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marysya
3 months ago
12

Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere.

Mathematics
1 answer:
Inessa [12.5K]3 months ago
3 0

Answer:

Wyjaśnienie krok po kroku:

Załóżmy, że prostokąt ma wierzchołki (x,y,z) w dodatnim oktancie.  Prostokątny pudełko musi być symetryczne względem wszystkich trzech osi.

Wtedy jego boki to

2x,2y,2z

Objętość = 8xyz

Maksymalizuj objętość przy założeniu

x^2+y^2+z^2 =1

tj. g(x,y,z) = x^2+y^2+z^2 -1=0

Użyj mnożników Lagrange'a, mamy

∇f(x,y,z)=λ∇g(x,y,z)przy maksimum

∇f(x,y,z)=λ∇g(x,y,z) =(8yz,8xz,8xy)\\∇g(x,y,z)=(2x,2y,2z)

8yz=2λx8xz=2λy8xy=2λz

Dzieląc otrzymujemy

\frac{y}{x} =\frac{x}{y} \\x^2=y^2

Podobnie y^2=z^2

Zatem otrzymujemy 3x^2 =1\\x = \frac{1}{\sqrt{3} }

Stąd wymiary to

(2x,2y,2z)

<pzatem wymiary="" to="">

\frac{2}{\sqrt{3} },\frac{2}{\sqrt{3} },\frac{2}{\sqrt{3} } )

</pzatem>
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