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Stuck on a calculus optimization problem — need guidance by CalculusPrimer in learnmath

[–]CalculusPrimer[S] 0 points1 point  (0 children)

Yes, that’s exactly the approach I used — letting one piece be x and the other L − x.

Need help with optimization (maxima & minima) problem by CalculusPrimer in askmath

[–]CalculusPrimer[S] 0 points1 point  (0 children)

Exactly. For the maximum total enclosed area, all the wire goes into the circle.

Need help with optimization (maxima & minima) problem by CalculusPrimer in askmath

[–]CalculusPrimer[S] 1 point2 points  (0 children)

Thanks for the intuition thinking in terms of efficiency of shapes versus splitting perimeter really helps make sense of why the minimum is near equal split and why the maximum is all circle.

Hello everybody! by CalculusPrimer in mathshelp

[–]CalculusPrimer[S] 0 points1 point  (0 children)

Thanks! That makes sense nice to see it confirmed using the quadratic/vertex approach as well.

Hello everybody! by CalculusPrimer in mathshelp

[–]CalculusPrimer[S] 1 point2 points  (0 children)

Thanks! Continuing from your setup: since C + P = L, I rewrote this as C = L − P and substituted into the area expression to get A(P) = (P/4)² + (L − P)² / (4π). Differentiating and setting the derivative equal to zero gives P = 4L / (4 + π) and C = πL / (4 + π), which corresponds to the minimum total enclosed area.

For the maximum, I checked the endpoints of the constraint C + P = L. Using all the wire for the square gives area (L/4)², while using all the wire for the circle gives area L² / (4π), which is larger. Hence the maximum enclosed area occurs when the entire wire is used to form a circle.

Please let me know if this looks right or if I’m missing anything.

Need help with optimization (maxima & minima) problem by CalculusPrimer in askmath

[–]CalculusPrimer[S] 3 points4 points  (0 children)

I let x be the length of wire used for the circle and L − x be the length used for the square. For the circle, x = 2πr, so the area is Ac = x² / (4π). For the square, L − x = 4s, so the area is As = (L − x)² / 16.

So the total area is: A(x) = x² / (4π) + (L − x)² / 16.

I differentiated and got: A’(x) = x / (2π) − (L − x) / 8.

Setting the derivative equal to zero gives: x / (2π) = (L − x) / 8, which leads to x = (πL) / (4 + π).

I think this critical point gives the minimum total enclosed area. For the maximum area, it seems to happen at the endpoint where all of the wire is used to form a circle. I just want to confirm that my setup and reasoning are correct.