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1 r2 , h 2 1 r2

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Now substitute the (positive) value of h into the formula for s: s[r_]= 2 o r h + 2 o r2 /. h 2 1 r2 2 r2 + 4 r 1 r2 Solve for the critical value of r : Solve[s'[r] 0,r] 1 1 r 10 5 + 5 , r 10 5 5

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Only the positive value of r is acceptable. We use it to compute the maximum surface area. s 1 5 + 5 //Simplify 10

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Sign s' 1 5 + 5 ' 10 1

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8.18 Find the points on the circle x 2 + y 2 2 x 4 y = 0 closest to and furthest from P(4, 4).

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SOLUTION

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First we draw a diagram. circle = ContourPlot[x2 + y2 2 x 4 y 0, {x, 5, 5}, {y, 2, 5}]; point = Graphics[{PointSize[Medium], Point[{4,4}]}]; Show[circle, point, Frame False, AspectRatio Automatic, Axes True]

5 4 3 2 1 4 2 1 2 2 4

Let (x, y) represent a point on the circle. First, we need to solve for y in terms of x. Solve[x 2 + y 2 2 x 4 y 0, y]//Simplify

{{y 2

4 + 2 x x2 , y 2 + 4 + 2 x x2

We shall minimize the square of the distance from (x, y) to (4, 4). We call this d2. It is clear from the picture that the point closest to P lies on the upper semicircle. y = 2 + 4 + 2 x x2; d2[x_]=(x 4)2 +(y 4)2; Solve[d2'[x] 0] 1 13+ 3 65 x 13

{x, y} /. x 1 13 + 3 65 //Simplify 13 5 , 2+ 2 5 1 + 3 13 13

%//N

{2.86052,

3.24035}

The point furthest from P lies on the lower semicircle. y = 2 4 + 2 x x2; d2[x_]=(x 4)2 +(y 4)2; Solve[d2'[x] 0]

Differential Calculus

x 1 (13 3 65 ) 13

{x,y} /. x 1 13 3 65 //Simplify 13

5 , 2 2 5 13 13

%//N { 0.860521, 0.759653}

8.19 A local telephone company wants to run a cable from point A on one side of a river 100 feet wide to point B on the opposite side, 500 feet along the shore from point C, which is opposite A. It costs three times as much money to run the cable underwater as on land. How should the company run the cable in order to minimize the cost of the project

SOLUTION

If we let a represent the cost per foot to run the cable on land, 3a is the cost to run a foot of cable underwater. The total cost is then c( x ) = 3a x 2 + 100 2 + a (500 x ) . Of course, 0 x 500.

500 C x 500 x B

100

x2 + 1002

c[x_]= 3 a x 2 + 1002 + a (500 x); Solve[c' [x] 0, x]

{{x 25 2}}

Now we must compare the cost corresponding to this solution with the cost at the endpoints of the interval. c[0] //N 800. a c[25 2] //N

For minimum cost, run the cable underwater to the point 25 2 feet from C, then on land to B.

782.843 a c[500] //N 1529.71 a

8.4 Power Series

The nicest functions to work with are polynomials. They are continuous and can easily be differentiated and integrated. If a difficult function is encountered in a problem, one approach is to approximate it by a polynomial.

Differential Calculus

If the value of the function and its derivatives are known at a single point, a, the function can often be represented by a power series. This, however, is usually an infinite series that must be truncated for practical application. The trick is to truncate it in such a way that it accurately approximates the given function, at least in some neighborhood of a. The following series, known as a Taylor series, gives a representation of an analytic1 function, f(x). If a = 0, the series is known as a Maclaurin series. f (x) =

k =0

f ( k ) (a) ( x a) k k!

f ( k ) (a) represents the kth derivative of f evaluated at a. If k = 0, it represents f(a).

If we truncate this infinite series by omitting all terms of degree greater than n, we obtain the nth degree Taylor polynomial of f about a. We shall represent this polynomial as pn ( x ). If a = 0, the polynomial is called a Maclaurin polynomial. Sum command or the symbol from the Basic Math Input palette. Here are three different ways the polynomial can be generated:

f[x_] = Exp[x]; (a) Sum[(D[f[x], {x, k}]/.x 0)/k! * xk, {k, 0, 5}]