Optimization

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Optimization

Steps for Solving Optimization Problems

Draw the picture and label variables.
Determine a constraint equation (if necessary) and a maximizing (minimizing) equation.
Use the constraint equation to solve for one of the variables and substitute it into the maximizing (minimizing) equation.
Take a derivative and set it equal to zero. Then solve.
Answer the question.

Examples

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?

    $Cylinder with height h, 0.1 cent for the top and bottom and 0.09 cents for the side and radius x$

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 + 0.18pr
                                         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.
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?
        $Picture of a lifeguard, the horizontal line x to a dot then 200-x to the end. 50 above the end is a person needing help. The line form the person to the line is shown, making a right triangle.$

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
        Time_Beach =
                               15

and

$ \text{Time}_{\text{Water}} = \frac{\sqrt{50^2 + (200 - x)^2}}{5} $

Hence
$ T = \frac{x}{15} + \frac{\sqrt{50^2 + (200 - x)^2}}{5} $

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.

$ T' = \frac{1}{15} - \frac{200 - x}{5\sqrt{50^2 + (200 - x)^2}} = 0 $

After a lot of algebra or using a computer we get

        x $ \approx $ 182.3 feet

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

Exercises

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?
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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