I have a max/min prob I have been trying to nut out and just can't seem to see it. Any help woudl be very appreciated.

thanks jacs

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- Jan 20th 2006, 05:50 AMjacsMaxima/minima problem
I have a max/min prob I have been trying to nut out and just can't seem to see it. Any help woudl be very appreciated.

thanks jacs - Jan 20th 2006, 09:48 AMThePerfectHackerQuote:

Originally Posted by**jacs**

Call side $\displaystyle AB=CD=x$ for $\displaystyle x>0$.

Then, $\displaystyle AD=BC=9/x$. (Becasue the area is 9).

But, $\displaystyle BD=\sqrt{CD^2+BC^2}$ by pythagorean theorem. Thus,

$\displaystyle BD=\sqrt{x^2+\frac{81}{x^2}$

Thus,

$\displaystyle BD=\sqrt{\frac{x^4+81}{x^2}}$

Thus,

$\displaystyle BD=\frac{\sqrt{x^4+81}}{\sqrt{x^2}}$

But, $\displaystyle \sqrt{x^2}=|x|=x$ because $\displaystyle x>0$.

Thus,

$\displaystyle BD=\frac{\sqrt{x^4+81}}{x}$

Thus, $\displaystyle L=x+\frac{\sqrt{x^4+81}}{x}$

This answers the first part.

Define a function $\displaystyle f(x)=x+\frac{\sqrt{x^4+81}}{x}$ $\displaystyle x>0$and attempt to find its minimum value.

This is done by taking the derivative of $\displaystyle f(x)$

Thus,

$\displaystyle f'(x)=1+\frac{(1/2)(4x^3)(x^4+81)^{-1/2}(x)-(x^4+81)^{1/2}(1)}{x^2}$

Simplify this mess,

$\displaystyle f'(x)=1+2\frac{x^2}{\sqrt{x^4+81}}-\frac{\sqrt{x^4+81}}{x^2}$

For simplicity sake let $\displaystyle z=\frac{x^2}{\sqrt{x^4+81}}$ thus,

$\displaystyle f'(x)=1+2z-\frac{1}{z}$

Now find the critical points (points where there is no deriviative or when it is zero). Since the derivative exists everywhere we find where it is zero, thus,

$\displaystyle f'(x)=0$

Thus,

$\displaystyle 1+2z-\frac{1}{z}=0$

Solve this quadradic,

$\displaystyle z+2z^2-1=0$

Thus, $\displaystyle z=-1,1/2$

But $\displaystyle z=\frac{x^2}{\sqrt{x^4+81}}>0$ thus,

$\displaystyle z=1/2$ as the only critical point.

Thus,

$\displaystyle \frac{x^2}{\sqrt{x^4+81}}=1/2$

Thus (squaring),

$\displaystyle \frac{x^4}{x^4+81}=\frac{1}{4}$

Thus,

$\displaystyle 4x^4=x^4+81$

Thus,

$\displaystyle 3x^4=81$

Thus, ($\displaystyle x>0$)

$\displaystyle x=\sqrt[4]{27}$.

Now we*prove*that $\displaystyle x=\sqrt[4]{27}$ is the minimum. We use the first-derivative test (there is no way I am going to use the second-derivative test :) )

Take any point to the left of $\displaystyle \sqrt[4]{27}$ say 1 and find the sign of $\displaystyle f'(1)$ is is negative-the function is decreasing. Take any point to the right of $\displaystyle \sqrt[4]{27}$ say 4, and find the sign of $\displaystyle f'(4)$ which is postive-the function is increasing. Thus, if a function switches from decreasing to increasing in a point that point is a relative minimum. Since the function on the interval $\displaystyle 0<x<\sqrt[4]{27}$ is decreasing thus $\displaystyle f(x)>L$. And because the function is increasing on $\displaystyle x>\sqrt[4]{27}$ thus, $\displaystyle f(x)>L$ thus, $\displaystyle L$ is minimized at

$\displaystyle x=\sqrt[4]{27}\approx 2.28$

In these long type of problems it is traditional for me to make a mistake somewhere, thus, if my answer does not agree with your books answer try to search where I made a mistake.

Q.E.D. - Jan 20th 2006, 04:46 PMjacsthanks
The PerfectHacker, the solution is perfect, no errors at all. :)

I can see where i messed up now, i magically dropped my x² in the denominator in my quotien rule, but even asssuming i didnt mess that up, i never would have recognized the hidden quadratic in the final equation.

and you are so right about not wanitng to use the second derivative test on that one

once again, many thanks

jacs