Optimization

Steps for Solving Optimization Problems

  1. Draw the picture and label variables.

  2. Determine a constraint equation  (if necessary) and a maximizing (minimizing) equation.

  3. Use the constraint equation to solve for one of the variables and substitute it into the maximizing (minimizing) equation.

  4. Take a derivative and set it equal to zero.  Then solve.

  5. Answer the question.


Examples

  1. You want to construct a can that holds 150 cubic inches of juice as cheaply as possible. The top and bottom costs .1 cents per square inch and the side costs .09 cents per square inch. What should the dimensions of the can be?

       

    Solution

    We use the volume formula to get

           
    150 = pr2h 

    and next calculate the cost

            Cost = 2pr2(0.1) + 2prh(0.09)        Cost of top and bottom + Cost of sides

    The volume equation gives us:

                      150
            h  =               
                      p r2

    so that

                                             150    
            C = 0.2pr2 + 0.18p              
                                             p r2

                                27
            =  0.2pr2 +          
                                  r

    To find the minimum cost we take the derivative and set it equal to 0:

                                 27
            C'  =  4pr -              =  0
                                  r2 

    So that

            4pr3  =  27

    or

                       27
            r3  =           
                      
    4p


            r = 2.14 in

    so that 

                        150
            h  =                        =  10.4 in
                       p(2.142

    The can should be constructed so that its radius is 2.14 inches and its height is 10.4 inches.

     

  2. A lifeguard swims at a rate of 5 feet per second and can run at a rate of 15 feet per second. Suppose that the lifeguard spots a drowning child in the ocean 200 feet down the shore and 50 feet out at sea. How far should the lifeguard run until she begins swimming?
           


    Solution

    Our goal is to minimize the total transit time.  The total transit time is 

            Total Transit Time (T) = Time Along the Beach + Time in the Water

    Using

                             Distance
            Time  =                         
                                Rate

    We have 

                                    x
            TimeBeach =             
                                   15

    and

           

    Hence

           

    Notice that the derivative of the inside of the square root sign is

            -2(200 - x)

    We can take the derivative of the transit time with respect to x by using the chain rule. Since we want a minimum, we set this derivative equal to zero.

           

    After a lot of algebra or using a computer we get

            x   @ 182.3 feet

    We can conclude that the lifeguard should run a little more than 182 feet before diving into the water.


Exercises

  1. A poster is to have an area of 120 square inches with one inch margins at the bottom and sides and a 2 inch margin at the top.  What dimensions will give the largest printed area?

  2. A quarter mile race track is to be designed by having a rectangle with semicircles on each end.  Find the dimensions that will make the area of the rectangle as large as possible.


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