Q.Let AP and BQ be two vertical poles at points A and B, respectively. If $AP = 16$ m, $BQ = 22$ m and $AB = 20$ m, then find the distance of a point R on AB from the point A such that $RP^2 + RQ^2$ is minimum.

Put $AR = x$ and write both squared distances in terms of $x$, then minimise the quadratic.

Set up the distances. Since the poles are vertical, $\triangle ARP$ is right-angled at $A$ and $\triangle BRQ$ is right-angled at $B$. With $AR = x$, we get $RB = 20 - x$.

Pythagoras.

Sum and simplify.

$S(x) = 256 + x^2 + 484 + (400 - 40x + x^2) = 2x^2 - 40x + 1140.$ …

The idea

We have two vertical poles $AP$ and $BQ$, and a point $R$ that can slide along the ground segment $AB$. As $R$ moves, the distances to the pole-tops $P$ and $Q$ change. We want the position that makes $RP^2 + RQ^2$ as small as possible. The trick is to write everything in terms of a single variable — the distance $AR$ — and then use calculus.

Setting up

Let $AR = x$ metres, so that $RB = AB - AR = 20 - x$.

Because the poles stand vertically on the ground, $\triangle ARP$ has its right angle at $A$ (pole $AP$ perpendicular to $AB$), and $\triangle BRQ$ has its right angle at $B$.

Applying Pythagoras

In $\triangle ARP$:

In $\triangle BRQ$:

Forming the function

Let $S(x) = RP^2 + RQ^2$:

Expand $(20 - x)^2 = 400 - 40x + x^2$:

Collecting the constants $256 + 484 + 400 = 1140$:

Minimising

Differentiate and set to zero: …

Method: Setting Up an Optimization Function With Right-Triangle (Pythagorean) Relationships

Used whenever a moving point creates right triangles with fixed perpendicular segments, and the goal is to optimise a combination of the resulting distances.

Steps

Step 1: Choose a single variable along the fixed segment

Let the distance from one fixed end to the moving point be $x$; the distance to the other end is then (total length $- x$) automatically, since the point lies on that fixed segment.

Step 2: Identify the right angles and apply Pythagoras

Wherever a perpendicular (vertical) segment meets the base, the triangle formed with the moving point has its right angle there, so each slant distance-squared equals (perpendicular length)$^2$ + (base segment)$^2$.

Step 3: Form the objective function and differentiate

Add up the required distances (or their squares, as needed), then differentiate with respect to $x$ and set the derivative to zero to find the critical value.

Step 4: Confirm the extremum type and check feasibility …

Common Mistakes

Mistake 1: Assuming the minimising point should be weighted toward the shorter or taller pole

Why it's wrong: since $AP=16$ m and $BQ=22$ m are different, it's tempting to think $R$ should sit closer to one pole, or that $AB$ should be split in the ratio $AP:BQ$, as in some other geometry problems with unequal segments. But expanding $S(x)=RP^2+RQ^2$ shows the pole heights only contribute fixed constants ($16^2$ and $22^2$) to the sum — they never multiply the variable $x$, so they cannot shift where the minimum occurs. Correct approach: always expand the full algebraic expression before guessing the answer from the picture; let the derivative, not intuition about the figure, decide where the minimum falls.

Mistake 2: Applying Pythagoras with the wrong right-triangle leg …