121616949-math.125

121616949-math.125 - 6.1 Optimization 111 either no width...

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6.1 Optimization 111 either no width or no height, so is not “really” a rectangle. But the problem is somewhat easier if we simply allow such rectangles, which have zero area.) Setting 0 = A ± ( x ) = 6 x 2 + 2 a we get x = p a/ 3 as the only critical value. Testing this and the two endpoints, we have A (0) = A ( a ) = 0 and A ( p a/ 3) = (4 / 9) 3 a 3 / 2 . The maximum area thus occurs when the rectangle has dimensions 2 p a/ 3 × (2 / 3) a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( h - R, r ) Figure 6.4 Cone in a sphere. EXAMPLE 6.10 If you ﬁt the largest possible cone inside a sphere, what fraction of the volume of the sphere is occupied by the cone? (Here by “cone” we mean a right circular cone, i.e., a cone for which the base is perpendicular to the axis of symmetry, and for which the cross-section cut perpendicular to the axis of symmetry at any point is a circle.) Let R be the radius of the sphere, and let r and h be the base radius and height of
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Unformatted text preview: the cone inside the sphere. What we want to maximize is the volume of the cone: πr 2 h/ 3. Here R is a ﬁxed value, but r and h can vary. Namely, we could choose r to be as large as possible—equal to R —by taking the height equal to R ; or we could make the cone’s height h larger at the expense of making r a little less than R . See the side view depicted in ﬁgure 6.4 . We have situated the picture in a convenient way relative to the x and y axes, namely, with the center of the sphere at the origin and the vertex of the cone at the far left on the x-axis. Notice that the function we want to maximize, πr 2 h/ 3, depends on two variables. This is frequently the case, but often the two variables are related in some way so that “really” there is only one variable. So our next step is to ﬁnd the relationship and use it to solve for one of the variables in terms of the other, so as to have a function of only one variable...
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