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10.14 - Section 11.7 Maximum and minimum values(continued...

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Section 11.7: Maximum and minimum values (continued) Go through proof on A25. [Omitted in 2011.] Find the maximum and minimum of f ( x , y ) = 3 x + 4 y on the set of points ( x , y ) with x 2 + y 2 4. Note that the gradient of f is never 0, so the maximum value must occur on the boundary. One way to find these maximum and minimum values is by patametrizing the boundary curve x 2 + y 2 = 4 by x ( θ ) = 2 cos θ and y ( θ ) = 2 sin θ , where θ is the angle made by the position vector x ( θ ), y ( θ ) with the x -axis, and then optimizing the function g ( θ ) = f ( x ( θ ), y ( θ )). g ( θ ) = ..?.. 6 cos θ + 8 sin θ (look familiar? ..?.. it’s that problem from yesterday!). The maximum value is 10, achieved when cos θ = 3/5 and sin θ = 4/5; i.e., when ( x , y ) = (6/5, 8/5). Trick: Every critical point for a function f is also a critical point for the function f n , for any integer n > 0.
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Individual work: What point on the surface z 2 = xy + x 2 + 1 is closest to the origin? ..?.. Derivation: The squared distance between (0,0,0) and ( x , y , z ) is ( x –0) 2 +( y –0) 2 +( z –0) 2 = x 2 + y 2 + z 2 , which in this case can be written as … ..?..
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