Download Name: Solutions Math 16B: Quiz 2 July 3, 2014 1. (4 points) Sketch

Name: Solutions
Math 16B: Quiz 2
July 3, 2014
1. (4 points) Sketch the level curves of the function f (x, y) = (y − x2 )3 at heights −2, 8
and 20 (on the same axes).
For level curve at height c, we have c = f (x, y) = (y − x2 )3 ⇒ c1/3 = y − x2 ⇒ y =
x2 + c1/3 . Plugging in the values of c: y = x2 − 21/3 , y = x2 + 2 and y = x2 + 201/3 .
20
15
y
10
c = 20
5
c=8
0
c = −2
−5
−4
−3
−2
−1
0
x
1
2
3
4
Figure 1
2 −y 2
2. (7 points) Find all the relative maxima and minima of f (x, y) = e4y−x
mine their nature.
and deter-
We find the partial derivatives of f and set them equal to zero. We have:
∂f
2
2
= (−2x)e4y−x −y = 0
∂x
∂f
2
2
= (4 − 2y)e4y−x −y = 0
∂y
̸ 0, the first equation implies that x = 0; from the second, we get
Since e4y−x −y =
4 − 2y = 0 ⇒ y = 2. Thus, the only possible extreme point of this function is at (0, 2).
2
2
2
To determine the nature of (0, 2), we use the second-derivative test. We have: ∂∂xf2 =
2
2
2
2
2
2
2
2
2
∂2f
−2e4y−x −y + 4x2 e4y−x −y , ∂∂yf2 = −2e4y−x −y + (4 − 2y)2 e4y−x −y and ∂x∂y
= (4 −
4y−x2 −y 2
4
4
2
8
2y)(−2x)e
. At (0, 2), we have D = (−2e )(−2e ) − (0) = 4e > 0. Moreover,
∂f
4
since ∂x2 = −2e < 0, f has a relative maximum at (0, 2).
1
3. (9 points) Find three nonnegative numbers whose sum is 12 and the sum of whose
squares is as small as possible. (Finding a relative minimum is fine; 5 bonus points if
you can show that the relative minimum is also an absolute minimum.)
Let the three numbers be x, y and z. We want to minimize S = x2 + y 2 + z 2 subject
to x + y + z = 12 ⇒ z = 12 − x − y. Thus, S = x2 + y 2 + (12 − x − y)2 . To find the
critical points, we set the partial derivatives of S equal to zero:
∂S
= 2x − 2(12 − x − y) = −24 + 4x + 2y = 0
∂x
∂S
= 2y − 2(12 − x − y) = −24 + 2x + 4y = 0
∂y
Add (−2) times the first equation to the second to get 24 − 6x = 0 ⇒ x = 4. Thus,
y = 4 and z = 4. At these values, S = 48.
2
To see whether or not this minimizes S, we use the Hessian test. We have: ∂∂xS2 = 4,
∂2S
∂2S
= 4 and ∂x∂y
= 2. Thus, D = (4)(4) − (2)2 = 16 − 4 = 12 > 0. Moreover, since
∂y 2
∂2S
∂x2
> 0, it follows that this choice of x, y and z relatively minimizes S.
Bonus part: Since z ≥ 0, the domain of S is the region given by x, y ≥ 0 and
x + y ≤ 12. See Figure 2 for the domain.
12
10
8
y
x + y = 12
6
4
2
0
0
2
4
6
x
8
10
12
Figure 2
(a) On the x = 0 boundary, we get S = y 2 + (12 − y)2 = 2y 2 − 24y + 144. This is a
concave-up parabola so it is minimized at its critical point. Differentiate this w.r.t
y and set it equal to zero to get 4y − 24 = 0 ⇒ y = 6. At y = 6, S = 36 + 36 = 72.
Thus, S ≥ 72 on this boundary.
(b) On the y = 0 boundary, we get S = x2 + (12 − x2 ). Same reasoning as above
shows that S never goes below 72 on this boundary.
(c) On x + y = 12, use y = 12 − x and plug into S to get S = x2 + (12 − x)2 . Same
reasoning as above shows that S never goes below 72 on this boundary.
Thus, comparing the value of S at the critical point and at the boundaries, we see that
S attains its least value of 48 at the critical point where x = y = z = 4.
2