Homework week 7
Solve problems 1,2 and 3. Problems 4 and 5, will help people who want to improve in their homework marks.
1. Linearise the function f x y at the given point P and find the error associated with the linearisation for
a region given by ∆x and ∆y around the given point.
a) f x y x4 2xy2 1 at 0 0 in the region subtended by ∆x 0 2 and ∆y 0 1. Solution:
we calculate the derivatives of the function to second order:
fx
4x3 2y2
fy
4xy
fxx
12x2
fyy
4x
fxy
4y
0
0
fx 0 0 fy 0 0 fxx 0 0 fyy 0 0 fxy 0 0 0
fxx 0 2 0 1 0 48
0 8
fxy 0 2 0 1 0 4
fyy 0 2 0 1 0
0
Very important to notice what we did, we want to approximate the function f by a plane in the
region ∆A ∆x∆y. Because the second derivatives of the function at the point 0 0 give us
all zero (second column) , we could be tempted to assume that the Error given by the formula
E 12 ∆x ∆y 2 M with M Maximum value of f xx fyy fxy is zero as well! This of
course is wrong since we now that f is quartic surface and not a plane, that is why in the third
column we calculate the values of the second derivatives on a different point, in fact we chose the
points where the values were greater and within the region ∆A.
The plane we are trying to linearise is
L x y f 0 0
fx 0 0 ∆x fy 0 0 ∆y
0
that is the plane parallel to the xy-plane at z 0. Next we calculate the error: Looking at the third
column above, we notice that M 0 8 therefore
1
0 2 0 1 2 0 8
2
0 036
E this is the error of the approximation, we conclude that for the area ∆A ∆x∆y around the point
0 0 the function
f x y L x y E b) f x y x3 5x2 xy2 1 at 0 0 in the region subtended by ∆x 0 1 and ∆y 0 3
2. Find the critical points of the function f x y xy x 2 y2 and then, using the Hessian find out if its a
Maximum a Minimum or a Saddle point.
1
a) f x y x3 x y y3 . Solution:
We want to calculate the local extremes. As we remember from calculus in one variable, the
maximum and the minimum are associated with second derivatives. We calculate them:
3x2 1
6x
fx
1 3y2
fy
fxx
fyy
fxy
6y
0
the extremes are when we require that the slopes of the tangent lines are zero and this is equivalent
to ask that f x 0 and fy 0:
3x2 1
1
3y
2
0
0
solving this system of equations we obtain that the maximum, minimum or saddle points are
located at P
points:
1 3
1
3
.
Later we will be interested in the sign of f xx evaluated at this
fxx 1
3
1
3
fxx 1
3
1
3
fxx 1
3
1
3
1
3
fxx 1
3
2
3
2
3
2
3
2
3
0
0
0
0
To now exactly what they are we calculate the Hessian
H
fxx
fyx
fxy
fyy
6x
0
0
6y
36xy
and evaluat it on the point where the extremum is
H H H H 1
3
1
3
1
3
1
3
1
3
1
3
36 1
3
36 1 3 1
3
36 1 3 2
1 3
1 3
4
1 3
4
36 1 3 1 3 1 3 4
0
4
0
0
0
according to the Hessian test (see footnote), the result is that at the points:
1
3
1
3
1
3
1
3
1
3
the function is a saddle point
1
3
the function is a maximum
1
3
the function is a maximum
1
3
the function is a saddle point
1
b) f x y xy x2 y2
3. Find the maximum and minimum values of the function f x y on the given constraint by using the
method of Lagrange Multipliers. Plot the level curves and the constraint equation together, and indicate
where the maximum occurs.
a) f x y 4x y constraint: x2 3y2 2. Solution
First we find the auxiliary function related to to the constraint
g x y
x 2 y2 2
secondly we calculate the gradient of both the function f and the function g:
4i j
∇f
∇g
2xi 2yj
the normal to this surfaces will coincide when ∇ f
λ∇g in this case give us the vectorial equation
4i j 2λxi 2λyj
the question is for which values of λ and x and y this equation is satified. Comparing x-components
with x-component and x-components with x-component we get: 4 2λx and 1 2λy or
λ
λ
4
2x
1
2y
so the point of coincidence is obtained by substituting one into another
x
1 Hessian
Test:
Local Maximum: H P 0 and fxx P
Local Minimum: H P 0 and fxx P
Saddle Point: H P 0
Inconclusive H P 0
0
0
3
8y
(0.1)
This result however is obtained by comparing the function f and the auxiliary function g x y x2 y2 2, we want now to specialise to the constraint of our problem, that is g x y 0, together
with the result (0.1)or
g x y
8y 2
y2 2 0
65y2 2 0
so
y
2
65
and therefore back into equation (0.1) we obtain 2 : x P2
P1
4
4
1
65
1
65
4
1
65
2
65
2
65
so the max-min points are at
we evaluate the function f x y 4x y at these points to get
f
f
4
4
1
65
1
65
2
65
2
65
4
4
4
4
1
65
1
65
2
65
2 16
2
65
2 16
so the maximum is at P2 and the minimum is at P1 .
b) f x y 5x y, constraint x2 y2
1
4. Using level curve technics plot the function f x y c
2, use the values x 2 1 5 1 0 5.
y 2 x2 . Hint: check the level curves c 0, c 1
5. Find the absolute maxima and minima of the function f x y x 2 y2 2x on the triangular plate in
the first quadrant bounded by the lines x 0 y 0 and y x 10.
2 calculator
says:
8 2
2
4