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.
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.
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.
Let $\displaystyle L \,=\,AB$ be the length of the rod.Code:| | A * | | * | y| * | | * | P D + - - - * - - - - - - - - | 4 : * | : * 8| 8: * | : * | : * * - - - + - - - - - * - - - - C 4 E x B
. . Then: .$\displaystyle L \;=\;\sqrt{(x+4)^2 + (y+8)^2}$ .[1]
We see that $\displaystyle \Delta PEB \sim \Delta ADP.$
. . Hence: .$\displaystyle \frac{x}{8} \,=\,\frac{4}{y} \quad\Rightarrow\quad y \:=\:\frac{32}{x}$
Substitute into [1], set $\displaystyle L' \,=\,0$, and solve for maximum $\displaystyle L.$
I get (substituting for y into the square of the length) the following as the critical value:
$\displaystyle x=4\sqrt[3]{2}$
I began with:
$\displaystyle L^2=(8+x)^2+(4+y)^2$
$\displaystyle L^2=(8+x)^2+\left(4+\frac{32}{x} \right)^2$
$\displaystyle L^2=(8+x)^2+16\left(1+\frac{8}{x} \right)^2$
Implicitly differentiating with respect to $\displaystyle x$, we have:
$\displaystyle 2L\cdot\frac{dL}{dx}=2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)$
Since we must have $\displaystyle 0<L$, to find the extrema, we may equate:
$\displaystyle 2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)=0$
$\displaystyle (8+x)-128\left(1+\frac{8}{x} \right)\left(\frac{1}{x^2} \right)=0$
$\displaystyle (8+x)-128\left(\frac{1}{x^2}+\frac{8}{x^3} \right)=0$
Multiplying through by $\displaystyle x^3$ and distributing, we find:
$\displaystyle x^4+8x^3-128x-1024=0$
Factor:
$\displaystyle (x+8)(x^3-128)=0$
Discarding the negative root, we have:
$\displaystyle x^3=128=2\cdot2^6$
$\displaystyle x=2^2\cdot\sqrt[3]{2}=4\sqrt[3]{2}$
I suspect you have set it up like Soroban did, so your critical value may be different from what I have.