Second Order Conditions: Practice Problems and Solutions

Mathematical Economics Problems and Solutions - Section 10

See Mr. Stolyarov's complete list ofMathematical Economics Problems and Solutions.

Note: Here, I will present solve problems typical of those offered in a mathematical economics or advanced microeconomics course. The problems were authored by Dr. Charles N. Steele and are reprinted with his generous permission. The solutions to the problems are my own work and not necessarily the only way to solve the problems.

For all problems here, find the extrema for each of the following functions, and identify whether the extrema are maxima or minima using second order conditions.

Problem 1.

z = x² + xy + 2y² + 23

Solution 1.

First order conditions (FOC):

z_x =2x + y = 0

z_y = x + 4y = 0

We can solve this system by row-reducing this matrix:

[ 2 1 │ 0

1 4 │0] with the following result:

[ 1 0 │ 0

0 1 │0], which means that an extreme point is (x*, y*) = (0, 0)

Second order conditions (SOC): z_xx = 2, z_xy = 1, z_yy = 4.

z_xx > 0, z_yy > 0, z_xxz_yy = 8, (z_xy)² = 1, so z_xxz_yy > (z_xy)² and (x*, y*) = (0, 0) is a minimum.

Problem 2.
z = - x² + xy - y² + 2x + y

Solution 2.

FOC:

z_x = -2x + y + 2 = 0 and so -2x + y = -2

z_y = x - 2y + 1 = 0 and so x - 2y = -1

We can solve this system by row-reducing this matrix:

[ -2 1 │ -2

1 -2│ -1] with the following result:

[ 1 0 │ 5/3

0 1 │4/3], which means that an extreme point is (x*, y*) = (5/3, 4/3)

SOC: z_xx = -2, z_yy = -2, z_xy = 1.

z_xx< 0, z_yy < 0, z_xxz_yy = 4, (z_xy) = 1, so z_xxz_yy > (z_xy)and (x*, y*) = (5/3, 4/3) is a maximum.

Problem 3.

z = 8x³ + 2xy - 3x² + y² + 1

Solution 3.

FOC:

z_x = 24x² + 2y - 6x = 0

z_y = 2x +2y = 0

2x +2y = 0 implies that y = -x and so 24x² + 2y - 6x = 0 can be written as 24x² - 8x = 0 and 8x(3x - 1) = 0. Thus, x = 0 or x = 1/3. If x = 0, then y = -0 = 0. If x = 1/3, then y = -1/3. So the two possible extrema are

(x*, y*) = (0, 0) and (x*, y*) = (1/3, -1/3)

SOC: z_xx = 48x - 6, z_yy = 2, z_xy = 2.

If (x*, y*) = (0, 0), then z_xx = -6 < 0, but z_yy = 2 > 0, so (x*, y*) = (0, 0) is neither a minimum nor a maximum; rather, (x*, y*) = (0, 0) is a saddle point.

If (x*, y*) = (1/3, -1/3), then z_xx = 10 > 0, z_yy = 2 > 0, z_xy² = 4, and z_xxz_yy = 20 > 4, so z_xxz_yy > z_xy² and

(x*, y*) = (1/3, -1/3) is a minimum.

Problem 4. z = e^2x - 2x + 2y² + 3

Solution 4.

FOC:

z_x = 2e^2x - 2 = 0

z_y = 4y = 0

4y = 0 implies that y = 0.

2e^2x - 2 = 0 implies that 2e^2x = 2 and e^2x = 1, so x = ln(1)/2 = 0. So an extreme point is (x*, y*) = (0, 0)

SOC: z_xx = 4e^2x, z_yy = 4 > 0, z_xy = 0. At (x*, y*) = (0, 0), z_xx = 4 > 0. Thus, z_xxz_yy = 16 > z_xy² = 0, so

(x*, y*) = (0, 0) is a minimum.

See Mr. Stolyarov's complete list ofMathematical Economics Problems and Solutions.