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Topos Uranos
English

Optimization problems

Summary

This lesson solves problems in which the best of many possible configurations must be found: the rectangle of largest area, the cylinder of least surface, the shortest path or the fastest one. The earlier lessons provided the tools for finding the absolute extrema of a given function; here the function is not given, and most of the work, and of the errors, lies in constructing it. The lesson sets out a method in five stages (the variable, the function, the domain, the absolute extrema with their justification, and the conclusion) and proves that a function and its composition with a strictly increasing function attain their extrema at the same points, which allows one, for instance, to minimize the square of a distance instead of the distance. It then proves that every point has a nearest point on the graph of a function continuous on a closed interval, and that, if this nearest point is not at the boundary, the given point lies on the normal line; and it deduces from the principle of least time the laws of refraction, Snell’s law, and of reflection, Heron’s law. The problems find the box of largest volume, the cylinder of least surface, the longest rod that can turn a corner, the place from which a painting is seen under the largest angle and the fastest way to cross a canal.

Learning objectives

  1. Translate a geometric or physical problem into the search for the absolute extremum of a function of one variable, on a precisely specified domain.
  2. Justify that the extremum found is absolute: by the Weierstrass theorem if the domain is a closed and bounded interval, and if it is not, by the sign of the derivative, convexity or the limits at the endpoints of the domain.
  3. Find the point of a graph nearest to a given point and recognize that the line joining them is normal to the graph.
  4. Deduce from the principle of least time the laws of reflection and of refraction, and recognize when the optimum lies at the boundary of the domain.

The method

The lessons on the Weierstrass theorem and on maxima and minima taught how to find the absolute extrema of a given function on a given interval. In an optimization problem, by contrast, what is asked for is the best configuration of a figure or of a motion, and the function to be studied, together with its domain, must be constructed from the statement. The method has five stages.

  1. The variable. Choose a quantity that determines the configuration (a length, an angle, a position) and express all the others in terms of it, by means of the conditions of the problem.
  2. The function. Write the quantity to be made maximal or minimal as a function of that single variable.
  3. The domain. Specify the set of values that the variable can take in the problem, which is usually an interval. If the function extends continuously to degenerate configurations, such as a rectangle of zero height or a box with no volume, it is convenient to admit them in order to obtain a closed interval, provided the extremum sought is not attained at them.
  4. The absolute extrema. Find them on the domain and justify that they are absolute. If the domain is a closed and bounded interval, the Weierstrass theorem guarantees that they exist, and the proposition on absolute extrema on a closed and bounded interval, from the lesson on maxima and minima, finds them by comparing values. If it is not, one uses the sign of the derivative (the theorem of the first derivative test for absolute extrema, or the corollary of a single critical point, from the same lesson), convexity (the corollary on the critical points of a convex function, from the lesson on concavity and convexity) or the limits at the endpoints of the domain (the proposition on absolute extrema on an open interval).
  5. The conclusion. Return to the problem: give the optimal configuration, not only the optimal value, and check that it makes sense, dimensionally and in particular cases.

The fourth stage is the one most often omitted: finding a point where the derivative vanishes is not enough. That point may be a minimum when a maximum is sought, a merely local extremum or a point outside the domain; and the absolute extremum may lie at an endpoint of the interval, where the derivative need not vanish. The lesson on maxima and minima showed, with 1x+x2\frac{1}{x} + x^2, that a single critical point guarantees nothing if the domain is not an interval. It is often convenient, moreover, to replace the function by a simpler one that attains its extrema at the same points.

LemmaExtrema and strictly monotonic functions

Let ff be a function defined on a set AA, cc a point of AA and φ\varphi a function strictly increasing on a set containing the values of ff. Then ff attains at cc its absolute maximum on AA if and only if φ∘f\varphi \circ f attains it there, and the same holds for the absolute minimum, for local extrema and for strict extrema. If φ\varphi is strictly decreasing, ff attains at cc a maximum if and only if φ∘f\varphi \circ f attains at cc a minimum of the same type, and conversely. A strictly increasing function does not alter the order among values, and extrema depend only on that order.

