Let f be a multivariable function defined by f(x, y) = x^3y – x^2y^2 where x and y are real
numbers.
Choose the same point to use as you work to complete parts A through D of the task.
I choose the point P = (1,0)
Requirements:
A. Explain how to find the direction of maximum increase for f at your chosen point, showing
all required work.
Solution:
Direction of maximum increase is f(P) = f(1,0)
Where f(x,y) = (fx,fy) = (3x2y-2xy2, x3-2x2y)
Then: f(1,0) = (0,1)
Answer: The direction of maximum increase for f at P=(1,0) is (0,1)
B. Explain how to find the direction of maximum decrease for f at your chosen point, showing
all required work.
Solution:
Direction of maximum decrease is -f(P) = -f(1,0)
Where f(x,y) = (fx,fy) = (3x2y-2xy2, x3-2x2y)
Then: -f(1,0) = -(0,1) = (0,-1)
Answer: The direction of maximum decrease for f at P=(1,0) is (0,-1)
C. Explain how to find the equation of the tangent plane to f at your chosen point, showing all
required work.
Solution:
The equation of the tangent plane to f at P = (1,0) is:
z-f(1,0) = fx(P)(x-1)+fy(P)(y-0)
z – 0 = 0(x-1)+1(y-0)
z=y
So the equation of the plane required is: y-z =0
Answer: y-z = 0
D. Explain how to find the equation of the normal line to f at your chosen point, showing all
required work.
Solution:
Since the equation of the tangent plane to f at the point P=(1,0) is:
y-z = 0 , a normal vector to that plane is: (0,1,-1)
Then the equation of a the normal line at P=(1,0) is:
(x,y,z) = (1,0,0) + (0,-1,1) ,  real
Answer:
x=1
y = -
z = 
real)
E. Demonstrate that the second derivative test for local extreme values of f is inconclusive for
all points on the y-axis.
Solution:
First we have to find the critical points of f solving this system:
fx(x,y) =3x2y-2xy2 = xy(3x-2y) = 0
fy(x,y) = x3-2x2y = x2(x-2y) =0
So the points on the y-axis ( points with the form (0,y)) are solutions of the system
So every point on the y-axis is a critical point.
Then we must compute D = fxxfyy –(fxy)2 at the given point (0,y)
fxx = 6xy -2y2 , then fxx(0,y) = -2y2
fyy = -2x2 = 0
fxy = 3x2-4xy = 0
D(0,y) = (-2y2)0-02 = 0
Since D(0,y) = 0 the second derivative test for local extreme values of f is inconclusive for all
points on the y-axis.