What are the extrema and saddle points of #f(x, y) = xy(1-x-y)#?
The points
We can expand
Clearly, To test the nature of these critical points, we find second derivatives: The discriminant is therefore: Plugging the first three critical points in gives: Plugging in the last critical point gives Below is a picture of the contour map (of level curves) of
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To find the critical points, solve ∇f(x, y) = 0. Differentiate f(x, y) with respect to x and y, then set both equal to zero. You'll get two critical points: (0, 0) and (1, 0). To determine their nature, evaluate the Hessian matrix at each critical point. The Hessian matrix for f(x, y) is [[2y - 6xy, 2x - 3x^2 - y], [2x - 3y - y^2, 2y - 3xy]]. Evaluate the determinant of the Hessian at each critical point. If it's positive, it's a minimum; if negative, a maximum; if zero, a saddle point. At (0, 0), the determinant is negative, so it's a saddle point. At (1, 0), the determinant is positive, so it's a minimum.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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