Chapter 4: Partial Differentiation
Section 4.9: Constrained Optimization
Example 4.9.1
Find the extreme values of fx,y=2 x2+3 y2 subject to the constraint gx,y≡x2+y2−1=0.
Solution
Mathematical Solution
Figure 4.9.1(a) shows the surface z=f and on it, the "lift" of the constraint curve g=0. From this figure, infer that there are four extrema, two maxima and two minima.
Figure 4.9.1(b) shows the level curves of f in black, and the curve defined implicitly by g=0, in red. The points of tangency between the constraint curve and a level curve of f defines a stationary point for the constrained optimization problem.
These stationary points occur where the gradients of f and g are collinear, that is, where ∇f is proportional to ∇g. The animation in Figure 4.9.1(c) shows these gradients traversing the constraint curve g=0. They become collinear at the four points of tangency that are the stationary points that the Lagrange multiplier method finds.
use plots in module() local p1,p2,p3,a; a:=1.37; p1:=plot3d(2*x^2+3*y^2,x=-a..a,y=-a..a,transparency=0,lightmodel=none); p2:=spacecurve([cos(t),sin(t),2*cos(t)^2+3*sin(t)^2],t=0..2*Pi,color=black,thickness=3); p3:=display(p1,p2,scaling=constrained,axes=frame,view=0..4,tickmarks=[3,3,5],labels=[x,y,z],orientation=[-35,70,0]); print(p3); end module: end use:
Figure 4.9.1(a) Constraint curve lifted to the surface f
use plots in E491:=module() local p1,p2,f,g; export p3; f:=2*x^2+3*y^2; g:=x^2+y^2-1: p1 := implicitplot(g=0,x=-1..1,y=-1..1, color=red): p2 := contourplot(f,x=-1.5..1.5,y=-1.2..1.2, contours=[1,2,3,4],color=black): p3:=display([p1,p2],tickmarks=[[-1,0,1],[-1,0,1]],scaling=constrained): print(p3); end module: end use:
Figure 4.9.1(b) f=c in black and g=0 in red
use Student:-VectorCalculus, plots in module() local p1,X; X:=proc(t) local Vf,Vg; Vf:=t->RootedVector(root=[cos(t),sin(t)],.7*<2*cos(t),3*sin(t)>); Vg:=t->RootedVector(root=[cos(t),sin(t)],<cos(t),sin(t)>); PlotVector([Vf(t),Vg(t)],color=[black,red]); end; p1:=animate(X,[t],t=0..2*Pi,frames=40,background=E491:-p3); print(p1); end module: end use:
Figure 4.9.1(c) ∇f and ∇g on g=0
The equations arising from ∇f−λ ∇g = 4 x6 y−λ 2 x2 y=00, and the constraint equation itself, are then
4 x−2 λ x=0,6 y− λ y=0,x2+y2=1
The first equation is satisfied if x=0 or λ=2. The second equation is satisfied if y=0 or λ=3.
If x=0, the third equation gives y=±1. If y=0, the third equation gives x=±1. Consequently, there are four solutions to these equations: 0,1,0,−1,1,0, and −1,0.
Maple Solution - Interactive
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Context Panel: Assign name
f=2 x2+3 y2→assign
Context Panel: Assign Name
g=x2+y2−1→assign
Optimization Assistant
The Optimization Assistant is no longer listed in Tools≻Assistants. It is now found in Tools≻Tutors≻Optimization
A numeric solution and some useful graphics are available via the , launched from the Context Panel on the sequence f,g=0.
Each of the four extrema are found separately.
The first extrema, the minimum of 2 at x,y=1,0, is obtained by clicking on the Solve button. If the Plot button is next clicked, Figures 4.9.1(e, f) can be obtained. Note the modification of the plotting ranges.
A second minimum can be obtained by adding initial values for the numeric search. Click the Edit button at the right of "Initial Values" and provide x,y=−3,1 for new initial values. The search will now find the minimum 2 at −1,0.
Figure 4.9.1(d) Optimization Assistant
Figure 4.9.1(e) Surface, constraint, & minimum
Figure 4.9.1(f) Constraint as a surface
The maxima can be found by again adjusting the initial values.
Implement the Lagrange multiplier method via first principles
F=f−λ g→assign
Write F and press the Enter key.
Context Panel: Student Multivariate Calculus≻ Differentiate≻Gradient
Context Panel: Conversions≻To List
Context Panel: Solve≻Solve
F
−x2+y2−1λ+2x2+3y2
→gradient
→to list
−x2−y2+1,−2λx+4x,−2λy+6y
→solve
λ=3,x=0,y=1,λ=3,x=0,y=−1,λ=2,x=1,y=0,λ=2,x=−1,y=0
The LagrangeMultipliers command in the Student MultivariateCalculus package is captured in the task template. Its use is illustrated in Table 4.9.1(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 4.9.1(a) The Lagrange Multiplier Method task template
Maple Solution - Coded
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Define f, the objective function.
f≔2 x2+3 y2:
Define g, the constraint function.
g≔x2+y2−1:
Implement the Lagrange multiplier method
Invoke the LagrangeMultipliers command from the Student MultivariateCalculus package.
LagrangeMultipliersf,g,x,y0,1,0,−1,1,0,−1,0 ⇒
Add the "detailed" option to the LagrangeMultipliers command.
Apply the print command and tilde, the element-wise operator, to print the four solutions one under the other.
S≔LagrangeMultipliersf,g,x,y,output=detailed: print~S
x=0,y=1,λ1=3,2x2+3y2=3
Implement the Lagrange multiplier method from first principles
Define F.
F≔f−λ g:
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 solutions of the equations in ∇f−λ ∇g=0.
solveEquateGradientF,x,y,λ,0,0,0,x,y,λ
x=0,y=1,λ=3,x=0,y=−1,λ=3,x=1,y=0,λ=2,x=−1,y=0,λ=2
The code in Table 4.9.1(b) will draw Figure 4.9.1(b).
use plots in p1 ≔ implicitplotg=0,x=−1..1,y=−1..1, color=red: p2 ≔ contourplotf,x=−1.5..1.5,y=−1.2..1.2, contours=1,2,3,4,color=black: displayp1,p2,tickmarks=−1,0,1,−1,0,1,scaling=constrained: end use:
Table 4.9.1(b) Code for generating Figure 4.9.1(a)
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