Đồ thị Parabol có dạng \(y=ax^2+bx+c\left(P\right)\)
\(\left(0;60\right)\in\left(P\right)\Leftrightarrow c=60\)
\(\left\{{}\begin{matrix}\left(-40;0\right)\in\left(P\right)\\\left(40;0\right)\in\left(P\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}40^2a+40b+60=0\\40^2a-40b+60=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}80a+2b+3=0\\80a-2b+3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{80}\\b=0\end{matrix}\right.\)
\(\Rightarrow\left(P\right):y=-\dfrac{3}{80}x^2+60\)
\(g\left(x\right)=\left(P\right)-\left(d\right)=-\dfrac{3}{80}x^2+60-\left(\dfrac{x^3}{25600}-\dfrac{3}{16}x+35\right)\)
\(\Rightarrow g\left(x\right)=-\dfrac{1}{25600}x^3-\dfrac{3}{80}x^2+\dfrac{3}{16}x+25\)
\(\Rightarrow g'\left(x\right)=-\dfrac{3}{25600}x^2-\dfrac{3}{40}x+\dfrac{3}{16}\)
\(g'\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2,49\\x=-642,49\end{matrix}\right.\)
\(\Rightarrow MN=g\left(2,49\right)=25,23\left(m\right)\)
\(PQ=g\left(-2,49\right)=24,3\left(m\right)\) (do đối xứng)
\(\Rightarrow PQ+MN=25,23+24,3\approx49,5\left(m\right)\)
