A marshy region used for agricultural drainage has become contaminated with selenium. It has been...

Question:

A marshy region used for agricultural drainage has become contaminated with selenium. It has been determined that flushing the area with clean water will reduce the selenium for a while, but it will then begin to build up again. A biologist has found that the percentage of selenium in the soil {eq}x {/eq} months after the flushing begins is given by

{eq}f(x) = \frac{x^2 +36}{2x} \space , \space x \in [1,12] {/eq}

When will the selenium be reduced to a minimum? What is the minimum percentage?

Differentiation:

A mathematical tool that used to break in infinitesimal small function or data to analyse the behaviour of input and output of the system is known as differentiation. It is widely used in Computational fluid dynamics to find the nature of the flow.

Given Data:

• The percentage of selenium after flushing begins is: {eq}f\left( x \right) = \dfrac{{{x^2} + 36}}{{2x}} \cdots\cdots\rm{(I)} {/eq}
• {eq}x \in \left[ {1,12} \right] {/eq}

Differentiate the expression of percentage of selenium after flushing begins with respect to x is

{eq}\begin{align*} \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} &= \dfrac{{d\left( {\dfrac{{{x^2} + 36}}{{2x}}} \right)}}{{dx}}\\ &= \dfrac{{2x\left( {2x} \right) - \left( {{x^2} + 36} \right)2}}{{{{\left( {2x} \right)}^2}}}\\ &= \dfrac{{4{x^2} - 2{x^2} - 72}}{{4{x^2}}}\\ &= \dfrac{{2{x^2} - 72}}{{4{x^2}}} \end{align*} {/eq}

The first derivative must be equates to zero to get minimum value of function

{eq}\begin{align*} \dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} &= 0\\ \dfrac{{2{x^2} - 72}}{{4{x^2}}} &= 0\\ 2\left( {{x^2} - 36} \right) &= 0\\ {x^2} - 36 &= 0\\ x &= \pm 6 \end{align*} {/eq}

The domain value is {eq}1 \le x \le 12 {/eq} so the critical value is 6

Substitute and solve the expression (I) at {eq}x=1 {/eq}

{eq}\begin{align*} f\left( 1 \right) &= \dfrac{{{1^2} + 36}}{{2\left( 1 \right)}}\\ &= 18.5 \end{align*} {/eq}

Substitute and solve the expression (I) at {eq}x=6 {/eq}

{eq}\begin{align*} f\left( 6 \right) &= \dfrac{{{6^2} + 36}}{{2\left( 6 \right)}}\\ &= 6 \end{align*} {/eq}

Substitute and solve the expression (I) at {eq}x=12 {/eq}

{eq}\begin{align*} f\left( {12} \right) &= \dfrac{{{{12}^2} + 36}}{{2\left( {12} \right)}}\\ &= 7.5 \end{align*} {/eq}

The critical value is {eq}6 {/eq} so substitute the value at {eq}x=5 {/eq} and {eq}x=7 {/eq} to see the at which point minimum percentage occur

Substitute and solve the expression (I) at {eq}x=5 {/eq}

{eq}\begin{align*} f\left( 5 \right) &= \dfrac{{{5^2} + 36}}{{2\left( 5 \right)}}\\ &= 6.1 \end{align*} {/eq}

Substitute and solve the expression (I) at {eq}x=7 {/eq}

{eq}\begin{align*} f\left( 7 \right) &= \dfrac{{{7^2} + 36}}{{2\left( 7 \right)}}\\ &= 6.07 \end{align*} {/eq}

Thus refer to above result the selenium be reduce to minimum at {eq}x = 6 {/eq} and it will be {eq}6% {/eq}

Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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