Chapter 4: Partial Differentiation
Section 4.9: Constrained Optimization
Example 4.9.10
Find the extrema of fx,y,z=2 x+3 y−5 z subject to the constraints 3 x2+2 y2=4 and 4 x+3 z=1.
Solution
Mathematical Solution
Figure 4.9.10(a) shows the constraint g=0 as the red cylinder, the constraint h=0 as the green plane, and the curve of intersection in black.
The constrained optimization problem consists in finding the extreme values of the function f along the black curve of intersection of the constraints.
Figure 4.9.10(b) shows the level surfaces of f at the two extrema on the intersection of the constraints. At each extrema, ∇f is drawn in black, ∇g, in red, and ∇h, in green.
At each extrema, ∇f lies in the plane determined by the vectors ∇g and ∇h.
use plots, Student:-VectorCalculus in module() local p1,p2,p3,p4,g,h; g:=3*x^2+2*y^2=4; h:=4*x+3*z = 1; p2:=implicitplot3d([g,h],x=-2..3,y=-2..2,z=-3..3,style=surface,color=[red,green],transparency=[0,.3]); p3:=intersectplot(g,h,x=-2..2,y=-2..2,z=-3..3,color=black,thickness=3); p4:=display(p2,p3,scaling=constrained,axes=frame,labels=[x,y,z],view=[-2..3,-2..2,-3..3],tickmarks=[5,7,9],orientation=[55,80,0]); print(p4); end module: end use:
Figure 4.9.10(a) Constraints g=0 and h=0
use plots, Student:-VectorCalculus in module() local p1,p2,p3,p4,g,h,M,GfM,GgM,GhM,m,Gfm,Ggm,Ghm,fM,fm,f,p1M,p1m; g:=3*x^2+2*y^2=4; h:=4*x+3*z = 1; f:=2*x+3*y-5*z; fM:=1/9*sqrt(9570)-5/3; fm:=-1/9*sqrt(9570)-5/3; M:=[52/4785*sqrt(9570),9/1595*sqrt(9570),-208/14355*sqrt(9570)+1/3]; m:=[-52/4785*sqrt(9570),-9/1595*sqrt(9570),208/14355*sqrt(9570)+1/3]; GfM:=RootedVector(root=M,<2,3,-5>); GgM:=RootedVector(root=M,<104/1595*sqrt(9570),36/1595*sqrt(9570),0>); GhM:=RootedVector(root=M,<4,0,3>); Gfm:=RootedVector(root=m,<2,3,-5>); Ggm:=RootedVector(root=m,<-104/1595*sqrt(9570),-36/1595*sqrt(9570),0>); Ghm:=RootedVector(root=m,<4,0,3>); p1M:=PlotVector([GfM,GgM,GhM],color=[black,red,green],width=.2); p1m:=PlotVector([Gfm,Ggm,Ghm],color=[black,red,green],width=.2); p2:=intersectplot(g,h,x=-2..2,y=-2..2,z=-3..3,color=black,thickness=3); p3:=implicitplot3d([f=fM,f=fm],x=-4..4,y=-3..3,z=-4..4,style=surface,transparency=.3); p4:=display(p1M,p1m,p2,p3,scaling=constrained,axes=frame,labels=[x,y,z],view=[-8..7,-3..4,-6..6],tickmarks=[8,7,9],orientation=[56,63,-1]); print(p4); end module: end use:
Figure 4.9.10(b) Gradients of f,g,h
The Lagrange multiplier method finds the point of tangency of the level surfaces of f with the set of points common to the two constraints. The constraints intersect in a closed curve. Along this curve, the extrema of f occur when such a plane is tangent to the curve. Such points of tangency are found by making ∇f a linear combination of the gradients of the two constraints, that is, by solving the equations in ∇f=λ ∇g+μ ∇h, along with the two constraint equations g=h=0. This is the set of five equations
−3x2−2y2+4
= 0
−4 x−3 z+1
−6 λ x−4 μ+2
−4 λ y+3
−3 μ−5
in the five unknowns x,y,z,λ,μ. The solutions of these equations are then
λ=1729570,μ=−53,x=5247859570,y=915959570,z=−208143559570+13
and
λ=−1729570,μ=−53,x=−5247859570,y=−915959570,z=208143559570+13
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign Name
f=2 x+3 y−5 z→assign
g=3 x2+2 y2−4→assign
Context Panel: Assign name
h=4 x+3 z−1→assign
Apply the LagrangeMultipliers command in the Student MultivariateCalculus package, captured in the task template. See Table 4.9.10(a).
