Q1
F (x, y) = 11+6x+x2 −4y+y 2 = (x2 +6x+9)−9+(x2 −4x+4)−4+11 = (x+3)2 +(y−2)2 −2
(x + 3)2 , (y − 2)2 ≥ 0 so F (x, y) ≥ −2 = F (−3, 2). Also, we can make F as big as we like by
making (x, y) big enough. So by inspection we have that F (x, y) has a global minimum at
(−3, 2) and no absolute maxima.
∇F (x, y) = (2x + 6, 2y − 4) = (0, 0) ⇒ (x, y) = (−3, 2)
So the only critical point is (-3,2).
2
∂ 2F ∂ 2F
∂ 2F
∂ 2F
∂ 2F
∂ 2F
=0⇒
= 2,
= 2,
−
=4>0
∂x2
∂y 2
∂x∂y
∂x2 ∂y 2
∂x∂y
So (−3, 2) is an extremum. In fact, since
∂2F
∂x2
> 0, (−3, 2) is a local minimum.
Q2
∇F = (2y−162x5 , 2x−2y) = (0, 0) ⇒ x = y, y−81x5 = 0 ⇒ x−81x5 = 0 ⇒ x = 0 or x = ±1/3
So the critical points are (0, 0), (1/3, 1/3) and (−1/3 − 1/3).
2
∂ 2F
∂ 2F
4 ∂ F
=
−810x
,
=
−2,
=2
∂x2
∂y 2
∂x∂y
(x0 , y0 )
(0, 0)
(1/3, 1/3)
(−1/3, −1/3)
∂2F
(x0 , y0 )
∂x2
∂2F
(x0 , y0 )
∂x∂y
∂2F
(x0 , y0 )
∂y 2
0
-10
-10
2
2
2
-2
-2
-2
D
-4
16
16
So (0, 0) is a saddle point and (1/3, 1/3) and (−1/3, −1/3) are local maxima.
Q3
∇F (x, y) = (4x3 , 4y 3 ) = (0, 0) ⇒ (x, y) = (0, 0)
So F has only one critical point at (0, 0).
∂ 2F
∂ 2F
∂ 2F
(0,
0)
=
0,
(0,
0)
=
0,
(0, 0) = 0 ⇒ D = 0
∂x2
∂y∂x
∂y 2
So the second partial test tells us nothing about the critical point. However, F (x, y) =
x4 + y 4 ≥ 0 = F (0, 0), so (0, 0) is a global minimum.
Q4
∇F (x, y) = (y − 1, x) = (0, 0) ⇒ (x, y) = (0, 1)
So the only critical point is (0, 1).
∂ 2F
∂ 2F
∂ 2F
(0,
1)
=
0,
(0,
1)
=
0,
(0, 0) = 1 ⇒ D = −1
∂x2
∂y 2
∂y∂x
So (0, 1) is a saddle point, and not a local maximum/minimum. So the function must achieve
it’s maximum and minimum on the boundary of R.
F (x, 0) = −x ⇒ max at (−1, 0) min at (1, 0)
1
F (x, 2) = x ⇒ max at (1, 2) min at (−1, 2)
F (−1, y) = 1 − y ⇒ max at (−1, 0) min at (−1, 2)
F (1, y) = y − 1 ⇒ max at (1, 2) min at (1, 0)
At each of the maximum points F (x, y) = 1 and at each of the minimum points F (x, y) = −1,
so the maxumum of F over R is 1 and the minumum of F over R is −1.
Q5 F (x, y) = xy(36 − x − y) and R = {(x, y) : x ≥ 0, y ≥ 0, x + y ≤ 36}. Note that
F (x, y) = 0 on the boundary of R, i.e. F (0, y) = F (x, 0) = F (x, 36 − x) = 0. So the
maximum is not on the boundary, since we can easily find (x, y) such that F (x, y) > 0, and
so the maximum must be at a critical point.
∇F (x, y) = (36y−2xy−y 2 , 36x−x2 −2xy) = (0, 0) ⇒ y 2 +(2x−36)y = 0, x2 +(2y−36)x = 0
y 2 + (2x − 36)y = 0 ⇒ y = 0 or y = 36 − 2x
We can ignore the y = 0 case by above.
⇒ x2 + (36 − 4x)x = 0 ⇒ x = 0 or x = 12
Again x = 0 can be ignored leaving x = 12. But y = 36 − 2x = 12. So the only critical point
inside R and not on the boundary is (12, 12). Now check that this is actually a maximum:
∂ 2F
∂ 2F
∂ 2F
(12,
12)
=
−24,
(12,
12)
=
−24,
(12, 12) = −12 ⇒ D = 432 > 0
∂x2
∂y 2
∂x∂y
So (12, 12) is the maximum, which means that the 3 numbers we were looking for are:(12, 12, 12).
