Quiz 10
Math 2850-007
Fall 2014
Name:
Let the function f : R2 −→ R1 be given by the formula
f (x, y) = x3 + 3xy 2 − 15x + y 3 − 15y
There is exactly one point P (x, y) where f has a local minimum. Find
this point. (There may or may not be points where f has local maxima
or saddle points, but I don’t care about these.)
Mark your answer by placing a check in the appropriate box below.
You are encouraged, although not required, to show your work.
(a) (2, 1)
(b) (−2, 1)
(c) (2, −1)
(d) (−2, −1)
√ (e) 0, 5
(f)
√ 0, − 5
solution
For each answer we need to check whether the given point is a critical
point and, if it is, whether the second derivative test verifies it is a
local minimum.
Find Critical Points
P is a critical point if fx (P ) = 0 and fy (P ) = 0.
fx (x, y) = 3x2 + 3y 2 − 15
fy (x, y) = 6xy + 3y 2 − 15
(a) fx (2, 1) = 0 and fy (2, 1) = 0
(b) fx (−2, 1) = 0 and fy (−2, 1) = −24; not a critical point
(c) fx (2, −1) = 0 and fy (2, −1) = −24; not a critical point
(d) fx (−2, −1) = 0 and fy (−2, −1) = 0
√
√
(e) fx (0, 5) = 0 and fy (0, 5) = 0
√
√
(f) fx (0, − 5) = 0 and fy (0, − 5) = 0
Analyze Critical Points
f has a local minimum at a critical point P if
fxx (P )fyy (P ) − fxy (P )2 > 0 and fxx (P ) > 0.
fxx (x, y) = 6x
fxy (x, y) = 6y
fyy (x, y) = 6x + 6y
2
∆(x, y) = fxx fyy − fxy
= 36(x2 + xy − y 2 )
So we check each critical point in turn:
(a) ∆(2, 1) = 36 · 5 > 0 and fxx (2, 1) = 12 > 0, so f has a local
minimum at (2, 1).
(b) not a critical point
(c) not a critical point
(d) ∆(−2, −1) = 36 · 5 > 0 and fxx (−2, −1) = −12 < 0, so f has a
local maximum at (−2, −1).
√
√
(e) ∆(0, 5) = −36 · 5 < 0, so f has a saddle point at (0, 5).
√
√
(f) ∆(0, − 5) = −36 · 5 < 0, so f has a saddle point at (0, − 5).