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Optimization≻Lagrange Multiplier Method
Method of Lagrange Multipliers
Enter objective function f
Enter constraints gk=0,k=1,…,entered as functions g1,g2,…
Enter coordinate variables, separated by commas:
Table 4.9.10(a) The Lagrange Multiplier Method task template
Converting the more cumbersome exact values in Table 4.9.10(a) to floating-point numbers results in the display to the right.
Obviously, one extrema is a maximum, and the other, a minimum.
x1.063-1.063y.5521-.5521z-1.0851.751λ11.359-1.359λ2-1.667-1.667f9.203-12.54
Optimization Assistant
The Optimization Assistant is no longer listed in Tools≻Assistants. It is now found in Tools≻Tutors≻Optimization
A numeric solution is available via the , launched from the Context Panel on the sequence f,g=0,h=0.
Figure 4.9.10(c) shows the Optimization Assistant finding the maximum of 9.2 at approximately 1.06,0.552,−1.084.
Figure 4.9.10(d) shows the Optimization Assistant finding the minimum of -12.54 at approximately −1.063,−0.552,1.751.
Figure 4.9.10(c) Constrained maximum
Figure 4.9.10(d) Constrained minimum
Implement the Lagrange multiplier method via first principles
F=f−λ g−μ h→assign
Write F and press the Enter key.
Context Panel: Student Multivariate Calculus≻Differentiate≻Gradient
Context Panel: Conversions≻To List
Context Panel: Solve≻Solve (explicit)
F
−3x2+2y2−4λ−4x+3z−1μ+2x+3y−5z
→gradient
→to list
−3x2−2y2+4,−4x−3z+1,−6λx−4μ+2,−4λy+3,−3μ−5
→solve
λ=1729570,μ=−53,x=5247859570,y=915959570,z=−208143559570+13,λ=−1729570,μ=−53,x=−5247859570,y=−915959570,z=208143559570+13
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define f, the objective function.
f≔2 x+3 y−5 z:
Define g, the first constraint function.
g≔3 x2+2 y2−4:
Define h, the second constraint function.
h≔4 x+3 z−1:
Implement the Lagrange multiplier method
Invoke the LagrangeMultipliers command from the Student MultivariateCalculus package. Use the print command and the tilde operator to list the solutions one beneath the other.
S≔LagrangeMultipliersf,g,h,x,y,z: print~S
5247859570,915959570,−208143559570+13
Add the "detailed" option to the LagrangeMultipliers command. Use the print command and the tilde operator to list the solutions one beneath the other.
Q≔LagrangeMultipliersf,g,h,x,y,z,output=detailed: print~Q
x=5247859570,y=915959570,z=−208143559570+13,λ1=1729570,λ2=−53,2x+3y−5z=199570−53
Implement the Lagrange multiplier method from first principles
Define F.
F≔f− λ g−μ h:
Use the Gradient command to obtain ∇f−λ ∇g.
Use the Equate command to equate each component of ∇f−λ ∇g to zero.
Use the solve command to obtain the real solution of the equations in ∇f−λ ∇g=0. Use the print command and the tilde operator to list the solutions one beneath the other.
R≔solveEquateGradientF,x,y,z,λ,μ,0,0,0,0,0,x,y,z,λ,μ,explicit: print~R
λ=1729570,μ=−53,x=5247859570,y=915959570,z=−208143559570+13
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