OptimizationA solution to nearly any problem in life involves some trade offs. In economics there is atrade off between selling more product at a lower price and selling less product at a higherprice: what is the ideal level of production? Before an exam there is a tradeoff betweenstudying more, and being well-rested and relaxed. In life in general there is the constantbattle between balancing work to make more money, and free time to be able to spend thatmoney. It is an interesting problem to try and find the ideal balance in one of these tradeoffs. In order to solve such a problem, we need to find a point in the tradeoff where a desiredquantity is maximized, or an undesired quantity is minimized. In life, we try to find a bal-ance between work and free time in order to maximize our happiness. In economics, we lookat maximizing profit. In manufacturing, we might want to minimize the amount of materialwe need to use. All of these problems are optimization problems, where the goal is to findan optimal solution, that maximizes or minimizes the quantity of interest. In order to solveoptimization problems, we need to construct a mathematical model for the situation, andsolve for the global mimimum or maximum.Example 1 Suppose you are trying to maximize your free time, in order to get the mostenjoyment out of life. In a given day, you have 24 hours, but you must take out time forwork and school, as well as eating and sleeping. Finally, if you do not sleep enough in agiven day, you will be tired and inefficient, wasting time. In order to try and find balance inyour life, you have developed the following model for your free timeF (t) = 24 − I − t −100(t + 1 − s)4To use your model, you first need to set the paramters I and s, where I represents the amountof time you must spend working and in school, and s represents the minimum amount oftime you need to sleep in order to be able to function. Once you have set these parameters,you must find a value of t for the amount of time you should spend sleeping and eating, inorder to maximize your free time F .Solution In order to maximize F , we need to look for a global maximum. The domain ofthis function must surely be smaller than (s−1, 24−I), because these extremes represent theminimum amount of time we could reasonably spend sleeping, and the maximum amount oftime we could possibly sleep in a day (if we slept the entire day way). These extremes areunlikely to be ideal, so we should seek a maximum on the interior of the domain, by lookingfor critical points.F0(t) = −1 +400(t + 1 − s)5= 0which implies thatt = (400)1/5+ s − 1Looking at the second derivative we findF00(t) = −2000(t + 1 − s)6< 0which means our function is concave down, and we have a maximum. For this value of t, wefind thatF ((400)1/5+ s −1) = 24 −I − (400)1/5− s + 1 −100(400)1/5+ s − 1 + 1 − s)4≈ 20.86 − I − sThus, for an individual with a work load of I = 10 and a minimum sleep tolerance of s = 7,the ideal amount of time to spend sleeping and eating ist = 9.3and such an individual hasF (9.3) = 20.86 − 10 − 7 = 3.86hours of free time each day.Example 2 The ideal dosage of a medication is a tradeoff between the efficacy of the drug,and the severity of its side effects. If the side effects of a treatment are too severe, it will haveto be discontinued, so it will fail to treat the patient. Similarly, if the dosage of the medicineis too low, its efficacy will not be sufficient to treat the patient. Suppose the probability ofan effective treatment that accounts for this trade off is given byP (x) =x1/21 + xwhere x is the dosage of the medication given to the patient. What is the ideal dosage?Solution To find the solution to this problem, we want to maximize P (x), so we are lookingfor a global maximum. First we will need to find and classify the critical points, and finallycompare them to the values of the endpoints.P0(x) =(1 + x)12x−1/2− x1/2(x + 1)2A fraction is only zero when its numerator is zero, so we need to solve12(1 + x)x−1/2− x1/2= 012(1 + x) − x = 012= x(1 −12)1 = xWe find that P (1) = 0.5. We find that P (0) = 0, and since the sign of the derivative canonly change at a critical point, and P (1) > P (0), we find that P0(x) > 0 for 0 < x < 1 (thederivative is not defined for x = 0). Similarly, to find the sign of the derivative for x > 1,we can choose a sample point, and the sign of the derivative of that point will give us thesign of the derivative for all x > 1. Choosing x = 4 (be cause the square root of 4 is easy toevaluate) we findP0(4) =(1 + 4)124−1/2− 41/2(4 + 1)2=52√4−√425=1.25 − 225< 0Thus, for x > 1, the function is decreasing, so we find that x is a local maximum. Sincethere is no right endpoint, we conclude that x = 1 is the global maximum.Example 3 An open-top box can be made by cutting out squares from a rectangular pieceof material, and folding up the resulting flaps. For a piece of material with length l andwidth w, what length for the sides of the cut-out squares will maximize the volume of theresulting box? For simplicity, assume w < l.Solution In this problem we want to maximize the volume of the box. In order to do so,we first need to find an equation for the volume of the box, and then search for a globalmaximum. The volume of the box is given by multiplying the length, width, and heightof the box together. Since we are removing squares of le ngth x from the original piece ofmaterial, the length and width of the resulting box will be l − 2x and w − 2x, respectively.The height of the box will simply be x. Thus, we find thatV (x) = x(l − 2x)(w − 2x) = 4x3− (2l + 2w)x2+ lwxThe domain of this function is [0, w/2], because we can begin by cutting no material, or wecan cut all the way through the width of the box. At both of these points we will clearlyhave 0 volume, and since this is a continuous function with positive volume on the interiorof its domain, we know that we will have a global maximum. In order to find this globalmaximum we differentiate, yieldingV0(x) = 12x2− 4(l + w)x + l wIn order to find the roots of this equation, we apply the quadratic formula, which tells usthatx =4(l + w) ±p16(l + w)2− 48lw24=4(l + w ±p(l + w )2− 3lw)24=l + w ±√l2+ w2− lw6How can we interpret these roots? Let us use the se cond derivative test to determine if theyare minima or maxima.V00(x) = 24x − 4(l + w)For the first of these roots, we haveV00(l …