Maximum value if this parabola is 7 when x=6 giving the coordinate (6,7) so 7 is your answer
So she used
![\frac{5}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B6%7D%20)
of the cord to make necklaces.
![1- \frac{5}{6} = \frac{1}{6}](https://tex.z-dn.net/?f=1-%20%5Cfrac%7B5%7D%7B6%7D%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20)
She splits the
![\frac{1}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B6%7D%20)
length into 4 equal lengths.
To divide a fraction by a fraction, you multiply it by its reciprocal.
![\frac{1}{6} ( \frac{1}{4})](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B6%7D%20%28%20%5Cfrac%7B1%7D%7B4%7D%29%20)
![=\frac{1}{24}](https://tex.z-dn.net/?f=%20%3D%5Cfrac%7B1%7D%7B24%7D%20)
She used
![\frac{1}{24}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B24%7D%20)
of the original chord to make each bracelet.
Word form - six.seven
Expand form - 6 + 0.7
Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.
![\bf \textit{volume of a rectangular prism}\\\\ V=lwh\quad \begin{cases} l = length\\ w=width\\ h=height\\ -----\\ w=l=x \end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\ -------------------------------\\\\ \textit{surface area}\\\\ S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h \\\\\\ \boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\ -------------------------------\\\\ V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20rectangular%20prism%7D%5C%5C%5C%5C%0AV%3Dlwh%5Cquad%20%0A%5Cbegin%7Bcases%7D%0Al%20%3D%20length%5C%5C%0Aw%3Dwidth%5C%5C%0Ah%3Dheight%5C%5C%0A-----%5C%5C%0Aw%3Dl%3Dx%0A%5Cend%7Bcases%7D%5Cimplies%20V%3Dxxh%5Cimplies%20%5Cboxed%7BV%3Dx%5E2h%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bsurface%20area%7D%5C%5C%5C%5C%0AS%3D4xh%2Bx%5E2%5Cimplies%20300%3D4xh%2Bx%5E2%5Cimplies%20%5Ccfrac%7B300-x%5E2%7D%7B4x%7D%3Dh%0A%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7B%5Ccfrac%7B75%7D%7Bx%7D-%5Ccfrac%7Bx%7D%7B4%7D%3Dh%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0AV%3Dx%5E2%5Cleft%28%20%5Ccfrac%7B75%7D%7Bx%7D-%5Ccfrac%7Bx%7D%7B4%7D%20%5Cright%29%5Cimplies%20V%28x%29%3D75x-%5Ccfrac%7B1%7D%7B4%7Dx%5E3)
so.. that'd be the V(x) for such box, now, where is the maximum point at?
![\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2 \\\\\\ \cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100 \\\\\\ x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}](https://tex.z-dn.net/?f=%5Cbf%20V%28x%29%3D75x-%5Ccfrac%7B1%7D%7B4%7Dx%5E3%5Cimplies%20%5Ccfrac%7BdV%7D%7Bdx%7D%3D75-%5Ccfrac%7B3%7D%7B4%7Dx%5E2%5Cimplies%200%3D75-%5Ccfrac%7B3%7D%7B4%7Dx%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B3%7D%7B4%7Dx%5E2%3D75%5Cimplies%203x%5E2%3D300%5Cimplies%20x%5E2%3D%5Ccfrac%7B300%7D%7B3%7D%5Cimplies%20x%5E2%3D100%0A%5C%5C%5C%5C%5C%5C%0Ax%3D%5Cpm10%5Cimpliedby%20%5Ctextit%7Bis%20a%20length%20unit%2C%20so%20we%20can%20dismiss%20-10%7D%5Cqquad%20%5Cboxed%7Bx%3D10%7D)
now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.
so, the volume will then be at