Math 225: Calculus III    Quiz 6    3/4&6/97

<<Calculus`Math225`

Let R be the triangular region bounded by the lines y==2+x, y==2-x, and y==0. Find the maximum and minimum values of f[x_,y_] = x^2 - x y + x on R and the points where they occur.

Solution

f[x_,y_] = x^2 - x y + x;

First find the critical points in R:

cp = CriticalPoints[f[x,y]]

f[0,1]

The Lagrange points on the boundary segments are:

lp1 = Lagrange[f[x,y], x + y == 2]

f[1/4,7/4]

lp2 = Lagrange[f[x,y], -x + y == 2]

lp3 = Lagrange[f[x,y], y == 0]

f[-1/2,0]

Of course, we must also check the "corner" points:

f[-2,0]

f[2,0]

f[0,2]

Therefore, the maximum is 6 at {2,0} and the minimum is -1/4 at {-1/2,0}.

ParametricPlot3D[{s(y-2),y,f[s(y-2),y]},{s,-1,1},{y,0,2},
PlotRange->{-1,6},ViewPoint->{3.011, 1.293, 0.845}]

Find the extreme values of f[x_,y_] = Sin[x y] subject to the constraint
x^2 + y^2 == 1.

Solution

f[x_,y_] = Sin[x y];

Plot3D[f[x,y],{x,-2,2},{y,-2,2}]

ParametricPlot3D[{Cos[t],Sin[t],f[Cos[t],Sin[t]]}, {t,0,2Pi}]

The Lagrange equations are:
    Cos[x y] y == 2 lambda x
    Cos[x y] x == 2 lambda y
    x^2 + y^2 == 1
Notice that if x==0 then the first eqn implies y==0 and then the third eqn cannot be satisfied. We reach the same conclusion if we first assume y==0. Therefore, we may assume neither x nor y is 0 and also that Cos[x y] is not 0. Solving for lambda we get
    Cos[x y] y/(2x) == Cos[x y] x/(2y)
or
    y^2 == x^2
When we plug this into the third eqn we get x == ±1/Sqrt[2],  y == ±1/Sqrt[2]. We now evaluate f[x,y] at the four resulting Lagrange points.

f[1/Sqrt[2], 1/Sqrt[2]]

f[1/Sqrt[2],-1/Sqrt[2]]

f[-1/Sqrt[2], 1/Sqrt[2]]

f[-1/Sqrt[2],-1/Sqrt[2]]

N[%]

The max is 0.479426 at the points ±{1/Sqrt[2], 1/Sqrt[2]} and the min is -0.479426 at the points ±{1/Sqrt[2],-1/Sqrt[2]}.