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265L12 - LESSON 12 Max-Min Problems READ Section 15.7 NOTES...

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LESSON 12 Max-Min Problems READ: Section 15.7 NOTES: Probably some of your fondest memories of Calculus I are the hours spent working on max-min problems. That must rank right up there with the time devoted to the infamous related rate problems. Now it’s time to try max-min problems for functions of several variables. Most of the terminology and methods are natural extensions of ideas familiar from the one variable case, so that ought to help a bit. Geometrically, the graph of z = f ( x, y ) represents a surface in 3-space. When we seek the local maximum and local minimum points on the surface, we are trying to determine where the peaks and valleys occur. The absolute maximum of the function is the height of the highest peak, and the absolute minimum of the function is the depth of the lowest valley. For a point ( x 0 , y 0 ) interior to the domain of the function z = f ( x, y ), there is a simple test that indicates if f might have a local max or min at that point. Here is a rough way to think of the test: suppose that f ( x 0 , y 0 ) is defined (in other words, suppose that both first partials of f exist at the point ( x 0 , y 0 )). If we are at a peak on the function, then no matter which direction we leave from ( x 0 , y 0 ), we cannot be heading uphill. That means f ( x 0 , y 0 ) must be 0 -→ ı + 0 -→ , since for any other gradient, we would be able to select a unit vector -→ u that would make f ( x 0 , y 0 ) · -→ u > 0, and so the directional derivative at ( x 0 , y 0 ) would be positive in some direction. We may conclude that a local maximum, if the first partials exist, they must have value 0. Likewise, if there is a valley bottom at the point ( x 0 , y 0 ), and the first partials exist there, they must have value 0. To say the same thing in other words, at a local max or min of z = f ( x, y ) , the first partials are both 0 , or at least one of them does not exist.
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