Question

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

Answer 1

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:

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

$C = \ 311.124$