Oct 21, 2013 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,775 Here is my solution using integral calculus: Spoiler Without loss of generality, I choose to let the trapezoid to be mapped to the first quadrant region bounded by the coordinate axes, the line $x=1$ and the line $y=-2\sqrt{2}x+8\sqrt{2}$. Next, choose a value $x=c$ such that: \(\displaystyle \int_0^c-2\sqrt{2}x+8\sqrt{2}\,dx=\int_c^1-2\sqrt{2}x+8\sqrt{2}\,dx\) Applying the FTOC, we find: \(\displaystyle -\sqrt{2}c^2+8\sqrt{2}c=-\sqrt{2}+8\sqrt{2}-\left(-\sqrt{2}c^2+8\sqrt{2}c \right)\) After simplifying, we have: \(\displaystyle 2c^2-16c+7=0\) Taking the root such that \(\displaystyle 0\le c\le1\) we obtain: \(\displaystyle c=4-\frac{5}{\sqrt{2}}\) Hence: \(\displaystyle \overline{MN}=y\left(4-\frac{5}{\sqrt{2}} \right)=-2\sqrt{2}\left(4-\frac{5}{\sqrt{2}} \right)+8\sqrt{2}=10\)
Here is my solution using integral calculus: Spoiler Without loss of generality, I choose to let the trapezoid to be mapped to the first quadrant region bounded by the coordinate axes, the line $x=1$ and the line $y=-2\sqrt{2}x+8\sqrt{2}$. Next, choose a value $x=c$ such that: \(\displaystyle \int_0^c-2\sqrt{2}x+8\sqrt{2}\,dx=\int_c^1-2\sqrt{2}x+8\sqrt{2}\,dx\) Applying the FTOC, we find: \(\displaystyle -\sqrt{2}c^2+8\sqrt{2}c=-\sqrt{2}+8\sqrt{2}-\left(-\sqrt{2}c^2+8\sqrt{2}c \right)\) After simplifying, we have: \(\displaystyle 2c^2-16c+7=0\) Taking the root such that \(\displaystyle 0\le c\le1\) we obtain: \(\displaystyle c=4-\frac{5}{\sqrt{2}}\) Hence: \(\displaystyle \overline{MN}=y\left(4-\frac{5}{\sqrt{2}} \right)=-2\sqrt{2}\left(4-\frac{5}{\sqrt{2}} \right)+8\sqrt{2}=10\)