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Differential Calculus
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points = {{x1, a[x1], N[a[x1]]}, {x2, a[x2], N[a[x2]]}, {x3, a[x3], N[a[x3]]}} //Together; TableForm[points, TableAlignments Center, TableSpacing {2, 5}, TableHeadings {None, {"x", "a[x] "N[a[x]]"}}] ",
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x 0 100 4+ 25 a [x] 2500 2500 4+ 625 N [a [x]] 795.775 350.062 625.
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The largest combined area occurs when x = 0 (all the wire is used to form the circle). The smallest area occurs when one side of the square is 100 (cut the wire 400 from one end). To further confirm that x = 100 gives a 4+ 4+ 4+ minimum area, we can apply the second derivative test. Sign a'' 100 4+ 1 Since the sign of the second derivative at the critical number is positive, A(x) has a relative minimum at 100 . 4+ Since this is the only relative extreme value, it must be the location of the absolute minimum.
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A function has a relative or local maximum at c if there exists an open interval, I, containing c such that f(x) f(c) for all x in I. In other words, there exists an open interval containing c such that f(c) is the largest value of f for all x in this interval. A similar definition holds for a relative minimum. Unlike an absolute maximum (minimum), a function may have several relative maxima (minima). If a numerical approximation of their location is all that is required, the Mathematica commands FindMinimum and FindMaximum offer an efficient and convenient procedure.
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FindMinimum[f[x], {x, x0}] finds the relative minimum of f(x) near x0. FindMaximum[f[x], {x, x0}] finds the relative maximum of f(x) near x0.
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As with FindRoot, the options AccuracyGoal and WorkingPrecision can be set if greater accuracy is desired. In addition, PrecisionGoal can be set to determine the precision in the value of the function at the maximum or minimum point. (Precision is the number of significant digits in the answer; accuracy is the number of significant digits to the right of the decimal point.)
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The function f(x) = x + sin(5 x) has three relative maxima and two relative minima in the interval [0, ]. A quick look at its graph gives good approximations to their locations. f[x_] = x + Sin[5 x] ; Plot[f[x], {x, 0, o}]
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Differential Calculus
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FindMinimum[f[x], {x, 1}]
{ 0.0775897, {x
0.902206}}
FindMinimum[f[x], {x, 2}]
{1.17905, {x
The value of the function comes rst, followed by the value of x.
2.15884}} 0.354431}} 1.61107}}
FindMaximum[f[x], {x, 0.4}]
{1.33423, {x
FindMaximum[f[x], {x, 1.5}]
{2.59086, {x
FindMaximum[f[x], {x, 3}]
{3.8475, {x
2.8677}}
The relative maximum points are (0.354431, 1.33423), (1.61107, 2.59086), and (2.8677, 3.8475). The relative minimum points are (0.902206, 0.0775897) and (2.15884, 1.17905). Note: Caution must be taken to examine the results of the calculation. The value obtained is not necessarily the one closest to the initial guess. For example, FindMaximum[f[x], {x, 2.8}]
{5.10414,{x
4.12434}}
but the value of x is not between 0 and .
SOLVED PROBLEMS
8.16 Find two positive numbers whose sum is 50, such that the square root of the first added to the cube root of the second is as large as possible.
SOLUTION
y = 50 x; f[x_]= x + 3 y; Plot[f[x],{x, 50}];
8 7 6 5
NSolve[f'[x] 0]
41.1553}}
y /. x 41.1553 8.8447 f[41.1553] 8.48329 The two numbers are x = 41.1553 and y = 8.8447. The maximum sum is 8.48329.
Differential Calculus
8.17 A right circular cylinder is inscribed in a unit sphere. (a) Find the largest possible volume. (b) Find the largest possible surface area.
SOLUTION
(a) We consider a two-dimensional perspective of the problem. Label the radius and height of the inscribed cylinder r and h, respectively. The volume of the inscribed cylinder is V = r 2 h and, by the Theorem 2 of Pythagoras, r 2 + h = 1. It is easily seen (even without Mathematica) that 2 2 2 h . Thus, the volume, as a function of h, becomes V (h) = 1 h h. 2 1 r = 1 h/2 2 2 r
v[h_]= o 1 ( h / 2)
) h;
Obviously only the positive value of h is appropriate. Since the sign of the second derivative at the critical point is negative, we have a relative maximum at 2 / 3 . Since this is the only relative extremum, it must be the absolute maximum.
Solve[v'[h] 0, h] 2 , h 2 h 3 3 vmax = v 2 / 3 4 3 3 sign[v' [2 / 3]] ' 1 (b) The surface area of the cylinder (including top and bottom) is S = 2 r h + 2 r 2. As in part (a), 2 r 2 + h = 1, but because r and r2 both appear in the equation for S, it is easier to solve for h in 2 terms of r. Solve[r 2 +(h /2)2 1, h]