# QUIZ 4 : MATH 251, Section 516

```QUIZ 4 : MATH 251, Section 516
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GRADE : . . . . . . . .
”An Aggie does not lie, cheat or steal, or tolerate those who do”
signature : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1. Let f (x, y, z) = zexy .
(a) [25pts] Find the direction in which f increases most rapidly at the point (0, 1, 2) .
(b) [25pts] Find the directional derivative of f at the point (0, 1, 2) in the direction &lt; 2, 1, −2 &gt;.
2. Let f (x, y) = 4x2 + y 2 − 4x + 2y.
(a) [25pts] Find the critical points of this function.
(b) [25pts] Classify these points ( say if these are maximum, minimum points or saddle points by
justifying).
1. (a) f is defined on Df = R3 . f is differentiable on Df . f increases most rapidly in the direction of
∇f (x, y, z) = hfx (x, y, z), fy (x, y, z), fz (x, y, z)i = hyzexy , xzexy , exy i.
So, at (0, 1, 2), we have ∇f (0, 1, 2) = h2, 0, 1i .
(b) Since f is differentiable on Df , its differential and its directional derivative coincides at (0, 1, 2) ∈
u
2 1 2
Df . So, denote u = h2, 1, −2i. The unit vector in the direction of u is :
=&lt; , , − &gt;. So,
|u|
3 3 3
2 1 2
2 1 2
2
.
Du f (0, 1, 2) = ∇f (0, 1, 2)• &lt; , , − &gt;= h2, 0, 1i• &lt; , , − &gt;=
3 3 3
3 3 3
3
2. f is a polynomial function. It admits second partial derivatives at any points in R2 .
(a) Determine the critical points of f i.e the sets of points (x, y) such that fx (x, y) = 8x − 4 = 0 =
fy (x, y) = 2y + 2. Solve the following system
8x − 4 = 0
⇔
2y + 2 = 0
(
1
2
y = −1,
x =
1
We obtain the only one critical point : ( , −1) .
2
(b) Now, classify this point. We have fxx (x, y) = 8, fyy (x, y) = 2, fxy (x, y) = 0 = fyx (x, y) (by the
Clairaut’s Theorem since the function is polynomial).
2
So, fxx ( 12 , −1)fyy ( 21 , −1) − fxy ( 12 , −1)
= 16 &gt; 0 and fxx ( 21 , −1) = 8 &gt; 0 so ( 12 , −1) is
a local minimum point.
```