help please! i'm stuck?

Question

another one of these:

"A kennel operator wishes to enclose a rectangular area against the wall of a building to form a dog run. A dog lover donated 120m of fencing to create a run of maximum area."

i have to find the dimensions of the 3-sided rec so that the area is maximized, and have to find the best dimensions for the run. this is a spreadsheet assignment for the length, width, and area. the max length has to be 120 and minimum 0, but i can't figure out how to calculate the dimensions on the spreadsheet (ie. i have to figure out the width and area of the 3-sided rec against the wall with its length being 120)

ty

Answer 1

What Shape Is The Kennel? :P

It could either be

b1 + b2 x height divided by 2 or L x W or B x H!

Answer 2

So, you want to maximize the area.

A = width * length

But you have a maximum amount of fencing as such:

120 = width + width + length

You can rewrite this to solve width in terms of length:

width = (120 - length) / 2.

Subsitute into the first equation to get:

Area = (120 - length) / 2 * length

But you can simplify this even better. Multiply both sides by 2:

2 Area = (120 - length) * length

This area is maximized when you have a square -- half the 120 == 60. So you want a length of 60 along the wall with 2-30-foot ends.

Answer 3

If your shape is rectangular, the most area you can enclose with a given perimeter is a square.

You can prove this to yourself on your spreadsheet by setting up your length, given two equal widths (these are your 3 sides - two widths x 1 length make up your perimeter). Put in a few possible values for the widths, calculate the corresponding length, and then determine the area. Plot the area verses with width and you will see a curve that maximizes when the widths are equal to the length.

Answer 4

Equation 1:

2w + l = 120

l = 120 - 2w

l * w = area

(120-2w)*w = 0 (solving for 0 will make this easier since this will give you critical points

120w - 2w^2 = 0 (divide both sides by -2)

w^2 - 60w = 0 (complete the square)

Complete the square

Take -60, divide by 2, then square (-60/2)^2 = 900

w^2 -60w + 900 = 0 (I know that this is wrong since I did not add 900 to both sides)

(w-30) ^2 = 0 width = 30, length = 60

If you do a quick check, you can see that this is true:

width = 29, length = 62, area = 1798
width = 30, length = 60, area = 1800
width = 31, length = 61, area = 1798

Answer 5

you got the length L and width W and you know that the total fence you have is 120m

so L+W+W = 120 (the perimiter minus the side on the house)

but the area is LxW

if you input the length L then you calc the W by W = (120 - L)/2

Answer 6

peri=L+2W----(1)

Area=L*W---(2)

we have the lenght so substitute W=area/lenght
then put that in eqn1 and the perimeter is 120m, u ll get the ans for the area then from are get the ans for W