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Chapter 6 · Application of Derivatives

Example 24: Poles of 16 m and 22 m: least RP2+RQ2RP^2 + RQ^2

ExamplesOptimisation · Section 6.4
The question

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 RP2+RQ2\text{RP}^2 + \text{RQ}^2 is minimum.

Exam answer

Let AR=x\text{AR} = x m. Then RB=20−x\text{RB} = 20 - x⁠, where 0≤x≤200 \le x \le 20⁠.

RP2=x2+162RQ2=(20−x)2+222\begin{aligned} \text{RP}^2 &= x^2 + 16^2 \\ \text{RQ}^2 &= (20 - x)^2 + 22^2 \end{aligned} S(x)=RP2+RQ2=2x2−40x+1140\begin{aligned} S(x) &= \text{RP}^2 + \text{RQ}^2 \\ &= 2x^2 - 40x + 1140 \end{aligned} S′(x)=4x−40,S′′(x)=4S'(x) = 4x - 40, \quad S''(x) = 4

S′(x)=0⇒x=10S'(x) = 0 \Rightarrow x = 10⁠, and S′′(10)=4>0S''(10) = 4 > 0⁠.

So SS has a local minimum at x=10x = 10⁠, its only critical point in [0,20][0, 20]⁠; hence SS is least there.

Hence the point R is at a distance of 10 m from A (the mid-point of AB).

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Each pole stands at a right angle to the ground, so RP and RQ are hypotenuses of right triangles. Pythagoras turns RP2+RQ2\text{RP}^2 + \text{RQ}^2 into a quadratic in the one unknown x=ARx = \text{AR}⁠. The heights of the poles only add constants, so they do not change where the minimum is: it is at the midpoint of AB.

Given, to find, and the plan
Given
AP = 16 m and BQ = 22 m, both perpendicular to AB; AB = 20 m; R is a point of the segment AB.
To find
AR when RP2+RQ2\text{RP}^2 + \text{RQ}^2 is least.
Plan
Let AR = xx with 0≤x≤200 \le x \le 20⁠. Use Pythagoras in triangles RAP and RBQ to write S(x)=RP2+RQ2S(x) = \text{RP}^2 + \text{RQ}^2⁠. Find the critical point, apply the second derivative test, and compare with the ends of the range.

Basics you need

1Pythagoras' theorem

In a right triangle, (hypotenuse)2^2 = sum of the squares of the other two sides. Basics: area and volume formulas

2Building the function

One variable (AR = xx⁠), its range, and the quantity to minimise written in terms of it. Basics: building the function to optimise

3Second derivative test (Theorem 4, Section 6.4)

S′(c)=0S'(c) = 0 and S′′(c)>0S''(c) > 0 make cc a point of local minima. Lesson: first and second derivative tests

4Least value on a closed interval (Section 6.4.1)

Working Rule: evaluate the function at the critical points inside the interval and at both end points; the smallest of these values is the absolute minimum. Lesson: absolute extrema on a closed interval

Solution

  1. Choose the variable and its range

    Let AR = xx m. Since AB = 20 m and R lies on AB, RB = (20−x)(20 - x) m and

    0≤x≤20.0 \le x \le 20.
    A B P Q R 16 m 22 m x 20 − x AB = 20 m RP RQ
    Like Fig 6.16, drawn to scale (1 m to 13 units) with R at the answer, x=10x = 10⁠.
    Where R is on AB is the only thing we are free to choose, so AR is the natural variable. The range is closed: R may be at A (x=0x = 0⁠) or at B (x=20x = 20⁠), so these two ends must be checked later.
    A point of a segment of length 20 splits it into parts xx and 20−x20 - x⁠. Basics: interval notation

  2. Pythagoras in the two right triangles

    The poles are vertical and AB is horizontal, so the angles at A and B are right angles:

    RP2=AR2+AP2=x2+162,RQ2=RB2+BQ2=(20−x)2+222.\begin{aligned} \text{RP}^2 &= \text{AR}^2 + \text{AP}^2 = x^2 + 16^2, \\ \text{RQ}^2 &= \text{RB}^2 + \text{BQ}^2 = (20-x)^2 + 22^2. \end{aligned}
    RP joins R on the ground to the top P of the first pole; RA and AP are the two legs of the right triangle RAP. The same holds for RQ in triangle RBQ.
    Pythagoras' theorem (Basic 1).

