Find the extremum of function f (x, y) = e ^ X-Y (x ^ 2-2y ^ 2)

Find the extremum of function f (x, y) = e ^ X-Y (x ^ 2-2y ^ 2)

Let one mediate, etc. 0;
y=e^x-yx^2-2y^3;
Y '= e ^ x-y'x ^ 2-2xy-6y ^ 2Y' is derived from X on both sides;
Let y '= 0
We get 2XY = e ^ X

Finding the extremum of function z = x + 2Y + 4x-8y + 2
Ditto. Urgent

There are 10 products, of which 8 are genuine and 2 are defective. Party A and Party B take one from each other. How about the probability that Party A has got the genuine product first?
Is there an answer to this?

The extremum of the function z = (x, y) determined by the equation 2x ^ 2 + 2Y ^ 2 + Z ^ 2 + 8yz-z + 8 = 0

First, the partial derivative is obtained by using the implicit equation derivation rule. Z for X: 4x + 2Z · partial Z / partial x + 8 (Z + X · partial Z / partial x) - partial Z / partial x = 0 → partial Z / partial x = - (4x + 8Z) / (2Z + 8x-1). Z for Y: 4Y + 2Z · partial Z / partial Y2 + 8x · partial Z / partial Y - partial Z / partial y = 0. → partial Z / partial y = - 4Y / (2Z + 8x-1). Then the directional derivative Z '= √ [(partial Z / partial x) ^ 2 +...]

Determine the monotone interval and extreme point of function f (x) = 2x & # 179; - 9x & # 178; + 12x-3 (thank you,

f'(x)=6x^2-18x+12=6(x^2-3x+2)=6(x-2)(x-1)
If f '(x) > 0, then x > 2, X