Question

show that z(x,y) =x³ + y³-3xy+1 is minimum at (1.1)

Answer 1

Z(x, y) = x³ + y³ - 3xy + 1

Differentiating Z(x, y) with respect to 'x' we get

Zx = 3x² - 3y

Differentiating Z(x, y) with respect to 'y' we get

Zy = 3y² - 3y

Differentiating Zx again with respect to 'x' we get

Zxx = A = 6x

Differentiating Zx again with respect to 'y' we get

Zxy = -3

Differentiating Zy again with respect to 'y' we get

Zyy = 6y

At (1, 1) we obtain Zx = 0 and Zy = 0

A = 0, B = -3, C = 6

We get AC - B² = 36 - 9 = 27

From this we observe that

A > 0 & AC - B² > 0

which is the necessary and sufficient condition for Minima.