C Roettger, Fall 14
Math 265 Quiz 7B – Solutions
Problem 1 Consider the function
f (x, y) = x2 y + 8x − y 3 + 2.
Find all critical points of f (x, y) in the plane.
Solution. First, compute the gradient of f (as always):
∇f (x, y) = (2xy + 8, x2 − 3y 2 )
This is the zero vector exactly if
2xy + 8 = 0
x2 − 3y 2 = 0.
AND
√
From the second equation, we get x = ± 3y. Substitute this into the first
to get
√
±2 3y 2 + 8 = 0.
Only the negative sign will give a solution, and for that case we can solve for
y:
s
2
4
= 2 · 3−1/4 .
y = ± √ = ±√
4
3
3
So we get two critical points (2 · 31/4 , 2 · 3−1/4 ) and (−2 · 31/4 , −2 · 3−1/4 ).