Lecture 17 optimization - section 4.6
-
Upload
njit-ronbrown -
Category
Education
-
view
132 -
download
1
description
Transcript of Lecture 17 optimization - section 4.6
![Page 1: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/1.jpg)
Optimization Problems4.6
![Page 2: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/2.jpg)
2
Example 1 – Maximizing Area
A farmer has 2400 ft of fencing and wants to fence off a
rectangular field that borders a straight river. He needs no
fence along the river. What are the dimensions of the field
that has the largest area?
![Page 3: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/3.jpg)
3
Example 1 – Solution
In order to get a feeling for what is happening in this
problem, let’s experiment with some special cases.
Figure 1 (not to scale) shows three possible ways of laying
out the 2400 ft of fencing.
Figure 1
![Page 4: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/4.jpg)
4
Example 1 – Solution
We see that when we try shallow, wide fields or deep,
narrow fields, we get relatively small areas. It seems
plausible that there is some intermediate configuration that
produces the largest area.
Figure 2 illustrates the general case. We wish to maximize
the area A of the rectangle.
cont’d
Figure 2
![Page 5: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/5.jpg)
5
Example 1 – SolutionLet x and y be the depth and width of the rectangle (in feet). Then we express A in terms of x and y:
A = xy
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x.
To do this we use the given information that the total length of the fencing is 2400 ft.
Thus
2x + y = 2400
From this equation we have y = 2400 – 2x, which gives
A = x(2400 – 2x) = 2400x – 2x2
cont’d
![Page 6: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/6.jpg)
6
Example 1 – Solution
Note that x 0 and x 1200 (otherwise A < 0). So the
function that we wish to maximize is
A(x) = 2400x – 2x2 0 x 1200
The derivative is A'(x) = 2400 – 4x, so to find the critical
numbers we solve the equation
2400 – 4x = 0
which gives x = 600.
The maximum value of A must occur either at this critical
number or at an endpoint of the interval.
cont’d
![Page 7: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/7.jpg)
7
Example 1 – Solution
Since A(0) = 0, A(600) = 720,000, and A(1200) = 0, the
Closed Interval Method gives the maximum value as
A(600) = 720,000.
[Alternatively, we could have observed that A'' (x) = –4 < 0
for all x, so A is always concave downward and the local
maximum at x = 600 must be an absolute maximum.]
Thus the rectangular field should be 600 ft deep and 1200 ft
wide.
cont’d
![Page 8: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/8.jpg)
8
Optimization Process1. Read the problem (several times)
2. Draw a figure
3. Identify variables
4. Equation for quantity to be optimized Reduce to 1 variable
5. Differentiate
6. Identify crit #’s (f ‘=0 or f ‘ undefined)
7. Test for max/min
8. Answer question, with units
![Page 9: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/9.jpg)
9
Max/Min Test with an IntervalFrom Section 4.2
Find crit #s
Test original function at crit #s & endpoints of interval
-largest is max value
-smallest is min value
![Page 10: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/10.jpg)
10
1st Derivative Test
![Page 11: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/11.jpg)
11
2nd Derivative TestIf second derivative is constant over the interval and
f ‘’ negative (f concave down) max
f ‘’ positive (f concave up) min
![Page 12: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/12.jpg)
12
Optimization - Exercise(6) Find the dimensions of a rectangle with area 1000 m^2 whose perimeter is as small as possible.
![Page 13: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/13.jpg)
13
Optimization - Exercise(10*) A box with an open top is to be constructed from a square piece of carboard, 12 inches on each side, but cutting a square from each of the 4 corners and bending up the sides. Find the largest volume that such a box can have.
![Page 14: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/14.jpg)
14
Optimization - Exercise(12*) A box with a square base and open top must have a volume of 32 cm^3. Find the dimensions of the box that minimize the amount of material used.
![Page 15: Lecture 17 optimization - section 4.6](https://reader035.fdocuments.in/reader035/viewer/2022081603/5591a1f21a28ab9a268b473f/html5/thumbnails/15.jpg)
15
Optimization - Exercise(18*) Find the dimensions of the rectangle of the largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola y = 12 – x2