Question

Examine the extreme values of the function
f(x, y) = x3 + y3 - 12x - 3y + 20​

Answer 1

Given :

f(x, y) = x³ + y³ - 12x - 3y + 20​

To find :

Extreme values of the function.

Solution :

∇f(x , y) = (3x² - 12 , 3y² - 3) = (0 , 0)

3x² - 12 = 0

3x² = 12

x² = 4

x = ±2

3y² - 3 = 0

3y² = 3

y² = 1

y = ±1

$\left[\begin{array}{ccc}f_x_x&f_x_y\\f_y_x&f_y_y\end{array}\right] = \left[\begin{array}{ccc}6x&0\\0&6y\end{array}\right]$

At (1 , 2) : D > 0 and fₓₓ > 0 , so it is a minimum point

(1 , -2) : D < 0 and fₓₓ > 0, so it is a saddle point.

(-1 , 2) : D < 0 and fₓₓ < 0, so it is a saddle point.

(-1 , -2): D > 0 and fₓₓ < 0, so it a maximum point.