Bonsoir.
Soit x le diametre du premier cercle en cm.
a/ Df = ]0 ; 10[.
b/ f(x) = π (x/2)² + π [ (10 - x) / 2 ]²
= π [ x²/4 + (10 - x)²/4 ]
= π [ x²/4 + (100 - 20x + x²)/4 ]
= π [ (2x² - 20x + 100) / 4 ]
= π (x²/2 - 5x + 25).
c/ Voir piece jointe.
f est decroissante, puis croissante sur ]0 ; 10[.
d/ π (x²/2 - 5x + 25) = π/2 (x² - 10x) + 25π
= π/2 (x - 5)² - 25π/2 + 25π
= π/2 (x - 5)² + 25π/2.
Extremum : le point S (5 ; 25π/2) ⇒ f(5) = 25π/2.
e/ f(x) - f(5) = π/2 (x - 5)²
⇔ π (x²/2 - 5x + 25) - 25π/2 = π/2 (x - 5)²
⇔ π/2 (x - 5)² + 25π/2 - 25π/2 = π/2 (x - 5)²
⇔ π/2 (x - 5)².
Aire minimale = f(5) = 25π/2 ≈ 39,3 cm².
Bonne nuit !