Q6 The volume of the box will be the product of the coordinates of the corner in the plane
x + y + z = 1 in the first octant. x + y + z = 1 ⇒ z = 1 − x − y so let F (x, y) = xy(1 − x − y).
To be in the first octant we need x, y, z ≥ 0 so the set we need to look in is R = {(x, y) :
x ≥ 0, y ≥ 0, x + y ≤ 1}. Again F (x, y) is 0 on the boundary of R so the maximum occurs
on the boundary.
∇F = (y−2xy−y 2 , x−x2 −2xy) = (0, 0) ⇒ y(y−1+2x) = 0 ⇒ y = 0(on boundary) or y = 1−2x
0 = x−x2 −2xy = x(1−x−2y) ⇒ x = 0(on boundary) or 0 = 1−x−2y = 1−x−2+4x ⇒ x = 1/3 ⇒ y =
So the only critical point of F is at (1/3, 1/3). R is closed and bounded so F must have a
maximum value. Since F is at a minimum on the boundary the maximum must occur at
a critical point. But F has only one critical point not on the boundary, which means that
(1/3, 1/3) is the maximum point. So the maximum volume is (1/3)(1/3)(1/3) = 1/(27)
Q7 Using the x-y data we get: x = 3.5, y = 12.7125, x2 = 17.5 and xy = 57.9625. So
M=
xy − xy
= 2.56547619048
x2 − x2
B = y − xM = 3.73333333332
⇒ L = mt + b where m = .256547619048, b = B + 60 − 193M = −431.40357142
So the prediction for 2010 is:
L = m(2010) + b = 84.257
2
Q8*
∇F (x, y) = (sin y, x cos y) = (0, 0) ⇒ sin y = 0 ⇒ y = kπ
where k ∈ Z.
0 = x cos y = x cos(kπ) = (−1)k x ⇒ x = 0
So (0, 0) is a critical point, as well as any point of the form (0, kπ).
∂ 2F
∂ 2F
∂ 2F
= cos y,
= 0,
= − sin y
∂x2
∂x∂y
∂y 2
⇒ D(0, kπ) = −(cos2 (kπ)) = −1
So all critical points are saddle points.
Q9* Let the verticesp
of the triangle be A = (ap
1 , a2 ), B = (b1 , b2 ) and C
p= (c1 , c2 ). We want
2
2
2
2
to miimise F (x, y) = (x − a1 ) + (y − a2 ) + (x − b1 ) + (y − b2 ) + (x − c1 )2 + (y − c2 )2 .
∂F
x − a1
x − b1
x − c1
=p
+p
+p
=0
2
2
2
2
∂x
(x − a1 ) + (y − a2 )
(x − b1 ) + (y − b2 )
(x − c1 )2 + (y − c2 )2
y − a2
y − b2
x − c1
∂F
=p
+p
+p
=0
2
2
2
2
∂y
(x − a1 ) + (y − a2 )
(x − b1 ) + (y − b2 )
(x − c1 )2 + (y − c2 )2
⇒ (0, 0) = ∇F (x, y) =
⇒
(x, y) − A
(x, y) − B
(x, y) − C
+
+
k(x, y) − Ak k(x, y) − Bk k(x, y) − Ck
A − (x, y)
B − (x, y)
C − (x, y)
+
+
= (0, 0)
kA − (x, y)k kB − (x, y)k kC − (x, y)k
So the sum of the unit vectors in the direction of A − (x, y), B − (x, y) and C − (x, y)
A−(x,y)
B−(x,y)
(i.e. the vectors going from each vertex to (x, y)) is 0. Let u = kA−(x,y)k
, v = kB−(x,y)k
and
w=
C−(x,y)
.
kC−(x,y)k
Then u + v + w = 0 and kuk = kvk = kwk = 1.
⇒ u = −v − w ⇒ 1 = kuk = k − v − wk =
p
(v1 + w1 )2 + (v2 + w2 )2
⇒ 1 = v12 + v22 + w12 + w22 + 2(v1 w1 + v2 w2 ) ⇒ v · w = −1/2
Let α be the angle between v and w. Then:
cos(α) =
v·w
1
2π
=− ⇒α=±
kvkkwk
2
3
So the angle between v and w is 120◦ . Similarly, we find that the angle between u and v is
also 120◦ . So the angles between (x, y) − A, (x, y) − B and (x, y) − C are all 120◦ as required.
3