Respuesta :
Answer:
6. 12 cm square by 6 cm deep
7. Tmax = 0.125k
Step-by-step explanation:
You want the dimensions of an open-top rectangular box with surface area 432 cm², and you want the maximum temperature T=kxyz² on the surface x² +y² +z² = 1.
Lagrangian
Each of these optimization problems can be solved using the method of Lagrange multipliers. First we define a Lagrangian function for maximizing f(x, y, z) subject to the constraint g(x, y, z) = 0. Then we set the partial derivatives with respect to the variables to zero and solve the resulting set of equations.
6. Open box
We want to maximize the volume f(x, y, z) = xyz, subject to the constraint g(x, y, z) = xy +2(xz +yz) -432 = 0. Then the Lagrangian is ...
ℒ(x, y, z, λ) = xyz +λ(xy +2(xz +yz) -432)
and the partial derivatives are ...
∂ℒ/∂x = yz +λ(y+2z) = 0
∂ℒ/∂y = xz +λ(x+2z) = 0
∂ℒ/∂z = xy +2λ(x+y) = 0
∂ℒ/∂λ = xy +2(xz +yz) -432 = 0
Solving the first two equations for λ gives ...
[tex]-\dfrac{yz}{y+2z}=\lambda=-\dfrac{xz}{x+2z}\quad\Longrightarrow\quad x=y[/tex]
Using this relation and solving the second and third equations for λ gives ...
[tex]-\dfrac{xz}{x+2z}=\lambda =-\dfrac{x^2}{4x}\quad\Rightarrow\quad 4x^2z=x^3+2x^2z\quad\Rightarrow\quad 2z=x[/tex]
Substituting these relations into the last equation, we find ...
(2z)² +2(2z² +2z²) -432 = 0
12z² = 432
z = √(432/12) = 6
x = y = 2z = 12
The dimensions of the box are 12 cm square by 6 cm deep.
7. Hot spot
We want to maximize the temperature T(x, y, z) = kxyz², subject to the constraint g(x, y, z) = x² +y² +z² -1 = 0. Then the Lagrangian is ...
ℒ(x, y, z, λ) = kxyz² +λ(x² +y² +z² -1)
and the partial derivatives are ...
∂ℒ/∂x = kyz² +λ(2x) = 0
∂ℒ/∂y = kxz² +λ(2y) = 0
∂ℒ/∂z = 2kxyz +λ(2z) = 0
∂ℒ/∂λ = x² +y² +z² -1 = 0
Using the same approach as for the previous problem, we can solve the first two equations for λ:
[tex]-\dfrac{kyz^2}{2x}=\lambda=-\dfrac{kxz^2}{2y}\quad\Longrightarrow\quad\dfrac{y}{x}=\dfrac{x}{y}\quad\Longrightarrow\quad x^2=y^2[/tex]
Solving the second and third equations for λ gives ...
[tex]-\dfrac{kxz^2}{2y}=\lambda=-\dfrac{2kxyz}{2z}\quad\Longrightarrow\quad z^2=2y^2[/tex]
Substituting these relations into the last equation, we find ...
x² + x² +2x² -1 = 0
x² = y² = 1/4
z² = 2y² = 1/2
T(x, y, z) = k(√0.25)(√0.25)(0.5) = 0.125k
The highest temperature on the surface of the unit sphere is 0.125k.
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Additional comment
In volume optimization, the open-top box is essentially half a cube. If weighting is applied to the areas (say, some cost function), then we find the pairs of opposite sides and the base all have the same cost.
In the temperature optimization we find the relative values of x², y², and z² in the sum (1, 1, 2) match the exponents in the objective function.
