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Math 251-copyright Joe Kahlig, 16A
1. False: You could have two maxima and a saddle point. Think about a pringle chip is the high
parts continue to a local maxima.
False: All paths (linear, quadratic, cubic,...) must go to the value of 2 for the limit to be 2. you
only know about the linear paths.
True: This is the definition of a relative extrema. Both partial derivatives are zero at this point.
True: if you compute fxy and fyx these are equal and both of these are continuous so there is a
fucntion that will generate these partials. f (x, y) = y ln(x2 + 1) + C where C is some constant.
2. (1, 2) local min
(1, 15) saddle point
(3, 8) saddle point
(4, 15) need more info since fxx fyy − (fxy )2 = 0
(2, 7) local max
3. f (x, x2 + 6) =
x2 + 6 − 5x
x2 − 5x + 6
=
x2 + 6 − 10
x2 − 4
Use L’Hopital rule to compute lim f (x, x2 + 6) =
x→2
−1
4
4. dx = 5.3 − 5 = 0.3 and dy = 0.6 − 21 = −0.4
1
−4
fx =
and fy =
x − 4y
x − 4y
dz = fx dx + fy dy
1
−4
∗ (0.3) +
∗ (−0.4) = 1.9
5 − 4(1)
5 − 4(1)
Thus f (5.3, 0.6) ≈ f (5, 1) + dz = 2 + 1.9 = 3.9
5. (a) ∇g = 2xy, 8y − 2 + x2
∇g(1, 3) = h6, 23i
(b) The vector from (1, 3) to (3, 4) is v = h2, 1i. make this into a unit vector
now compute dz at the point (5, 1): dz =
1
u = √ h2, 1i
5
1
35
Du g(1, 3) = h6, 23i · √ h2, 1i = √
5
5
(c) h−6, −23i
√
(d) | h6, 23i | = 565
x3
so that given surface is considered the level surface when F = 0.
z
Thus ∇F will the the normal vector to the tangent plane at the point (2, 1, 4), and this vector
will be the direction vector for the normal line at this point. Note any vector parallel to this
vector will work as the direction vector.
6. Let F = 3y 4 z − 10 −
∇F (2, 1, 4) = h−3, 48, 3.5i
x = 2 − 3t
y = 1 + 48t
z = 4 + 3.5t
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Math 251-copyright Joe Kahlig, 16A
7. fy = x2 − 4y + 8 and fx = 2xy − 6x = 2x(y − 3) and fx = 0 when x = 0 or y = 3
if x = 0 then fy = 0 gives 0 = −4y + 8 or y = 2
if y = 3 then fy = 0 gives 0 = x2 − 4 or x = ±2
points (0, 2), (2, 3) and (−2, 3)
8. Let F = y 2 z 5 + x4 y − xz then zx =
−Fx
−(4x3 y − zxz−1 )
=
fz
5y 2 z 4 − xz ln(x)
9. The normal vector for the surface,z = f (x, y) = 3x2 + y 2 , at any point is given by
n1 = hfx , fy − 1i = h6x, 2y, −1i
The normal vector for the plane is n2 = h120, 50, −5i
The point we want is where n1 = cn2 Easiest method is to divide n2 by 5 so that the z
components are equal. Then
6x = 24 and 2y = 10 or x = 4 and y = 5
The point on the surface is (4, 5, 73)
10. step 1 find fx and fy and solve for any critical points in the interior. you get the point (1, 5)
note this in not in the region so ignore it.
Boundary 1: x-axis.
x = x and y = 0 for 0 ≤ x ≤ 2.
g(x) = f (x, 0) = 3x2 − x.
g 0 = 6x − 1 set equal to zero and solve
has a critical value at x = 1/6
Boundary 2: vertical line x=2
x=2, y = y for 0 ≤ y ≤ 8
g(y) = f (2, y) = 10 − y.
since g 0 = −1 no critical values.
Boundary 3:
x= x, and y = 4x for 0 ≤ x ≤ 2
g(x) = f (x, 4x) = −x2 + 3x
take derivative and find critical value x = 1.5
critical points function value
(0,0)
0
10
(2.0)
(2,8)
2
(1/6, 0)
-1/12
(1.5,6)
2.25
abs max = 10
abs min = -1/12