Answer:
The minimum possible length of such a line is 8 cm
Step-by-step explanation:
If we had a rectangle, we can name each side "a" and "b".
The area of the rectangle will be:
[tex]S=a\cdot b = 64[/tex]
Note: This is the constraint of our optimiztion problem.
Applying the Pitagoras theorem, the line, as in the figure attached, will have a length of:
[tex]L=\sqrt{(a/2)^2+b^2}=\sqrt{a^2/4+b^2[/tex]
We can replace "a" as a function of "b":
[tex]ab=64\\\\a=64/b[/tex]
Then,
[tex]L=\sqrt{\frac{1}{4}(\frac{64}{b} )^2 +b^2}=\sqrt{\frac{1024}{b^2} +b^2[/tex]
To calculate the minimum length, we derive and equal to zero:
[tex]dL/db=\frac{d}{db} [(\frac{1024}{b^2}+b^2)^{\frac{1}{2}} ]\\\\dL/db=\frac{1}{2} (\frac{1024}{b^2}+b^2)^{(-\frac{1}{2})}\cdot \frac{d}{db} [\frac{1024}{b^2}+b^2]\\\\ dL/db=\frac{2b+1024\cdot(-2)\cdot b^{-3}}{2\sqrt{(\frac{1024}{b^2}+b^2)}} \\\\\\ dL/db=\frac{2b-2048\cdot b^{-3}}{2\sqrt{(\frac{1024}{b^2}+b^2)}}[/tex]
[tex]dL/db=\frac{2b-2048\cdot b^{-3}}{2\sqrt{(\frac{1024}{b^2}+b^2)}}=0\\\\\\2b-2048b^{-3}=0\\\\2b=\frac{2048}{b^3}\\\\b^4=\frac{2048}{2} =1024\\\\b=\sqrt[5]{1024}\approx5.66[/tex]
Now, we know that one side is 5.66 cm.
Then, the other side should be:
[tex]a=64/b=64/5.66=11.31[/tex]
The length of the line for this side dimensions will be:
[tex]L=\sqrt{\frac{1024}{b^2} +b^2}=\sqrt{\frac{1024}{5.66^2} +5.66^2}\\\\L=\sqrt{\frac{1024}{32} +32}=\sqrt{32+32}=\sqrt{64}=8[/tex]
