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Find the extrema of the function f(x,y)=x^2+y^2

User Matthew FL
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Answer:

The function f(x,y) = x^2 + y^2 represents the equation of a paraboloid with its vertex at the origin. Since the function increases as we move away from the origin, there are no maximum points, but there is a minimum point at the origin.

To find this minimum point, we can take the partial derivatives of f(x,y) with respect to x and y, and set them equal to zero:

df/dx = 2x = 0

df/dy = 2y = 0

Solving these equations simultaneously gives us x = 0 and y = 0, which is the only critical point of the function. To determine whether this point is a minimum or maximum, we can use the second partial derivative test.

Taking the second partial derivatives, we have:

d^2f/dx^2 = 2

d^2f/dy^2 = 2

d^2f/dxdy = 0

At the point (0,0), both second partial derivatives are positive, which indicates that the point is a minimum. Therefore, the function f(x,y) = x^2 + y^2 has a minimum value of 0 at the point (0,0).

User Whoppa
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