4 4 modeling and optimization
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4.4 Modeling and Optimization. Extreme Value Theorem. Reminder:. The Extreme Value Theorem states that every continuous function on a closed interval has both a maximum and a minimum value on that interval. (The max and min MAY occur at the endpoints.). The “Box” Problem.

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4.4 Modeling and Optimization

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4 4 modeling and optimization

4.4Modeling and Optimization


Extreme value theorem

Extreme Value Theorem

  • Reminder:

The Extreme Value Theorem states that every continuous function on a closed interval has both a maximum and a minimum value on that interval.

(The max and min MAY occur at the endpoints.)


The box problem

The “Box” Problem

  • An 11 by 8.5 inch piece of cardboard is to be made into a rectangular open-topped box by cutting out a square at each corner. What dimensions will produce a box with the maximum volume?

x

x

x

x

x

x

x

x


The box problem1

The “Box” Problem

  • An 11 by 8.5 inch piece of cardboard is to be made into a rectangular open-topped box by cutting out a square at each corner. What dimensions will produce a box with the maximum volume?

Length of box = 11 – 2x

Width = 8.5 – 2x

Height = x

Volume = x(11 – 2x)(8.5 – 2x)

Now what???

Volume = 93.5x – 39x2 + 4x3

Finding max, so take derivative

V’= 93.5 – 78x + 12x2= 0

Quadequ

x = 4.915 x = 1.585

(critical points)


The box problem2

The “Box” Problem

  • An 11 by 8.5 inch piece of cardboard is to be made into a rectangular open-topped box by cutting out a square at each corner. What dimensions will produce a box with the maximum volume?

V’= 93.5 – 78x + 12x2= 0

x = 4.915 x = 1.585

To find max/min critical point, use the First Derivative Test.

Is there a restricted interval I must keep my “testing” numbers in between?

0 ≤ x ≤ 4.25

f’(1)

f’(2)

Dec.

Inc.

0

1.585

4.25

MAX


The box problem3

The “Box” Problem

  • An 11 by 8.5 inch piece of cardboard is to be made into a rectangular open-topped box by cutting out a square at each corner. What dimensions will produce a box with the maximum volume?

Since x = 1.585 produces a max, we need to cut out a 1.585 by 1.585 square from each corner.

Length of box = 7.83 in.

Width = 5.33

Height = 1.585

Volume = 66.148 in3


Process for optimization problems

Process for Optimization Problems

  • Identify all given quantities and quantities to be determined.

  • Write an equation for the quantity to be maximized or minimized.

    • Be sure the equation is reduced to have only one independent variable.

  • Take the derivative and find critical points.

  • Determine the feasible domain of the equation so you will know which of the critical points is/are important.

  • Use First Derivative Test to find absolute max/min.

    • BE SURE TO CHECK YOUR ENDPOINTS…many optimization problem solutions occur at the endpoints.


Examples

Examples

  • Which points on the graph of y = 4 – x2 are closest to the point (0, 2)?

  • A rectangular page is to contain 24 inches of print. The margins at the top and bottom of the page are to be 1.5 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the page be so that the least amount of paper is used?


Examples1

Examples

  • 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 total area?

  • A utilities company wants to deliver gas from a source S to a plant P located across a straight river 3 miles wide, then downstream 5 miles. It costs $4 per foot to lay the pipe in the river but only $2 per foot to lay it on land. How can the pipe be laid most economically?


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