Maxima/minima problem

Quote:

Originally Posted by jacs

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

Understand that $L$ is the total length of the two strips. Thus, $L=EF+BD$
Call side $AB=CD=x$ for $x>0$ .
Then, $AD=BC=9/x$ . (Becasue the area is 9).
But, $BD=\sqrt{CD^2+BC^2}$ by pythagorean theorem. Thus,
$BD=\sqrt{x^2+\frac{81}{x^2}$
Thus,
$BD=\sqrt{\frac{x^4+81}{x^2}}$
Thus,
$BD=\frac{\sqrt{x^4+81}}{\sqrt{x^2}}$
But, $\sqrt{x^2}=|x|=x$ because $x>0$ .
Thus,
$BD=\frac{\sqrt{x^4+81}}{x}$
Thus, $L=x+\frac{\sqrt{x^4+81}}{x}$
This answers the first part.

Define a function $f(x)=x+\frac{\sqrt{x^4+81}}{x}$ $x>0$ and attempt to find its minimum value.
This is done by taking the derivative of $f(x)$
Thus,
$f'(x)=1+\frac{(1/2)(4x^3)(x^4+81)^{-1/2}(x)-(x^4+81)^{1/2}(1)}{x^2}$
Simplify this mess,
$f'(x)=1+2\frac{x^2}{\sqrt{x^4+81}}-\frac{\sqrt{x^4+81}}{x^2}$
For simplicity sake let $z=\frac{x^2}{\sqrt{x^4+81}}$ thus,
$f'(x)=1+2z-\frac{1}{z}$
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,
$f'(x)=0$
Thus,
$1+2z-\frac{1}{z}=0$
Solve this quadradic,
$z+2z^2-1=0$
Thus, $z=-1,1/2$
But $z=\frac{x^2}{\sqrt{x^4+81}}>0$ thus,
$z=1/2$ as the only critical point.
Thus,
$\frac{x^2}{\sqrt{x^4+81}}=1/2$
Thus (squaring),
$\frac{x^4}{x^4+81}=\frac{1}{4}$
Thus,
$4x^4=x^4+81$
Thus,
$3x^4=81$
Thus, ( $x>0$ )
$x=\sqrt[4]{27}$ .

Now we prove that $x=\sqrt[4]{27}$ 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 $\sqrt[4]{27}$ say 1 and find the sign of $f'(1)$ is is negative-the function is decreasing. Take any point to the right of $\sqrt[4]{27}$ say 4, and find the sign of $f'(4)$ 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 $0<x<\sqrt[4]{27}$ is decreasing thus $f(x)>L$ . And because the function is increasing on $x>\sqrt[4]{27}$ thus, $f(x)>L$ thus, $L$ is minimized at
$x=\sqrt[4]{27}\approx 2.28$

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.