Question

A marshy region used for agricultural drainage has become contaminated with a particular element. It has been determined that flushing the area with clean water will reduce the element for a​ while, but it will then begin to build up again. A biologist has found that the percent of the element in the soil x months after the flushing begins is given by the function below. When will the element be reduced to a​ minimum? What is the minimum​ percent? ​f(x)equalsStartFraction x squared plus 64 Over 2 x EndFraction ​, 1less than or equalsxless than or equals12

Answer 1

Answer:

8.49%

Step-by-step explanation:

x months

$f(x) =\frac{x^2+64}{2x}$ 1 ≤ x ≤ 12

minimum: $f'(x) = 0$

$f'(x) = (\frac{x^{2} +64}{2x})' = \frac{(x^{2} +64)'.2x - (x^{2} +64).(2x)'}{(2x)^2}= \frac{2x.2x-(x^{2}+64).2}{4x^{2} } = \frac{4x^2 - 2x^2-64}{4x^2}=\frac{2x^2-64}{4x^2}$

$\frac{x^2-32}{2x^2} = 0 \\x^{2} -32 = 0\\x^{2} =32\\x = \sqrt{32}\\ x = 4\sqrt{2}$

$f(4\sqrt{2} ) = \frac{(4\sqrt{2})^2 +64}{2.4\sqrt{2} } = \frac{32+64}{8\sqrt{2} } = \frac{96}{8\sqrt{2} }=\frac{12}{\sqrt{2} } = 6\sqrt{2}$ ≅8.49%