/3)/√3 = 6.7527297 m

Answer 2

Let s = length of a side of the triangle
Let h = height of prism

Volume of an equilateral triangular prism
V = Area of equilateral triangle * height of prism
= (Sqrt(3)/4)*s^2*h

Surface area of an equilateral triangular prism
SA = 2*Area of equilateral triangle + 3*Area of rectangle
= (Sqrt(3)/2)*s^2 + 3*s*h

Since we know V = 400, we can easily solve for h in terms of s using the first equation:
400 = (Sqrt(3)/4)*s^2*h
h = 400/[(Sqrt(3)/4)*(s^2)]

Substituting this h into the second equation we get:
SA = (Sqrt(3)/2)*s^2 + 4800/[Sqrt(3)*s]

Since we are trying to minimize the material (i.e. minimize the surface area), we would take the derivative of SA with respect to s, and then find critical points.
SA' = Sqrt(3)*s - 4800/[Sqrt(3)*(s^2)]
= [3s^3-4800]/[Sqrt(3)*(s^2)]

Finding critical points:
SA' = 0 when 3s^3 - 4800 = 0 => s = +/- (1600)^(1/3)
SA' DNE when s = 0

The answers s = 0 and s = -(1600)^(1/3) do not make sense, so we will consider s = 1600^(1/3). By the first derivative test, s = 40 is the answer that produces minimum surface area. (i.e. note that the derivative is negative when s < 1600^(1/3) and the derivative is positive when s > 1600^(1/3)).

Plugging in s = 1600^(1/3) into the third equation, we get that h = 1600^(1/3)/Sqrt(3)

So the dimensions to minimize the amount of materials used for packaging would be a triangle side length of 1600^(1/3) meters (or 11.696 meters) and a prism height of 1600^(1/3)/Sqrt(3) meters (6.753 meters).

That's a huge piece of chocolate.

Answer 3

Let b be the length of the base and h be the height of the prism.

v = Bh = (√3a^2/4)h = 400

area = s = 2B + 3ah = 2(√3a^2/4) + 3[1600/(√3 a)]

Solving ds/da = 0 for a gives,

a = (1600)^(1/3) = 11.69607 m
h = 6.76273 m
s(min) = 355.4 m^2