Proof

  1. u≤w⇔φ(u)≤φ(w),u<w⇔φ(u)<φ(w)u \leq w \Leftrightarrow \varphi(u) \leq \varphi(w), \qquad u < w \Leftrightarrow \dato{orden}{\varphi(u) < \varphi(w)}

    Let uu and ww be in the domain of φ\varphi. If u<wu < w, strict monotonicity gives φ(u)<φ(w)\varphi(u) < \varphi(w); if u=wu = w, the values coincide; and if u>wu > w, it gives φ(u)>φ(w)\varphi(u) > \varphi(w). Hence the inequalities between the numbers and between their images are equivalent.

  2. f(x)≤f(c)⇔φ(f(x))≤φ(f(c))f(x) \leq f(c) \Leftrightarrow \resaltar{\varphi\left(f(x)\right) \leq \varphi\left(f(c)\right)}

    With u=f(x)u = f(x) and w=f(c)w = f(c), for each xx in AA, or for each xx in AA whose distance from cc is less than some δ\delta, the inequalities that define the absolute maximum, the local one and the strict ones, according to the lesson on maxima and minima and the definition of local extremum in the lesson on the theorems of Fermat, Rolle, Lagrange and Cauchy, hold for ff if and only if they hold for φ∘f\varphi \circ f. The minimum is handled in the same way, and if φ\varphi is strictly decreasing, the inequalities between the images are reversed.

The most frequently used cases are four. The square root is strictly increasing on [0,∞)\ico{0}{\infty}: a distance and its square are minimized at the same points, and the square contains no square roots. Squaring is strictly increasing on [0,∞)\ico{0}{\infty}: a nonnegative quantity and its square are maximized at the same points. The logarithm is strictly increasing on (0,∞)\ioo{0}{\infty}: a product of positive powers and its logarithm, which is a sum, are as well. And a positive function attains its maximum where its reciprocal attains its minimum, because 1u\frac{1}{u} is strictly decreasing on (0,∞)\ioo{0}{\infty}.

ExampleThe rectangle inscribed in a semicircle

Among the rectangles inscribed in a semicircle of radius RR, with one side on the diameter, find the one of largest area. The reasoning follows the stages of the method.

Proof

  1. A(x)=2xR2−x2A(x) = 2x\sqrt{R^2 - x^2}

    The variable and the function. Place the center at the origin and the diameter on the horizontal axis. The rectangle is determined by xx, half of its base: its upper vertices lie on the semicircle, at (±x,R2−x2)\left(\pm x, \sqrt{R^2 - x^2}\right), and its area is the following function.

  2. F(x)=(A(x))2=4x2(R2−x2),x∈[0,R]F(x) = \left(A(x)\right)^2 = \resaltar{4x^2\left(R^2 - x^2\right)}, \qquad x \in \icc{0}{R}

    The domain. The genuine rectangles correspond to xx in (0,R)\ioo{0}{R}; we also admit 00 and RR, which give degenerate rectangles, of zero area, and we obtain a closed and bounded interval on which AA is continuous: by the Weierstrass theorem, the maximum exists. Since AA is not negative, by the lemma it suffices to maximize its square, which is a polynomial and, unlike AA, is differentiable at RR as well.

  3. F′(x)=8x(R2−2x2)=0, 0<x<R→x=R2F'(x) = 8x\left(R^2 - 2x^2\right) = 0, \ 0 < x < R \rightarrow \resaltar{x = \frac{R}{\sqrt{2}}}

    The candidates. The derivative of FF vanishes at a single point of (0,R)\ioo{0}{R}.

  4. F(0)=F(R)=0<F(R2)=R4,A(R2)=R2F(0) = F(R) = 0 < F\left(\frac{R}{\sqrt{2}}\right) = R^4, \qquad \resaltar{A\left(\frac{R}{\sqrt{2}}\right) = R^2}

    The comparison and the conclusion. By the proposition on absolute extrema on a closed and bounded interval, the maximum of FF is the largest of its values at 00, at RR and at the critical point. The optimal rectangle has base 2R\sqrt{2}R and height R2\frac{R}{\sqrt{2}}, half the base, and area R2R^2.

Reflecting the optimal rectangle in the diameter yields a square of side 2R\sqrt{2}R inscribed in the full circle, of area 2R22R^2, which is, by the same calculation, the rectangle of largest area inscribed in it. The degenerate rectangles did not alter the result: the maximum is attained at a genuine rectangle.