Question

Let AD and BC be two vertical poles at A and B respectively on a horizontal ground. If AD = 8 m, BC = 11 m and AB = 10 m; then the distance (in meters) of a point M on AB from the point A such that \(MD^2+MC^2\) is minimum is _______.

Hint:

Let AM = x, express \(MD^2+MC^2\) using right triangles at A and B, and minimise the quadratic in x.

Answer 1

Correct Answer: 5

Let \(AM=x\), so \(MB=10-x\). In the right triangles at A and B: \(MD^2=x^2+8^2\) and \(MC^2=(10-x)^2+11^2\).
\(S(x)=x^2+64+(10-x)^2+121\). \(S'(x)=2x-2(10-x)=4x-20=0\Rightarrow x=5\).
\(S''(x)=4\gt0\), so it is a minimum.
Distance from A \(=5\) m.
Answer: 5