Respuesta :
Answer:
x = ∛(2V/7)
y = ∛(2V/7)
z = 3.5 [∛(2V/7)]
{x,y,z} = { ∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)] }
Step-by-step explanation:
The aquarium is a cuboid open at the top.
Let the dimensions of the base of the aquarium be x and y.
The height of the aquarium is then z.
The volume of the aquarium is then
V = xyz
Area of the base of the aquarium = xy
Area of the other faces = 2xz + 2yz
The problem is to now minimize the value of the cost function.
The cost of the area of the base per area is seven times the cost of any other face per area.
With the right assumption that the cost of the other faces per area is 1 currency units, then, the cost of the base of the aquarium per area would then be 7 currency units.
Cost of the base of the aquarium = 7xy
cost of the other faces = 2xz + 2yz
Total cost function = 7xy + 2xz + 2yz
C(x,y,z) = 7xy + 2xz + 2yz
We're to minimize this function subject to the constraint that
xyz = V
The constraint can be rewritten as
xyz - V = 0
Using Lagrange multiplier, we then write the equation in Lagrange form
Lagrange function = Function - λ(constraint)
where λ = Lagrange factor, which can be a function of x, y and z
L(x,y,z) = 7xy + 2xz + 2yz - λ(xyz - V)
We then take the partial derivatives of the Lagrange function with respect to x, y, z and λ. Because these are turning points and at the turning point, each of the partial derivatives is equal to 0.
(∂L/∂x) = 7y + 2z - λyz = 0
λ = (7y + 2z)/yz = (7/z) + (2/y) (eqn 1)
(∂L/∂y) = 7x + 2z - λxz = 0
λ = (7x + 2z)/xz = (7/z) + (2/x) (eqn 2)
(∂L/∂z) = 2x + 2y - λxy = 0
λ = (2x + 2y)/xy = (2/y) + (2/x) (eqn 3)
(∂L/∂λ) = xyz - V = 0
We can then equate the values of λ from the first 3 partial derivatives and solve for the values of x, y and z
(eqn 1) = (eqn 2)
(7/z) + (2/y) = (7/z) + (2/x)
(2/y) = (2/x)
y = x
Also,
(eqn 1) = (eqn 3)
(7/z) + (2/x) = (2/y) + (2/x)
(7/z) = (2/y)
z = (7y/2)
Hence, at the point where the box has minimal area,
y = x,
z = (7y/2) = (7x/2)
We can then substitute those into the constraint equation for y and z
xyz = V
x(x)(7x/2) = V
(7x³/2) = V
x³ = (2V/7)
x = ∛(2V/7)
y = x = ∛(2V/7)
z = (7x/2) = 3.5 [∛(2V/7)]
The values of x, y and z in terms of the volume that minimizes the cost function are
{x,y,z} = {∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)]}
Hope this Helps!!!
The dimensions of the aquarium that minimize the cost of the materials will be {x,y,z} = {∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)]}.
How to compute the dimension?
From the information, the aquarium is a cuboid open at the top. Let the dimensions be x, y, and z. The volume of the aquarium will be V = xyz
The problem is to minimize the value of the cost function. Since the cost of the area of the base per area is seven times the cost of any other face per area, the cost of the base of the aquarium per area would then be 7 currency units.
- Cost of the base of the aquarium = 7xy
- cost of the other faces = 2xz + 2yz
- Total cost function = 7xy + 2xz + 2yz
C(x,y,z) = 7xy + 2xz + 2yz and we are to minimize this function subject to the constraint. The constraint can be rewritten as xyz - V = 0
Using Lagrange multiplier, L(x,y,z) = 7xy + 2xz + 2yz - λ(xyz - V)
Taking the partial derivatives,
(∂L/∂x) = 7y + 2z - λyz = 0
λ = (7y + 2z)/yz = (7/z) + (2/y) (eqn 1)
(∂L/∂y) = 7x + 2z - λxz = 0
λ = (7x + 2z)/xz = (7/z) + (2/x) (eqn 2)
(∂L/∂z) = 2x + 2y - λxy = 0
λ = (2x + 2y)/xy = (2/y) + (2/x) (eqn 3)
(∂L/∂λ) = xyz - V = 0
Equating the equations
(7/z) + (2/y) = (7/z) + (2/x)
(2/y) = (2/x)
y = x
(7/z) + (2/x) = (2/y) + (2/x)
(7/z) = (2/y)
z = 7y/2
Since xyz = V, x(x)(7x/2) = V
(7x³/2) = V
x³ = (2V/7)
x = ∛(2V/7)
y = x = ∛(2V/7)
z = (7x/2) = 3.5 [∛(2V/7)]
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