Minimizing the Calculus in Optimization Problems Teylor Greff Mathematics Department Whitman College May 2016 Abstract: Do we actually need calculus to solve maximum/minimum problems? Optimization problems are explored and solved using the AM/GM inequality and Cauchy Schwarz inequality, while simultaneously finding trends and evolutions in these optimization problems as we look at a textbooks ranging from 1902 − 2015. Contents 1 Introduction 3 2 The AM/GM Inequality 3 2.1 Diagram Proof . .3 2.2 Algebra Proof . .5 2.3 Formal Proof . .5 3 Optimization Problems and the AM/GM Inequality 7 3.1 Fence Problems . .7 3.1.1 Derivative Approach . .8 3.1.2 AM/GM Approach . .8 3.1.3 Similar Problem Examples . .9 3.2 Number Problems . .9 3.2.1 Incorrect Application . 10 3.2.2 Correct Application . 10 3.2.3 Similar Problem Examples . 11 3.3 Geometry Problems . 11 3.3.1 Application . 11 3.3.2 Similar Problem Examples . 13 3.4 Three Dimensional Problems . 13 3.4.1 Application . 13 3.4.2 Similar Problem Examples . 14 4 Non-AM/GM Generalized Problems 14 4.1 \Triathlon" Problems . 14 4.1.1 Specific Solution . 15 4.1.2 Generalized Solution . 16 4.1.3 Similar Problem Examples . 17 4.2 Pipe Corridor Problem . 18 4.2.1 Generalized Solution . 18 1 4.2.2 Specific Solution . 19 4.2.3 Similar Problem Examples . 19 4.3 Post Problems . 19 4.3.1 Generalized Solution . 20 4.3.2 Specific Solution . 21 4.3.3 Similar Problem Examples . 21 5 Problems that Have Evolved 21 5.1 Number Problem ! Box Problem . 22 5.1.1 Similar Problem Examples . 23 5.2 Graph Problem ! Pipe Problem ! Ladder Problem . 23 5.2.1 Similar Problem Examples . 24 6 The CS Inequality 25 6.1 Formal Proof - CS . 25 7 Optimization Problems and the CS Inequality 26 7.1 Typical Problem . 26 7.2 Optimization Problem . 27 8 Historical Bit 29 8.1 Economic/Medical/Physical Problems . 29 9 Conclusion 30 2 1 Introduction The focus of this paper is optimization problems in single and multi-variable calculus spanning from the years 1900 − 2016: The main goal was to see if there was a way to solve most or all optimization problems without using any calculus, and to see if there was a relationship between this discovery and the published year of the optimization problems. The primary method used to accomplish this was the Arithmetic Mean/Geometric Mean inequality, but upon further exploration, the Cauchy Schwarz inequality and other generalized solutions proved to be incredibly helpful as well. This paper lays out how these problems are solved, gives examples of similar problems, and also explores the historical evolution of these problems. 2 The AM/GM Inequality The Arithmetic Mean/Geometric Mean inequality was the main method when it came to solving optimization problems without differentiation. Before looking at applications of the inequality, a firm grasp on the inequality concept is key. The result states that the Arithmetic Mean is always greater than or equal to the Geometric Mean. The AM/GM inequality is stated below with three proofs. Two for intuition and one formal proof. We will begin by looking at a diagram proof and an algebraic proof for two positive numbers a and b before seeing the AM/GM applied to the nth case. 2.1 Diagram Proof The following proof for the AM/GM inequality is useful for two numbers. First, note that the arithmetic mean and geometric mean for two numbers are as follows: a + b p AM = ; GM = ab: 2 Using a semicircle where AB is the diameter representing a + b and P is the point distinguishing a from b, we can illustrate the AM/GM. Since the diameter of the circle is a + b, its radius is (a + b)=2. The red line represents the arithmetic mean of a and b. 3 A C P B a b We now draw an altitude from point P to point D on the semicircle and line segments from point D to points A and B. Thus, since ∆ADP is similar to ∆DBP it follows that ∆ADB is a right triangle, since an angle inscribed in a semicircle is a right angle, or: a h p = ; h = ab: h b D h A C P B a b Therefore, the altitude is the geometric mean shown in blue, and is clearly less than the hypotenuse. 