Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #108

Prove that a rectangle of greatest perimeter inscribed in a circle of radius r>0 is a square of side r√2.

✅ Step-by-Step Solution

Final Answer: Square of side r√2; Pmax=4r√2.

Let its half-sides be x,y≥0. The diagonal is a circle diameter, so x²+y²=r².
The perimeter is P=4(x+y)=4[x+√(r²−x²)], 0≤x≤r.
For 0<x<r, P′=4[1−x/√(r²−x²)]=0 gives x=r/√2.
P′ is positive before and negative after this point, and endpoint perimeters 4r are smaller. Thus y=r/√2 and this is the global maximum.
The full sides are 2x=2y=r√2, so the rectangle is a square and Pmax=4r√2.

[1 mark] Let its half-sides be x,y≥0. The diagonal is a circle diameter, so x²+y²=r².
[1 mark] The perimeter is P=4(x+y)=4[x+√(r²−x²)], 0≤x≤r.
[1 mark] For 0<x<r, P′=4[1−x/√(r²−x²)]=0 gives x=r/√2.
[1 mark] P′ is positive before and negative after this point, and endpoint perimeters 4r are smaller. Thus y=r/√2 and this is the global maximum.
[1 mark] The full sides are 2x=2y=r√2, so the rectangle is a square and Pmax=4r√2.

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