Math 241, Quiz 7. 10/15/12.
Name:
• Read problems carefully. Show all work. No notes, calculator, or text.
• There are 15 points total.
1. §14.7, #11 (7 points): Find the local maximum and minimum values and saddle point(s) of
f (x, y) = x3 − 12xy + 8y 3 .
Solution: We first compute the critical points. We solve fx = 3x2 − 12y = 0 ⇐⇒
x2 − 4y = 0 and fy = 24y 2 − 12x = 0 ⇐⇒ 2y 2 − x = 0 ⇐⇒ x = 2y 2 together.
We substitute the second equation in the first to obtain 0 = x2 − 4y = (2y 2 )2 − 4y =
4y 4 − 4y ⇐⇒ y(y 3 − 1) = 0 ⇐⇒ y = 0, 1. When y = 0, we see that x = 0; when
y = 1, the second equation gives x = 2. Hence, the critical points are (0, 0) and (2, 1). We
compute fxx = 6x, fyy = 48y, and fxy = −12. It follows that
0 −12
12 −12
= −144, D(2, 1) = D(0, 0) = −12 48 = 12·48−12·12 = 12·36 = 144·3 = 432 > 0.
−12 0 The Second Derivative Test implies that (0, 0) is a saddle, while (2, 1) is a local minimum
since fxx (2, 1) = 12 > 0.
2. §14.7, #32 (8 points): Find the absolute maximum and minimum values of
f (x, y) = 4x + 6y − x2 − y 2 on the set D = {(x, y) | 0 ≤ x ≤ 4, 0 ≤ y ≤ 5}.
To expedite things, I computed the values of f (x, y) at the four corners of D, which you may
use:
f (0, 0) = 0, f (4, 0) = 0, f (4, 5) = 5, f (0, 5) = 5.
Solution: We first compute the critical points. We solve fx = 4 − 2x = 0 ⇐⇒ x = 2 and
fy = 6 − 2y = 0 ⇐⇒ y = 3. Hence (2, 3) ∈ D is a critical point. Next, we find critical
points on the boundary.
L1 : y = 0. We have f (x, 0) = 4x−x2 . We solve f 0 (x, 0) = 4−2x = 0 ⇐⇒ 2x = 4 ⇐⇒ x = 2
to obtain the critical point (2, 0).
L2 : x = 4. We have f (4, y) = 16 + 6y − 16 − y 2 = 6y − y 2 . We solve f 0 (4, y) = 6 − 2y = 0 ⇐⇒
2y = 6 ⇐⇒ y = 3 to obtain the critical point (4, 3).
L3 : y = 5. We have f (x, 5) = 4x + 30 − x2 − 25 = 4x − x2 + 6. We solve f 0 (x, 5) = 4 − 2x = 0
to obtain the critical point (2, 5).
L4 : x = 0. We have f (0, y) = 6y − y 2 . We solve f 0 (0, y) = 6 − 2y = 0 to obtain the critical point
(0, 3).
To conclude, we have the following points to test:
{(2, 3), (2, 0), (4, 3), (2, 5), (0, 3), (0, 0), (4, 0), (4, 5), (0, 5)}.
It remains to compute f (x, y) at the 5 critical points:
f (2, 3) = 13, f (2, 0) = 4, f (4, 3) = 9, f (2, 5) = 9, f (0, 3) = 9.
Comparing to the given values at the corners, we find that the absolute maximum is f (2, 3) =
13 while the absolute minimum is f (0, 0) = f (4, 0) = 0.