Optimization  Techniques

    An optimization  procedure is a step-by-step procedure to solve an application problem using calculus; and in particular, using derivatives. Here is a typical optimization procedure:

    (i) Introduce mathematical notation.
    
    (ii) Express information as equations.
    
    (iii) Formulate the problem mathematically.
    
    (iv) Solve the mathematical problem.
    
    (v) Answer the original problem.

We will illustrate this optimization procedure in the following examples.

Example (Optimizing Area) A woman plans to fence off a rectangular garden whose area is 64 optimization techniques _gr_1.gif] What should be the dimensions of the garden if she wants to minimize the amount of fencing used?

    Solution. Let optimization techniques _gr_2.gif] and optimization techniques _gr_3.gif] be the dimensions of the rectangular plot. The fencing (perimeter) is optimization techniques _gr_4.gif] and the area is optimization techniques _gr_5.gif] with domain optimization techniques _gr_6.gif] We want to minimize optimization techniques _gr_7.gif] so we write optimization techniques _gr_8.gif] as a function of one variable, say

optimization techniques _gr_9.gif]

Since

optimization techniques _gr_10.gif]

we see that optimization techniques _gr_11.gif] with optimization techniques _gr_12.gif] when optimization techniques _gr_13.gif] and optimization techniques _gr_14.gif] also since  

optimization techniques _gr_15.gif]

we see that optimization techniques _gr_16.gif] Thus, optimization techniques _gr_17.gif] is a minimum; and the dimensions of the garden should be optimization techniques _gr_18.gif] ft by optimization techniques _gr_19.gif] ft.   optimization techniques _gr_20.gif]

Example (Optimizing Volume) Find the dimensions of the right circular cylinder of largest volume that can be inscribed in a sphere of radius optimization techniques _gr_21.gif]

    Solution. Let optimization techniques _gr_22.gif] be the radius of the sphere and optimization techniques _gr_23.gif] the radius of the cylinder with height optimization techniques _gr_24.gif] so that optimization techniques _gr_25.gif] Since optimization techniques _gr_26.gif] the volume of the cylinder is given by

optimization techniques _gr_27.gif]

Since

optimization techniques _gr_28.gif]

optimization techniques _gr_29.gif] when optimization techniques _gr_30.gif] Since optimization techniques _gr_31.gif] and optimization techniques _gr_32.gif] give minima, we see by the Extreme Value Theorem that optimization techniques _gr_33.gif] must be a maximum. The dimensions are optimization techniques _gr_34.gif] and optimization techniques _gr_35.gif] optimization techniques _gr_36.gif]

Example (Optimizing an Angle) The bottom of an 8-ft-high mural painted on a vertical wall is 13 ft above the ground. The lens of a camera fixed to a tripod is 4 ft above the ground. How far from the wall should the camera be placed to photograph the mural with the largest possible angle?

    Solution. Let the horizontal distance from the camera to the wall be optimization techniques _gr_37.gif] Let optimization techniques _gr_38.gif] be the angle of elevation from the camera lens to the top of the mural and optimization techniques _gr_39.gif] the angle of elevation from the camera to the bottom of the mural. Also, let optimization techniques _gr_40.gif] Then
    
optimization techniques _gr_41.gif]

Since

optimization techniques _gr_42.gif]

optimization techniques _gr_43.gif]

optimization techniques _gr_44.gif]

we see that optimization techniques _gr_45.gif] and optimization techniques _gr_46.gif] when optimization techniques _gr_47.gif] Since the endpoint optimization techniques _gr_48.gif] is obviously a minimum we have the maximum when optimization techniques _gr_49.gif] optimization techniques _gr_50.gif] or approximately optimization techniques _gr_51.gif] feet. optimization techniques _gr_52.gif]

Example (Optimizing Distance) A truck is 250 mi due east of a sports car and is traveling west at a constant speed of 60 mi/h. Meanwhile, the sports car is going north at 80 mi/h. When will the truck and the car be closest to each other? What is the minimum distance between them?

