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help-optimization ?

0 votes

asked May 4, 2015 in CALCULUS by anonymous

3 Answers

0 votes

(1)

Step 1:

The total material is required to fence a rectangular field is image ft.

Let the rectangular field has length ft and breadth ft.

The perimeter of the rectangular field is image.

image

image

image.

The area of the rectangular field is image.

Substitute image in image.

image

image.

The area of the rectangular field is always positive.

image

image

image and image.

image is  positive on the interval image.

answered May 4, 2015 by Sammi Mentor
edited May 4, 2015 by bradely
0 votes

Contd...

Step 2:

Find the dimensions of the rectangular field, that will enclose the maximum area.

image

Apply derivative on each side with respect to image.

image

image.

To find the critical numbers by equating image.

image

image

image ft.

answered May 4, 2015 by Sammi Mentor

Contd...

Step 3:

The maximum value of image occurs at either at critical number or at end points of the interval image.

Substitute image in image.

image.

Substitute image in image.

image.

Substitute image in image.

image.

The area maximum at length of rectangular field is image ft.

Substitute image in image.

image

image ft.

The dimensions of the rectangular field are image ft and image ft.

Solution:

The dimensions of the rectangular field are image ft and image ft.

 

0 votes

(2)

Step 1:

The rectangular field area is square ft.

Let the rectangular field has length ft and breadth ft.

The area of the rectangular field is .

.

The perimeter of the rectangular field is .

Substitute in .

.

The perimeter of the rectangular field is always positive.

image

Perimeter is positive on the interval .

Step 2:

Find the dimensions of the rectangular field, that will require least amount of fencing.

Apply derivative on each side with respect to .

.

Find the critical numbers by equating .

ft.

answered May 4, 2015 by Sammi Mentor

Contd...

Step 3:

The minimum value of occurs at either at critical number or at end point of the interval .

image is negative for .

image is positive for .

Since is decreasing for all to the left of the critical number and increasing for all to the right, must give rise to an absolute minimum.

Substitute in .

.

The dimensions of the rectangular field are ft and ft.

Solution:

The dimensions of the rectangular field are ft and ft.

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