Instruction

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*Necessary condition of extremum is equal to zero partial derivatives*

**of a function**in x and y. Point M0(x0, y0), which in turn zero, both partial derivatives is called a stationary point of a

**function**z=f(x, y).

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Remark. Partial derivatives

**of the function**z=f(x, y) may not exist at the point of extremum, so the points of extremum are not only stationary points, but the points at which partial derivatives do not exist (they correspond to the edge surface of the graphics**function**).3

*Now you can go to sufficient conditions for the existence of the extremum. If a differentiable function has an extremum, then it can only be at a stationary point. Sufficient conditions are formulated as follows: suppose that in some neighborhood of the stationary point (x0, y0) the function f(x, y) has continuous partial derivatives of second order. For example: (cm. Fig.2)*

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Then: a) if Q>0, the point (x0, y0) the function has an extremum, with f’(x0, y0)0) - local minimum; b) if Q

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For finding the extremum of

**functions**of two**variables**can suggest the following scheme: first, are the stationary points**of the function**. Then these points are checked sufficient conditions. If the function in some points does not have partial derivatives in these points also can be an extremum, but sufficient conditions are no longer applicable.6

*Example. Find extrema*

**of the function**z=x^3+y^3-xy.Solution. Find the stationary points

**of the function**(see Fig. 3):

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*The latter system gives the stationary point (0, 0) and (1/3, 1/3). It is now necessary to check that sufficient conditions. Find the second derivative and the stationary points Q(0,0 ) and Q(1/3, 1/3) (see figure 4):*

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Since Q(0, 0)0, consequently, at the point (1/3, 1/3) is the extremum. Given the fact that the second derivative (xx) to (1/3, 1/3) is greater than zero, you must make a decision that this point is a minimum.