
Let AM=xandS=MD2+MC2
So, MD2+MC2=64+x2+121+(10−x)2
⇒S=285−20x+2x2
dxdS=−20+4x⇒dxdS=0⇒x=5
dx2d2S=4=+ve
JEE Main 2020 — Mathematics Calculus
Let AD and BC be two vertical poles at AandB respectively on a horizontal ground. If AD=8m, BC=11m, AB=10m; then the distance (in meters) of a point M lying in between AB from the point A such that MD2+MC2 is minimums, is__
Held on 6 Sept 2020 · Verified 6 Jul 2026.
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