So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
I think this question is a part of another question; you can have any ratio that can add up to 114 cm.
For me personally, the easiest way to do this is by isolating the x² term, and finding the square root of both sides. The hardest way (well actually, the longest way) would be to use the quadratic formula. It just complicates things unnecessarily.
Answer:
6g
Step-by-step explanation:
I think it is this answer