    Solution. Draw a figure with the car at the origin of a Cartesian coordinate system and the truck at optimization techniques _gr_53.gif] At time optimization techniques _gr_54.gif] (in hours) the truck is at position optimization techniques _gr_55.gif] while the car is at optimization techniques _gr_56.gif] Let optimization techniques _gr_57.gif] be the distance that separates them. Then optimization techniques _gr_58.gif] and optimization techniques _gr_59.gif] so that optimization techniques _gr_60.gif] and optimization techniques _gr_61.gif] We will minimize the square of the distance,
    
optimization techniques _gr_62.gif]

optimization techniques _gr_63.gif]

Since optimization techniques _gr_64.gif] the derivative of the distance squared is 0 when optimization techniques _gr_65.gif] Substituting into the equation for optimization techniques _gr_66.gif] produces the shortest distance:

optimization techniques _gr_67.gif]

Thus,

optimization techniques _gr_68.gif]

and so optimization techniques _gr_69.gif] which is the minimum distance (because there is no maximum distance and the Extreme Value Theorem applies). optimization techniques _gr_70.gif]

Example (Optimizing Time) A jeep is on the desert at a point optimization techniques _gr_71.gif] located 40 km from a point optimization techniques _gr_72.gif], which lies on a long straight road. The driver can travel at 45 km/h on the desert and 75 km//h on the road. The driver will win a prize if he arrives at the finish line at point optimization techniques _gr_73.gif], 50 km from optimization techniques _gr_74.gif], in 84 minutes or less. What route should he travel to minimize the time of travel? Does he win the prize?

    Solution.  Suppose that the driver heads for a point optimization techniques _gr_75.gif] located optimization techniques _gr_76.gif] km down the road from optimization techniques _gr_77.gif] towards his destination. We want to minimize the time. We will need to remember the formula optimization techniques _gr_78.gif] or in terms of time optimization techniques _gr_79.gif] Since the distance between optimization techniques _gr_80.gif] and optimization techniques _gr_81.gif] is optimization techniques _gr_82.gif] and the distance between optimization techniques _gr_83.gif] and optimization techniques _gr_84.gif] is optimization techniques _gr_85.gif] the total time is given by

optimization techniques _gr_86.gif]

Since

optimization techniques _gr_87.gif]

we find that optimization techniques _gr_88.gif] is the only critical number of optimization techniques _gr_89.gif] To find the extreme values we evaluate optimization techniques _gr_90.gif] at the endpoints, we find

optimization techniques _gr_91.gif]

optimization techniques _gr_92.gif]

optimization techniques _gr_93.gif]

Therefore, the driver can minimize the total driving time by heading for a point that is optimization techniques _gr_94.gif] from the point optimization techniques _gr_95.gif] and then traveling on the road to point optimization techniques _gr_96.gif] He wins the prize because the minimal route is only 83 minutes. optimization techniques _gr_97.gif]

    Marginal analysis is concerned with the way quantities such as price, cost, revenue, and profit vary with small changes in the level of production. The demand function optimization techniques _gr_98.gif] is defined to be the price that consumers will pay for each unit of the commodity when optimization techniques _gr_99.gif] units are brought to market. Then optimization techniques _gr_100.gif] is the total revenue function derived from the sale of the optimization techniques _gr_101.gif] units and optimization techniques _gr_102.gif] is the total profit function where optimization techniques _gr_103.gif] is the total cost function for producing optimization techniques _gr_104.gif] units.  

Example (Optimizing Profits) A toy manufacturer produces an inexpensive doll (Dolly) and an expensive doll (Polly) in units of optimization techniques _gr_105.gif] hundred and optimization techniques _gr_106.gif] hundred, respectively. Suppose it is possible to produce the dolls in such a way that

optimization techniques _gr_107.gif]

and that the company receives twice as much for selling a Polly doll as for selling a Dolly doll. Find the level of production for both optimization techniques _gr_108.gif] and optimization techniques _gr_109.gif] for which total revenue derived from selling these dolls is maximized. What vital assumption must be made about sales in the model?

Example (Optimizing Revenue) A business manager estimates that when optimization techniques _gr_110.gif] dollars are charged for every unit of a product, the sales will be optimization techniques _gr_111.gif] units. At this level of production, the average cost is modelled by

optimization techniques _gr_112.gif]

(a) Find the total revenue and total cost functions, and express the profit as a function of optimization techniques _gr_113.gif]

(b) What price should the manufacturer charge to maximize profit? What is the maximum profit?

Example (Optimizing Costs) Suppose the total cost (in dollars) of manufacturing optimization techniques _gr_114.gif] units of a certain commodity is

optimization techniques _gr_115.gif]

(a) At what level of production is the average cost per unit the smallest?

(b) At what level of production is the average cost per unit equal tot he marginal cost?

(c) Graph the average cost and the marginal cost on the same set of axes, for optimization techniques _gr_116.gif]

Cite this as:
Optimization Techniques
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
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