# Math Help - Application of maxima and minima

1. ## Application of maxima and minima

Find the length of the longest rod which can be carried horizontally around a corner from a corridor 8m wide into one 4m wide. (Without involving angles if possible)
VIEW the attachment for the figure.

2. ## Re: Application of maxima and minima

Let:

$L^2=(8+x)^2+(4+y)^2$

By similarity, we have:

$\frac{y}{8}=\frac{4}{x}$

Can you proceed from here?

3. ## Re: Application of maxima and minima

I already did but it gave me a wrong answer. Maybe something is wrong with my solution. I'll check it.

4. ## Re: Application of maxima and minima

What did you get for your critical value?

5. ## Re: Application of maxima and minima

4*2^(2/3) on my second soltn and 8 on the first.

Btw, w/c function must I use? The one above or this one (L=a+b):
L(x)=a+b=sqrt(x^2+64)+sqrt(y^2+16) ??

6. ## Re: Application of maxima and minima

Hello, Kaloda!

This is a classic problem . . .

Find the length of the longest rod which can be carried horizontally around a corner
from a corridor 8m wide into one 4m wide.

Code:
      |       |
A *       |
| *     |
y|   *   |
|     * | P
D + - - - * - - - - - - - -
|   4   : *
|       :   *
8|      8:     *
|       :       *
|       :         *
* - - - + - - - - - * - - - -
C     4   E     x     B
Let $L \,=\,AB$ be the length of the rod.

. . Then: . $L \;=\;\sqrt{(x+4)^2 + (y+8)^2}$ .[1]

We see that $\Delta PEB \sim \Delta ADP.$

. . Hence: . $\frac{x}{8} \,=\,\frac{4}{y} \quad\Rightarrow\quad y \:=\:\frac{32}{x}$

Substitute into [1], set $L' \,=\,0$, and solve for maximum $L.$

7. ## Re: Application of maxima and minima

Okay. That's exactly how I solve the problem but I obtain a wrong answer! Maybe I'm missing something.
And ow, the book hints that this length (max) is the minimum of certain lengths.

8. ## Re: Application of maxima and minima

I get (substituting for y into the square of the length) the following as the critical value:

$x=4\sqrt[3]{2}$

I began with:

$L^2=(8+x)^2+(4+y)^2$

$L^2=(8+x)^2+\left(4+\frac{32}{x} \right)^2$

$L^2=(8+x)^2+16\left(1+\frac{8}{x} \right)^2$

Implicitly differentiating with respect to $x$, we have:

$2L\cdot\frac{dL}{dx}=2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)$

Since we must have $0, to find the extrema, we may equate:

$2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)=0$

$(8+x)-128\left(1+\frac{8}{x} \right)\left(\frac{1}{x^2} \right)=0$

$(8+x)-128\left(\frac{1}{x^2}+\frac{8}{x^3} \right)=0$

Multiplying through by $x^3$ and distributing, we find:

$x^4+8x^3-128x-1024=0$

Factor:

$(x+8)(x^3-128)=0$

Discarding the negative root, we have:

$x^3=128=2\cdot2^6$

$x=2^2\cdot\sqrt[3]{2}=4\sqrt[3]{2}$

I suspect you have set it up like Soroban did, so your critical value may be different from what I have.

9. ## Re: Application of maxima and minima

Oh yeahhh.! TY for the replies. I got the correct answer but I didn't realized that it just has a different form from the one given by the book.
Didn't know it til I checked their numerical values. I'm such a f*ol!