11.23 - Section 4.5, continued: Problem: Which rectangle...

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Section 4.5, continued: Problem: Which rectangle with area 1 has smallest perimeter? Goal: Minimize P = 2 x + 2 y subject to xy = 1 Write y in terms of x : ..?. . y = 1/ x P ( x , y ) = P ( x ) = 2 x + 2/ x (for x > 0) Critical points: P = 2 – 2/ x 2 which is undefined up at x = 0 (which we’ve excluded from the domain) and vanishes at … ..?. . x 2 = 1 x = 1 (ignore extraneous root –1) y = 1/1 = 1 What kind of critical point? P ′′ = 4/ x 3 = 4 > 0 local minimum (the only one) 0.5 1.0 1.5 2.0 - 2 0 2 4 6 8 10 So the best rectangle is a square. Can we use the First Derivative test here?
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..?. . Yes. 2 – 2/ x 2 is negative for x < 1 (with x > 0) and positive for x > 1, so the function P ( x ) has a local minimum at x = 1. Here’s a geometrical way to see the link between the problem “Maximize A ( x , y ) = xy subject to 2 x +2 y = p ” and “Minimize P ( x , y ) = 2 x +2 y subject to xy = a ”. The lines below are the loci of 2 x +2 y = p with p = 2, 4, and 6, and the curves are the loci of xy = a with a = 1/4, 1, and 9/4: 0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 1.5 2.0 2.5 3.0 Maximizing xy with 2 x +2 y fixed is asking “Among all the given curves, what’s the highest one that crosses a particular line?”, and the answer is, it’s the one that’s tangent to the line. Minimizing 2 x +2 y with xy fixed is asking “Among all the given lines, what’s the lowest one that crosses a particular curve?”, and the answer is, it’s the one that’s tangent to the curve.
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11.23 - Section 4.5, continued: Problem: Which rectangle...

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