4 D AM GM A C P B a b Therefore, no matter what the lengths of a and b, AM ≥ GM, and equality holds whenever a = b. 2.2 Algebra Proof Another way to easily reason thep inequalityp with two positive numbers a and b, is by first recognizing 0 ≤ ( a − b)2; then it follows easily: p p 0 ≤ a − 2 a b + b; p a + b ab ≤ : 2 2.3 Formal Proof Now that we have proved the AM/GM inequality for two numbers, we prove the inequality for n numbers. To prove this inequality, we use the following lemma. Lemma 1. Let n ≥ 2 be an integer. Suppose that b1; b2; b3; :::; bn are positive real numbers that are not all equal. If b1b2b3 ··· bn = 1, then b1 + b2 + b3 + ··· + bn > n. Proof. We will use the Principle of Mathematical Induction. For the case n = 2 we know that b1 6= b2 and b1b2 = 1. It follows that, p p 2 p 0 < ( b1 − b2) = b1 − 2 b1b2 + b2 = b1 − 2 + b2 and thus b1 + b2 > 2; 5 showing that the result is true when n = 2. Now suppose the result is valid for some positive integer p ≥ 2. Let b1; b2; :::; bp; bp+1 be positive real numbers that are not all equal and satisfy b1b2 ··· bpbp+1 = 1. Without loss of generality, we may assume that the numbers are in increasing order, that is, b1 ≤ b2 ≤ · · · ≤ bp ≤ bp+1. By the assumptions on these numbers, we must have b1 < 1 < bp+1. Since the conclusion of the lemma is assumed to be true when n = p, we consider the product (b1bp+1)b2 ··· bp = 1, which is a product of p numbers. If all of these numbers are equal (and thus all equal 1), then b2 + b2 + ··· + bp = p − 1 and b1 + bp+1 > 2 (the inequality follows from the first part of the proof) and it follows that b1 + b2 + ··· + bp + bp+1 > p + 1: If the numbers b1bp+1; b2; :::; bp are not all equal, then b1bp+1 + b2 + ··· + bp > p by the induction hypothesis. Since the quantity (bp+1 − 1)(1 − b1) is positive, we find that b1 + b2 + ··· + bp+1 = (b1bp+1 + b2 + ··· + bp) + 1 + (bp+1 − 1)(1 − b1) > p + 1 + (bp+1 − 1)(1 − b1) > p + 1: This shows that the result holds when n = p + 1. By the Principle of Mathematical Induction, the conditional statement given in the lemma is valid for all integers n ≥ 2. [7] Theorem 2 (Arithmetic Mean/Geometric Mean Inequality). Let n be a positive integer. If a1; a2; : : : ; an are nonnegative real numbers, then 1 a1 + a2 + ··· + an (a a ··· a ) n ≤ : 1 2 n n Equality occurs if and only if a1 = a2 = ··· = an. Proof. The equality clearly holds when a1 = a2 = ··· = an. If n = 1, then the result is trivial. Also, if any of the ak = 0, the result is trivial as well. 6 Now, suppose for n ≥ 2, all ak's are positive, and that not all of ak's are 1 equal to each other. If we let r = (a1a2:::an) n , we get the following result, a a a a a ··· a 1 2 ··· n = 1 2 n = 1: r r r rn By the lemma, it follows that, a a a 1 + 2 + ··· + n > n: r r r Multiplying through by r and dividing by n, we find that, a1 + a2 + ··· + an 1 > r = (a a ··· a ) n : [7] n 1 2 n This completes the proof. 3 Optimization Problems and the AM/GM Inequality The following problems are typical problems seen in most calculus textbooks. These can all be solved using the AM/GM inequality, and are categorized into a few different types of problems that often appear in maximum/minimum sections of calculus textbooks. Starting with a simple example, the derivative approach is used, then a solution is shown using the AM/GM inequality. The rest of the problems that follow, are solved using the AM/GM inequality. 3.1 Fence Problems Fence problems are extremely popular in calculus textbooks, and perhaps the most basic. There are many variations of this problem, but all can be solved using the AM/GM inequality and provide an easy starting place to see an application of the inequality. Example 3. A landowner wishes to use 2 miles of fencing to enclose a rectangular region of maximum area. However, one side runs along a stream, so only three sides must be fenced in. Find the lengths of the sides of the rectangular region having the largest area that can be enclosed. [5] 7 river x x y 3.1.1 Derivative Approach We first solve the problem using calculus. Suppose that x is the width of the rectangle, and y is its length. We want to maximize the area A, where A = xy. The problem of maximizing the area of the region can be reduced to the following: maximize xy subject to 2x + y = 2: Noticing that y = 2 − 2x and thus, A(x) = x(2 − 2x) = 2x − 2x2; we can take the derivative and set it equal to zero to find our maximal points: 1 0 = 2 − 4x and hence x = : 2 We have thus shown for the fencing to have maximum area, it will have a width of 1=2 miles, and length of 1 mile.