2 After this lesson, you should be able to: Be able to model the real-life problem to mathematical problem.Solve applied maximum and minimum problems.
3 Optimization Problems One of the most valuable aspects of differential calculus is the ability to find where something is maximized or minimized. When you think about your own personal finances, aren't two of the most important questions you can ask "When/where do I have the least amount of cash and when/where do I have the most?" What factors led to that? At what point is this occurring and why?
4 Optimization Problems Manufacturers want to maximize their profit while minimizing their costs. Thanks to calculus, we've learned that our derivative tells us plenty about our function. And when our derivative is zero, WE WANT TO PAY CLOSE ATTENTION TO WHAT IS HAPPENING TO OUR FUNCTION. Remember that max's and min's can occur where our derivative is zero or undefined, so applying this concept is not too difficult. Now, let's define the steps in analyzing optimization problems.
5 ExampleExample A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume?SolutionLet the length of the base be x inch,hand the height be h inch.Then the volume of the box in terms of x and h inch isxxPrimary EquationNotice that “surface area of 108 in2 ” providesSecondary Equation
6 ExampleExample A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume?SolutionFrom the secondary equation, we havehThen the volume of the box in terms of x in the primary equation isxxNow, before we aim on V(x) and try to find its maximum volume while x has certain length, we should find the feasible domain of x.
7 Notice thatandThis meansFeasible DomainThen taking the derivative of V(x) and set to zero. We haveThe critical number within the feasible domain isandCompute the endpointsandTherefore, V(6, 3) is maximum when x = 6 and h = 3. The box dimension is 6 x 6 x 3.
9 Guidelines for Solving Applied Minimum and Maximum Problems
10 ExampleExample You are designing an open box to be made of cardboard that is 10 inches by 15 inches. The box will be formed by making the cuts at each corner. How long should you make the cuts? What is the maximum volume?1015
11 xxxxExample10SolutionLet the side of square to be cut be x inch.xxxx15Then the dimensions of the boxd can be interpreted asLength =Width =Height =Therefore, the volume isNotice thatandThenFeasible Domain
12 ExampleThen taking the 1st derivative of V(x) and set to zero. We haveThe critical number within the feasible domain is
13 Example Then taking the 2nd derivative of V(x), we have Therefore, is maximumvolume when we cut 4 squares of side
14 ExampleExample Which points on the graph of y = 4 – x2 are closest to the point (0, 2)?SolutionThe distance between the point (0, 2) to any point (x, y) on the graph is(x, y)dNotice that y = 4 – x2, thenLet D(x) = d2(x), then
15 ExampleThen taking the 1st derivative of D(x) and set to zero. We haveThe critical number within the feasible domain isandThen taking the 2nd derivative of D(x), we haveandThis verifies that x = 0 yields a relative maximum andyield a minimum distance to the given point.
16 Notice that in this example, we can not find the absolute maximum, but only the relative maximum. However, we do have absolute minimum.
17 ExampleExample A man is in a boat 2 miles from the nearest point R on the coast. He is trying to go to a point Q, located 3 miles down the coast and 1 mile inland. He can row at 2 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time?SolutionMLet the point on the coast toward by the man be P, and PR = x, then PS = 3 – x. Then23 – xPR3Sx1Q
18 Example M 2 3 – x P Then the total time from point M to Q will be R 3 Sx1QThen taking the 1st derivative of T(x) and set to zero. We haveThe feasible domain for x is [0, 3]. The solution for x within the domain is x = 1.
19 ExampleThen computeWe conclude that when x = 1 yields the minimum time.
20 ExampleExample Four feet of wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum area?xSolutionxLet the side of the square be x and the radius of the circle be r. ThenPrimary EquationSince the total length of wire is 4 feet, thenrSecondary EquationSo,
21 xExamplexrThe feasible domain of x is restricted by the square’s perimeterTaking the 1st derivative of A(x) and set to zero. We haveThe critical number within the feasible domain is
22 ExampleThen computeWe conclude that when yields the maximum area. That means all the wire is used to form circle.
23 ExampleExample A square fountain has a water pool of d feet wide around. A repairman tried to fix a water pipe in the fountain. He used 2 pieces of wood of length L(<d) to access the water pipe in the center. What is the minimum length of the wood board? How did he place these 2 pieces of wood?SolutionWe simplify the problem to the following question: in a square of side length of d, find the 2 perpendicular segments of length L and when L can reach its minimum.dLdL
24 dTExampleLLet the acute angle between one wood board and outer edge of the fountain be θ. Extending TS to R.dLSThen in TRQ, RQ = d tanRPQAnd PQ = d – LcosTherefore RP = RQ – PQ = d tan – (d – Lcos )= d (tan – 1) + LcosIn PRS, RS = RP sin =[d (tan – 1) + Lcos ]sinIn TRQ, RS + ST = d/cosOr, [d (tan – 1) + Lcos ]sin + L = d/cosSolve for L, we get
25 Example Now we seek what value will yields the minimum L. Or, We only need what value will yields the minimum value to f( ). We notice that 0 < < /2 is the feasible domain.Taking the 1st derivative of f( ), we haveThe critical number within the feasible domain isandare not in the 0 < < /2 )
26 ExampleWe are going to take the 2nd derivative of f( ). The derivative of the numerator is
27 ExampleWe simplify the 2nd derivative of f( ). We get
28 Example When =/4, we know that all the factors are positive. So When =/4, f( ) yields the minimum value. Therefore,