NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Miscellaneous Exercise Question 9

✅ Step-by-Step Solution

1. Put the legs on the positive axes and the point P=(a,b). If the hypotenuse makes angle θ with the horizontal, its two parts have lengths a sec θ and b cosec θ, so L=a sec θ+b cosec θ, 0<θ<π/2.
2. L′=a sec θ tan θ−b cosec θ cot θ. The stationary equation is a sin³θ=b cos³θ, i.e. tan θ=(b/a)^(1/3).
3. L′ has sign a tan³θ−b, so it changes from negative to positive exactly once. Also L tends to infinity at each endpoint.
4. At this angle cos θ=a^(1/3)/√(a^(2/3)+b^(2/3)), sin θ=b^(1/3)/√(a^(2/3)+b^(2/3)); substitution gives the stated minimum.