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Lelu [443]
3 years ago
6

A rectangular parking area measuring 5000 ft squared is to be enclosed on three sides using​ chain-link fencing that costs ​$5.5

0 per foot. The fourth side will be a wooden fence that costs ​$7 per foot. What dimensions will minimize the total cost to enclose this​ area, and what is the minimum​ cost?
Mathematics
1 answer:
kaheart [24]3 years ago
5 0

Answer:

Dimensions: 75.3778 ft and 66.3325 ft

Minimum price: $1658.31

Step-by-step explanation:

Let's call the length of the parking area 'x', and the width 'y'.

Then, we can write the following equations:

-> Area of the park:

x * y = 5000

-> Price of the fences:

P = 2*x*5.5 + y*5.5 + y*7

P = 11*x + 12.5*y

From the first equation, we have that y = 5000/x

Using this value in the equation for P, we have:

P = 11*x + 12.5*5000/x = 11*x + 62500/x

To find the minimum of this function, we need to take its derivative and then make it equal to zero:

dP/dx = 11 - 62500/x^2 = 0

x^2 = 65000/11

x = 250/sqrt(11) = 75.3778 ft

This is the x value that gives the minimum cost.

Now, finding y and P, we have:

x*y = 5000

y = 5000/75.3778 = 66.3325

P = 11*x + 62500/x = $1658.31

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<h3>Answer:  b. either 2.5 inches or about .716 inches</h3>

=====================================================

Work Shown:

I have attached two images below. The first shows figures 1 through 4. The second image shows figure 5.

Start with figure 1. Work your way left to right, top to bottom til you get to figure 4. This is the basic process to turn a flat piece of cardboard into a 3D folded box.

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Figure 4 is the completed open-top-box. In other words, this box does not have a lid.

The green portions in figures 3 and 4 indicate where the flaps used to be in reference to figure 2.

-------------------

Since those dashed lines in figure 1 were x inches long, this means that the distance from A to B is 11-x-x = 11-2x inches long.

This makes segment BD to be 8-2x inches long. We are subtracting off 2 copies of x each time for each dimension.

------------------

length = AB = 11-2x

width = BD = 8-2x

Since we want the length to be positive, this means 11-2x > 0 which solves to x < 5.5

Since we want the width to be positive, this means 8-2x > 0 which solves to x < 4

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Furthermore, x is some length itself, so it cannot be negative either. Therefore x > 0 or 0 < x

Combine 0 < x and x < 4 to get 0 < x < 4

The restriction 0 < x < 4 will be used later.

------------------

For now, let's compute the volume.

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