Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #71

A right triangle has fixed sum k>0 of its hypotenuse and one leg. Prove its area is greatest when the angle between these two sides is π/3.

✅ Step-by-Step Solution

Final Answer: Angle π/3; chosen leg k/3 and hypotenuse 2k/3.

Let the chosen leg be x and hypotenuse k−x; the other leg is √(k²−2kx), with 0<x<k/2.
Since area A>0, maximize F=A²=x²(k²−2kx)/4.
F′=(2k²x−6kx²)/4=kx(k−3x)/2, giving interior critical point x=k/3.
F′ is positive before and negative after k/3, and the limiting endpoint areas are zero. Thus A is maximized there.
The hypotenuse is 2k/3, so cosθ=(k/3)/(2k/3)=1/2 and θ=π/3.

[1 mark] Let the chosen leg be x and hypotenuse k−x; the other leg is √(k²−2kx), with 0<x<k/2.
[1 mark] Since area A>0, maximize F=A²=x²(k²−2kx)/4.
[1 mark] F′=(2k²x−6kx²)/4=kx(k−3x)/2, giving interior critical point x=k/3.
[1 mark] F′ is positive before and negative after k/3, and the limiting endpoint areas are zero. Thus A is maximized there.
[1 mark] The hypotenuse is 2k/3, so cosθ=(k/3)/(2k/3)=1/2 and θ=π/3.

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