Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #77

Prove that the least perimeter of an isosceles triangle circumscribing a circle of radius r>0 is 6√3r.

✅ Step-by-Step Solution

Final Answer: 6√3r, attained by the equilateral triangle.

Let half the base be b>r, triangle height H and equal side L=√(H²+b²). Since area=rs, bH=r(b+L).
Squaring L=bH/r−b and simplifying gives H=2b²r/(b²−r²) and L=b(b²+r²)/(b²−r²).
Thus perimeter P(b)=2(b+L)=4b³/(b²−r²); P′=4b²(b²−3r²)/(b²−r²)².
For b>r, P′<0 below √3r and P′>0 above √3r. Therefore P is globally minimized at b=√3r.
Substitute to get Pmin=4(3√3r³)/(2r²)=6√3r; H=3r and L=2√3r, so all three sides are equal.

[1 mark] Let half the base be b>r, triangle height H and equal side L=√(H²+b²). Since area=rs, bH=r(b+L).
[1 mark] Squaring L=bH/r−b and simplifying gives H=2b²r/(b²−r²) and L=b(b²+r²)/(b²−r²).
[1 mark] Thus perimeter P(b)=2(b+L)=4b³/(b²−r²); P′=4b²(b²−3r²)/(b²−r²)².
[1 mark] For b>r, P′<0 below √3r and P′>0 above √3r. Therefore P is globally minimized at b=√3r.
[1 mark] Substitute to get Pmin=4(3√3r³)/(2r²)=6√3r; H=3r and L=2√3r, so all three sides are equal.

Practise interactive questions, flashcards, visual mind maps, and timed board tests for Applications Of Derivatives.