Class 12 Applications Of Derivatives · Very Long Answer5 Marksconstrained_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #75

A circle and a square have total perimeter k>0. Prove that their combined area is least when the square side is twice the circle radius.

✅ Step-by-Step Solution

Final Answer: Square side a=2r; Amin=k²/[4(π+4)].

Let square side a>0 and circle radius r>0. The constraint 4a+2πr=k gives r=(k−4a)/(2π), 0<a<k/4.
Their total area is A(a)=a²+(k−4a)²/(4π).
A′=2a−2(k−4a)/π. Setting this to zero gives πa=k−4a=2πr, so a=2r.
A″=2+8/π>0, proving this feasible stationary point is the unique global minimum.
Here a=k/(π+4), r=k/[2(π+4)], giving Amin=k²/[4(π+4)].

[1 mark] Let square side a>0 and circle radius r>0. The constraint 4a+2πr=k gives r=(k−4a)/(2π), 0<a<k/4.
[1 mark] Their total area is A(a)=a²+(k−4a)²/(4π).
[1 mark] A′=2a−2(k−4a)/π. Setting this to zero gives πa=k−4a=2πr, so a=2r.
[1 mark] A″=2+8/π>0, proving this feasible stationary point is the unique global minimum.
[1 mark] Here a=k/(π+4), r=k/[2(π+4)], giving Amin=k²/[4(π+4)].

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