Two vertical light poles of height 22 m and 16 m stand on the opposite sides of a 20 m wide road as shown in the figure. Two ladders of length $l_1$ and $l_2$ are placed from a common point R on the road at a distance of $x$ m from the smaller pole. Based on the above information, answer the following questions: (i) Express $p(x) = l_1^2 + l_2^2$ in terms of $x$. (1) (ii) Find $p'(x)$. (1) (iii) (a) Find the value of $x$ for which $l_1^2 + l_2^2$ is minimum. (2) OR (iii) (b) If the 22 m long pole is also replaced by a 16 m long pole, at what distance from either pole should the ladders be kept so that the sum of squares of lengths of ladders needed to reach the top of the pole is minimum? (2)

Two vertical light poles of height 22 m and 16 m stand on the opposite sides of a 20 m wide road as shown in t

Question

l_1

(i) Express

p(x) = l_1^2 + l_2^2

in terms of

x

(1)

(ii) Find

p'(x)

(1)

(iii)

(a) Find the value of

x

for which

l_1^2 + l_2^2

is minimum. (2)

(OR)

(iii)

(b) If the 22 m long pole is also replaced by a 16 m long pole, at what distance from either pole should the ladders be kept so that the sum of squares of lengths of ladders needed to reach the top of the pole is minimum? (2)

CBSECBSE Class XII Board 2026Subjective· 4mImportance★★★★★

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Part (a)Concept understanding — Minimizing Sum of Squares

Minimizing the Sum of Squares

A classic optimisation problem in this chapter asks: split a fixed quantity into parts so that the sum of their squares is as small as possible. For instance, "Find two positive numbers whose sum is $S$ and the sum of whose squares is a minimum." The derivative handles it neatly.

Setting it up

Suppose two numbers add up to a fixed total $S$ . Call them $x$ and $S - x$ . The quantity we want to minimise is the sum of their squares:

f(x) = x^2 + (S - x)^2.

By writing the second number in terms of the first, we have turned a two-variable question into a single-variable function — exactly what makes it a maxima–minima problem.

Applying the derivative

Differentiate and set to zero:

f'(x) = 2x - 2(S - x) = 4x - 2S.

f'(x) = 0 \ \Rightarrow\ x = \frac{S}{2}.

The second derivative is $f''(x) = 4 > 0$ , which is positive everywhere, so this critical point is a genuine minimum. Hence the two numbers are equal, each $\dfrac{S}{2}$ , and the least possible sum of squares is

\left(\frac{S}{2}\right)^2 + \left(\frac{S}{2}\right)^2 = \frac{S^2}{2}.

Note

The result matches intuition: squaring punishes large numbers heavily, so making the parts as balanced as possible — all equal — keeps the total of the squares down. Pushing the split to extremes (one number near $S$ , the other near $0$ ) makes the sum of squares largest, not smallest.

The general recipe

Whenever a problem says "minimise the sum of squares subject to a fixed sum":

Use the constraint to express everything in one variable. …

Part (b)Concept understanding — Minimizing Sum of Squares

Minimizing the Sum of Squares

Setting it up

Suppose two numbers add up to a fixed total $S$ . Call them $x$ and $S - x$ . The quantity we want to minimise is the sum of their squares:

f(x) = x^2 + (S - x)^2.

By writing the second number in terms of the first, we have turned a two-variable question into a single-variable function — exactly what makes it a maxima–minima problem.

Applying the derivative

Differentiate and set to zero:

f'(x) = 2x - 2(S - x) = 4x - 2S.

f'(x) = 0 \ \Rightarrow\ x = \frac{S}{2}.

\left(\frac{S}{2}\right)^2 + \left(\frac{S}{2}\right)^2 = \frac{S^2}{2}.

Note

The general recipe

Whenever a problem says "minimise the sum of squares subject to a fixed sum":

Use the constraint to express everything in one variable. …

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(i) $p(x)=2x^2-40x+1140$ ; (ii) $p'(x)=4x-40$ . (iii)(a) minimum at $x=10$ m. (iii)(b) with two 16 m poles the minimum is again at $x=10$ m (midpoint).

Each ladder is the hypotenuse of a right triangle whose legs are the pole height and the horizontal distance from R to that pole's base. R is $x$ m from the $16$ m pole and, since the road is $20$ m wide, $(20-x)$ m from the $22$ m pole.

(i) $p(x)=l_1^2+l_2^2$

l_1^2=16^2+x^2=256+x^2,\qquad l_2^2=22^2+(20-x)^2=484+(20-x)^2.

p(x)=256+x^2+484+(400-40x+x^2)=2x^2-40x+1140.

(ii) $p'(x)$

p'(x)=4x-40.

Part (a) — Minimising $p(x)$ …

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(i) p(x)=l12+l22p(x)=l_1^2+l_2^2p(x)=l12​+l22​

(ii) p′(x)p'(x)p′(x)

Part (a) — Minimising p(x)p(x)p(x) …

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(i) $p(x)=l_1^2+l_2^2$

(ii) $p'(x)$

Part (a) — Minimising $p(x)$ …

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