CBSE Class 12 Mathematics 2026 Main Paper 65/3/2 · Q384 Marksgeometry_optimization
CBSE Class 12 Mathematics 2026 Main Paper 65/3/2 · Question 38 (Applications Of Derivatives)
Two vertical poles, 22 m and 16 m high, stand on opposite sides of a 20 m wide road. Two ladders of lengths l₁ and l₂ extend from a point R on the road to the tops of the taller and shorter poles respectively. R is x m from the shorter pole.
Part (1) 1 Marks
Express p(x)=l₁+l₂ in terms of x.
Answer: √((20−x)²+484)+√(x²+256), for 0≤x≤20.
- The horizontal distances to the poles are 20−x and x.
- Use the Pythagorean theorem on each ladder and add the resulting lengths.
Part (2) 1 Marks
Find p′(x).
Answer: (x−20)/√((20−x)²+484)+x/√(x²+256).
- Differentiate each square-root term by the chain rule.
- Add the derivatives to get (x−20)/√((20−x)²+484)+x/√(x²+256).
Part (3) 2 Marks
At what value of x is l₁²+l₂² minimized?
Answer: x=10 m.
- The sum is (20−x)²+484+x²+256.
- Complete the square to write it as 2(x−10)²+940.
- Since 10 lies on the road, the minimum occurs at x=10 m.
Part (4) 2 Marks
If the taller pole is replaced with a second 16 m pole, how far from each pole should R be to minimize the sum of squared ladder lengths?
Answer: 10 m from each pole.
- The squared lengths now sum to (20−x)²+x²+512.
- Complete the square to obtain 2(x−10)²+712.
- The minimum occurs at x=10, so R is 10 m from each pole.