009A Sample Final 1, Problem 10 Detailed Solution

If a resistor of  $R$  ohms is connected across a battery of  $E$  volts with internal resistance  $r$  ohms, then the power (in watts) in the external resistor is

$P={\frac {E^{2}R}{(R+r)^{2}}}.$

If  $E$  and  $r$  are fixed but  $R$  varies, what is the maximum value of the power?

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Background Information:
Recall
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}$

Solution:

Step 1:
Let  $f(x)={\frac {1+x}{3x}}.$
Using the limit definition of derivative, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {({\frac {1+(x+h)}{3(x+h)}})-({\frac {1+x}{3x}})}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {({\frac {1+x+h}{3x+3h}})-({\frac {1+x}{h}})}{h}}.}\end{array}}$

Step 2:
Now, we get a common denominator for the fractions in the numerator.
Hence, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\frac {(1+x+h)3x}{(3x+3h)(3x)}}-{\frac {(1+x)(3x+3h)}{(3x+3h)(3x)}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {\frac {3x+3x^{2}+3xh-(3x+3h+3x^{2}+3hx)}{(3x+3h)(3x)}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {-3h}{h(3x+3h)(3x)}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {-3}{(3x+3h)(3x)}}}\\&&\\&=&\displaystyle {\frac {-3}{(3x)(3x)}}\\&=&\displaystyle {-{\frac {1}{3x^{2}}}.}\end{array}}$

Final Answer:
${\frac {dy}{dx}}=-{\frac {1}{3x^{2}}}$

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