1. Quotient Rule Word Problem

I am having difficulty remembering how to multiply the square root portion of this problem.

A study indicates that spending money on pollution control is effective up to a point but eventually becomes wasteful. Suppose it is known that when x million dollars is spent on controlling pollution, the percentage of pollution removed is given by:

P(x) = 100(square root x)/0.03x^2+9

Determine the expenditure that results in the largest percentage of pollution removal. What is this maximum percentage?

According to the quotient rule, I have (0.3x^2+9)(100square root x)-(100square root x)(0.06x)/(0.03x^2+9)^2

Multipling out (3x+900square root x)-(6x^3/2)/0.03x^2+9)

If I am right so far, can you give me a nudge on what to do with all these square roots?

Thank You

2. Originally Posted by becky

P(x) = 100(square root x)/0.03x^2+9

Determine the expenditure that results in the largest percentage of pollution removal. What is this maximum percentage?
To avoid using constants I will do,
$\displaystyle P=\frac{Ax^{1/2}}{Bx^2+C}$
Where,
$\displaystyle A=100,B=.03,C=9$

To find the maximimumumumumum you need the critical points.
Begin by finding the derivative,
$\displaystyle P'=\frac{(Ax^{1/2})'(Bx^2+C)-(Ax^{1/2})(Bx^2+C)'}{(Bx^2+C)^2}$
Thus,
$\displaystyle P'=\frac{\frac{1}{2}Ax^{-1/2}(Bx^2+C)-(Ax^{1/2})(2Bx)}{(Bx^2+C)^2}$
Since, the derivative exists everywhere we need when the derivative is zero, that is,
$\displaystyle \frac{1}{2}Ax^{-1/2}(Bx^2+C)-(Ax^{1/2})(2Bx)=0$
Since $\displaystyle x^{-1/2}\not = 0$ divide both sides by it,
$\displaystyle \frac{1}{2}Ax^{-1/2-(-1/2)}(Bx^2+C)-Ax^{1/2-(-1/2)}(2Bx)=0$
Simplify,
$\displaystyle \frac{1}{2}Ax^0(Bx^2+C)-(Ax^1)(2Bx)=0$
Thus,
$\displaystyle \frac{1}{2}A(Bx^2+C)-2ABx^2=0$
Open sesame,
$\displaystyle \frac{1}{2}ABx^2+\frac{1}{2}AC-2ABx^2=0$
Multiply by 2 and divide by "A":
$\displaystyle Bx^2+C-4Bx^2=0$
Thus,
$\displaystyle Bx^2-4Bx^2=-C$
Swictch signs,
$\displaystyle 4Bx^2-Bx^2=C$
Thus,
$\displaystyle 3Bx^2=C$
Thus,
$\displaystyle x^2=\frac{C}{3B}$
Thus, (substitute values back inside)
$\displaystyle x=\pm \sqrt{\frac{C}{3B}}=\pm \sqrt{\frac{9}{3(.03)}}}=\pm \sqrt{\frac{3}{\frac{3}{100}}}=\pm \sqrt{100}=\pm 10$
Reject the negative thus,
$\displaystyle x=10$.

(Now the trick is to show that 10 is a maximum )

3. Hello, becky!

You differentiated incorrectly . . .

A study indicates that spending money on pollution control is effective up to a point
but eventually becomes wasteful.
Suppose it is known that when $\displaystyle x$ million dollars is spent on controlling pollution,
the percentage of pollution removed is given by: .$\displaystyle P(x) \:= \:\frac{100\sqrt{x}}{0.03x^2+9}$

Determine the expenditure that results in the largest percentage of pollution removal.
What is this maximum percentage?

We have: .$\displaystyle P(x)\;=\;\frac{100x^{\frac{1}{2}}}{0.03x^2 + 9}$

Then: .$\displaystyle P'(x)\;=\;\frac{(0.03x^2 + 9)\cdot100\cdot\frac{1}{2}x^{-\frac{1}{2}} - 100x^{\frac{1}{2}}(0.06x)}{(0.03x^2 + 9)^2}$

. . . . . $\displaystyle P'(x) \;= \;\frac{100\cdot\frac{1}{2}x^{-\frac{1}{2}}(0.03x^2 + 9) - 6x^{\frac{3}{2}}}{(0.03x^2+9)^2}$

Multiply top and bottom by $\displaystyle 2x^{\frac{1}{2}}:\;P'(x) \;= \;\frac{100(0.03x^2 + 9) - 12x^2}{2x^{\frac{1}{2}}(0.03x^2+9)^2}$

. . $\displaystyle P'(x) \;= \;\frac{3x^2 + 900 - 12x^2}{2\sqrt{x}(0.03x^2+9)^2}\;=\;\frac{900 - 9x^2}{2\sqrt{x}(0.03x^2+9)^2}$ $\displaystyle \;= \;\frac{9(100 - x^2)}{2\sqrt{x}(0.03x^2+9)^2}$