Respuesta :
Answer:
[tex]x = y \approx 14.142\,m[/tex]
Step-by-step explanation:
Let be x and y the length and width of the rectangular field, whose area and perimeter are, respectively:
[tex]A = x\cdot y[/tex]
[tex]p = 2\cdot (x + y)[/tex]
Building cost is minimum when perimeter is minimum, then equation system must be reduced and critical values have to be found:
[tex]x \cdot y = 200[/tex]
[tex]p = 2\cdot (x + y)[/tex]
Then,
[tex]y = \frac{200}{x}[/tex]
[tex]p = 2\cdot \left(x + \frac{200}{x} \right)[/tex]
Now, critical values are determined by First and Second Derivative Tests:
FDT
[tex]p' = 2 \cdot \left(1 -\frac{200}{x^{2}}\right)[/tex]
[tex]2\cdot \left(1-\frac{200}{x^{2}} \right) = 0[/tex]
[tex]x^{2} - 200 = 0[/tex]
[tex]x = \sqrt{200}[/tex]
[tex]x \approx 14.142\,m[/tex] (As length is a positive number)
SDT
[tex]p'' = 2\cdot \left(\frac{400}{x^{3}} \right)[/tex]
[tex]p'' = 0.283[/tex] (which means that critical point leads to a minimum).
Now, the value of the other variable is:
[tex]y = \frac{200\,m^{2}}{14.142\,m}[/tex]
[tex]y = 14.142\,m[/tex]
Lastly, the cost of building the fence is:
[tex]C = 2 \cdot [\$ 3 + \$ 8 ]\cdot (14.142\,m)[/tex]
[tex]C = \$ 311.124[/tex]
The dimensions of the field will be 75.77 and 28.41 feet.
How to calculate the dimensions?
It should be noted that 200m² is the same as 2152.78 ft². In order to build the fence, the total cost will be:
= (2x × 3) + (2y × 8)
C = 6x + 16y
In this case, xy = 2152.78. Therefore, y = 2152.78/x.
C = 6x + 34444.48/x
dc/dx = 6 - 34444.48/x²
x = 75.77
Therefore, y will be:
= 2152.78/75.77
= 28.41
In conclusion, the dimensions are 75.77 and 28.41 feet.
Learn more about dimensions on:
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