Question

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

Answer 1

Step 1:Illustrating the figure using the given data

Let $AD$ and $BC$ be two vertical poles at $A$ and $B$ respectively on horizontal ground.

$AD=8m$

$BC=11m$

$AB=10m$

Let's consider $AM=x;MB=10-x$

Step 2: Finding $M{D}^2+M{C}^2$

Applying Pythagoras theorem in $△DAM$

$$\begin{gathered} M{D}^2=A{D}^2+A{M}^2 \\ M{D}^2=8^2+x^2 \\ =x^2+64 \\ \text{Applying Pythagoras theorem in △CBM} \\ M{C}^2=M{B}^2+B{C}^2 \\ M{C}^2={\left(10-x\right)}^2+{11}^2 \\ =100-20x+x^2+121 \\ =x^2-200x+221 \end{gathered}$$

Step 3: Finding the minimum value

The function $y=f\left(x\right)$ has a minimum at $a$ if $f^{\text{'}}\left(a\right)=0$ and $f^{\text{'}\text{'}}\left(a\right)>0$.

$$\begin{gathered} f\left(x\right)=M{D}^2+M{C}^2 \\ =2x^2-20x+285 \\ \Rightarrow f^{\text{'}}\left(x\right)=4x-20 \end{gathered}$$

equating $f^{\text{'}}\left(x\right)=0$

$$\begin{gathered} 4x-20=0 \\ 4x=20 \\ x=5 \end{gathered}$$

Check $f^{\text{'}\text{'}}\left(a\right)<0$

$$\begin{gathered} f^{\text{'}}\left(x\right)=4x-20 \\ \Rightarrow f^{\text{'}\text{'}}\left(x\right)=4>0 \end{gathered}$$

Thus, at $x=5$ we get the point of minima

Therefore, the distance of a point $M=5m$ on $AB$ from the point $A$ such that $M{D}^2+M{C}^2$ is minimum.