Critical Points of a Differentiable Function
Author: Joseph Lo

Consider the function \[f(x,y) = xy - \dfrac{1}{9} (x^3+y^3).\] The critical points (\((0,0)\) and \((3,3)\)) occur when \(\nabla f = \langle f_x, f_y \rangle = \langle 0, 0 \rangle\). There are no other critical points since \(f(x,y)\) is differentiable for all \(x\) and \(y\).

The diagram shows the surface of f(x,y) and its level curve containing the point \(P(x,y)\). As \(P\) approach the critical points, \(\nabla f \to langle 0,0 \rangle\)

Maximum

Saddle

∇f

Show maximum

Show saddle point

Show plane

Maximum

Show ∇f

Show tangent in dir of ∇f

f(x, y) = ((-x^(3))) / 9 - y^(3) / 9 + (x * y)
fx = Derivative(f(x, y), x)
fy = Derivative(f(x, y), y)
Curves = Sequence(f(x, y) = i, i, -10, 10)
x0 = 2 (slider from -2 to 5, step 0.1)
y0 = 3 (slider from -1 to 4.5, step 0.1)
P: (x0, y0)
fP: f(x0, y0)
Curve: f(x, y) = fP
X: (x0, y0, 0)
Z: (x0, y0, fP)
Plane = Plane((0, 0, fP), (1, 0, fP), (0, 1, fP))
fxP: fx(x0, y0)
fyP: fy(x0, y0)
gradf = Vector(P, (x0 + fxP, y0 + fyP))
Max = (3, 3)
Saddle = (0, 0)
Saddle3D: (0, 0, f(0, 0))
lengthgf: sqrt(fxP^(2) + fyP^(2))
B: (x0 + ((1 + 0.2 / lengthgf) * fxP), y0 + ((1 + 0.2 / lengthgf) * fyP))
maxcheck = false
saddlecheck = false
poly1, c, d, e, g = Polygon((-9.854378668830853, 5.046649757418296), (10.145621331169147, 5.046649757418296), (10.145621331169147, 11.04664975741829), (-9.854378668830853, 11.04664975741829))
zplanecheck = true
Max3D: (3, 3, f(3, 3))
Curves2 = Sequence(f(x, y) = i, i, -20, 5)
text2: "Point P = (" + x0 + "," + y0 + ")"
text3: "∇f(" + x0 + "," + y0 + ") = <" + fxP + "," + fyP + ">"
text4: "f(" + x0 + "," + y0 + ") = " + fP + ""
gradientcheck = true
text5: "z = f(" + x0 + "," + y0 + ")"
tangentline = Line((x0, y0, fP), (x0 + fxP / lengthgf, y0 + fyP / lengthgf, fP + lengthgf))
tangentcheck = true
text6: "|∇f(" + x0 + "," + y0 + ")| = " + lengthgf + ""