Quiz 14.7 & 14.8

M-273 - Multivariable Calculus

Gustavo Barrionuevo

March 19, 2015

1. Let f(x, y) = xe(x2y2) with domain R2. Find all local extrema, saddles, and points

where the second derivative test fails. To help you along, the Hessian Determinant H(x, y),also known as the Discriminant D(x, y), is given by:

H(x, y) = D(x, y) = 4e(2x22y2)(2x4 + 2x2y2 3x2 + y2)

SOLUTION

Taking the partial derivatives of f(x, y) with respect to x, y and setting them equal to zero,we find

f

x= e(x

2y2)(1 2x2) = 0

f

y= 2xye(2x22y2) = 0

Since e(x2y2) 6= 0, the first equation gives 1 2x2 = 0, that is, x =

12, and, therefore,

2(

12

)ye(12y

2) = 0

since e(12y2) 6= 0, we have y = 0. Thus, the critical points are

(12, 0),(

12, 0). The

discriminant of theses points is

D

(

12, 0)= D

(12, 0)

= 4e[

2

(1

2

)2

] [2(

12

)4 3

(12

)2]

= 4e1 > 0

1

The second derivative test, 2f

x2= 2xe(x2y2)(3 2x2), for

(12, 0)and

(

12, 0), gives

2f

x2

(12, 0)= 2

2e

1

2 < 0

2f

x2

(

12, 0)= 2

2e

1

2 > 0

2f

x2= 0. Thus,

(12, 0)is a local maximum and

(

12, 0)is a local minimum.

2. An ellipse is given by

(x

a

)2+(y

b

)2= 1

with constants a, b > 0. Using the Lagrange Multipliers, find the point (x, y) on the ellipsewith y > 0 such that the area of the right triangle (a, 0), (x, 0), (x, y) is maximized (seediagram below). Note that the point (x, y) may depend on a and b.

(x, 0)

(x, y)

(a, 0)

b

bb

SOLUTION

According to the figure, the area of the triangle is given by A(x, y) = y(x+a)2

. Since A 0and A(x, 0) = 0, the maximum point of A occurs with y > 0. Since A is continuous on

the segment(

xa

)2+(

y

b

)2= 1, y 0, which is a closed and bounded set in R2, then A

has minimum and maximum values on the segment. The minimum value is 0 (obtainedat (a, 0) and (a, 0)), therefore the maximum value occurs for x > 0. By the method ofLagrange multipliers, we must find the maximum value of the function A(x.y) = y(x+a)

2

under the constraint g(x, y) =(

xa

)2+(

y

b

)2= 1. The gradient vectors are A = y

2, x+a

2 and

g = 2xa2, 2y

b2. The Lagrange Condition gives A = g, which is the same as A g.

So, we can useA

y, A

x

g

x, g

y

= 0 that gives

x aa

,y

2

2x

a2,2y

b2

= 0

or

(x a2

)(2x

a2

)+(y

2

)(2y

b2

)= 0

(y

b

)2=

x2 + ax

a2

2

We now substitute(

y

b

)2= x

2+axa2

in the equation of the constraint condition and solve for x:

(x

a

)2+

(x2 + ax

a2

)2= 1 x = a

2

We find y using the relation y = ba

x2 + ax:

y =b

a

(a

2

)2+a2

2=

3

2b

The critical point is thus

x0 =a

2, y0 =

3

2b (1)

The critical point (1) corresponds to the maximum value. Thus, we conclude the maximum

value of A(x, y) = y(x+a)2

on(

xa

)2+(

y

b

)2= 1, y > 0 is

A

(a

2,

3b

2

)=

1

2

3b

2

(a

2+ a

)=

33

8ab.

3