  3. The function to minimise

    S(x)=x2+256+(20−x)2+484=x2+256+400−40x+x2+484=2x2−40x+1140,0≤x≤20.\begin{aligned} S(x) &= x^2 + 256 + (20 - x)^2 + 484 \\ &= x^2 + 256 + 400 - 40x + x^2 + 484 \\ &= 2x^2 - 40x + 1140, \quad 0 \le x \le 20. \end{aligned}
    Adding the two lines of Step 2 gives exactly the quantity in the question. Collecting terms: 256+400+484=1140256 + 400 + 484 = 1140⁠.
    (20−x)2=400−40x+x2(20 - x)^2 = 400 - 40x + x^2⁠. Basics: building the function to optimise

  4. Critical point

    S′(x)=4x−40,S′(x)=0  ⟺  x=10.S'(x) = 4x - 40, \qquad S'(x) = 0 \iff x = 10.

    x=10x = 10 lies inside [0,20][0, 20]⁠. SS is a polynomial, differentiable everywhere, so there is no other critical point.

    An interior least value can only occur where S′=0S' = 0 (Theorem 2). The answer must therefore be x=10x = 10 or one of the ends.
    Power rule: ddx(2x2−40x+1140)=4x−40\frac{d}{dx}(2x^2 - 40x + 1140) = 4x - 40⁠.

  5. Second derivative test

    S′′(x)=4>0 for all x, so S′′(10)>0.S''(x) = 4 > 0 \text{ for all } x, \text{ so } S''(10) > 0.

    By the second derivative test, x=10x = 10 is a point of local minima of SS⁠.

    This is the book's argument. It shows that S(10)S(10) is smaller than the values of SS at nearby points. To be sure it beats every allowed position of R, including the ends, we compare values in the next step.
    Theorem 4(ii), Section 6.4. Lesson: first and second derivative tests

  6. Compare with the ends: the least value

    xxPosition of RS(x)=2x2−40x+1140S(x) = 2x^2 - 40x + 1140
    0at A11401140
    10midpoint of AB200−400+1140=940200 - 400 + 1140 = 940
    20at B800−800+1140=1140800 - 800 + 1140 = 1140

    The smallest value is 940940 m2^2⁠, at x=10x = 10⁠. So by the Working Rule of Section 6.4.1, SS is least when AR = 10 m. (Also S(x)=2(x−10)2+940≥940S(x) = 2(x - 10)^2 + 940 \ge 940⁠, with equality only at x=10x = 10⁠.)

    On a closed interval the absolute minimum is the smallest among the values at the interior critical points and at the two ends. Here the ends both give 1140, well above 940.
    Working Rule, Section 6.4.1 (Basic 4). Lesson: absolute extrema on a closed interval
Result

R must be at the midpoint of AB: AR = 10 m. Then RP2+RQ2=356+584=940\text{RP}^2 + \text{RQ}^2 = 356 + 584 = 940 m2^2⁠, the least possible value.

Check it with numbers

At x=10x = 10⁠: RP=356=18.8679623\text{RP} = \sqrt{356} = 18.8679623 m and RQ=584=24.1660919\text{RQ} = \sqrt{584} = 24.1660919 m, so RP2+RQ2=940\text{RP}^2 + \text{RQ}^2 = 940⁠. At x=9x = 9⁠: 81+256+121+484=94281 + 256 + 121 + 484 = 942⁠. At x=11x = 11⁠: 121+256+81+484=942121 + 256 + 81 + 484 = 942⁠. Both neighbours are larger.

Common mistakes
  1. Minimising RP + RQ instead. The question asks for the sum of the squares; the sum of lengths is a different problem with a different answer.
  2. Writing RB as x−20x - 20⁠. A length is positive: RB = 20−x20 - x for 0≤x≤200 \le x \le 20⁠.
  3. Ignoring the ends. R may be at A or at B; compare SS there too.
What to take away
  1. Right angles in a figure usually mean Pythagoras builds the function for you.
  2. For x2+(L−x)2+constantsx^2 + (L - x)^2 + \text{constants}⁠, the minimum is always at x=L2x = \frac{L}{2}⁠, whatever the constants.
  3. On a closed range, finish with a table of values at the critical point and the ends.

Every fact used, and where it comes from

FactUsed inSource
Pythagoras' theoremStep 2Geometry; Fig 6.16 (Basics: formulas)
(20−x)2=400−40x+x2(20-x)^2 = 400 - 40x + x^2Step 3Algebra
Extrema only at critical pointsStep 4Theorem 2, Section 6.4
Second Derivative TestStep 5Theorem 4, Section 6.4 (Lesson: tests)
Working Rule for a closed intervalStep 6Section 6.4.1 (Lesson: absolute extrema)
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