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AURORKA [14]
3 years ago
7

We are going to fence in a rectangular field that encloses 200m2. If the cost of thematerial for of one pair of parallel sides i

s $3 per meter and cost of the material for theother pair of parallel sides is $8 per meter determine the dimensions of the field thatwill minimize the cost to build the fence around the field.
Mathematics
1 answer:
Crazy boy [7]3 years ago
5 0

Answer:

x = y \approx 14.142\,m

Step-by-step explanation:

Let be x and y the length and width of the rectangular field, whose area and perimeter are, respectively:

A = x\cdot y

p = 2\cdot (x + y)

Building cost is minimum when perimeter is minimum, then equation system must be reduced and critical values have to be found:

x \cdot y = 200

p = 2\cdot (x + y)

Then,

y = \frac{200}{x}

p = 2\cdot \left(x + \frac{200}{x} \right)

Now, critical values are determined by First and Second Derivative Tests:

FDT

p' = 2 \cdot \left(1 -\frac{200}{x^{2}}\right)

2\cdot \left(1-\frac{200}{x^{2}} \right) = 0

x^{2} - 200 = 0

x = \sqrt{200}

x \approx 14.142\,m (As length is a positive number)

SDT

p'' = 2\cdot \left(\frac{400}{x^{3}} \right)

p'' = 0.283 (which means that critical point leads to a minimum).

Now, the value of the other variable is:

y = \frac{200\,m^{2}}{14.142\,m}

y = 14.142\,m

Lastly, the cost of building the fence is:

C = 2 \cdot [\$ 3 + \$ 8 ]\cdot (14.142\,m)

C = \$ 311.124

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