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OpenStudy (anonymous):

If the total surface area of a sphere and a cube is 1 square meter, what dimensions should the two solids be to make the total volume of them a maximum?

OpenStudy (anonymous):

Intuitively the largest volume is reached when all the surface area is given to the sphere

OpenStudy (anonymous):

what math class is this problem from?

OpenStudy (anonymous):

calculus 1

OpenStudy (anonymous):

find a formula for the sum of the surface areas of the sphere and cube

OpenStudy (anonymous):

Ac=6x², As=4*pi*r²

OpenStudy (anonymous):

you are trying to maximize volume. you are given Ac+As=1

OpenStudy (anonymous):

so maybe I can show by cases that the volume is optimized by giving all the area to the sphere?

OpenStudy (anonymous):

find the side of the cube as a function of the radius of the sphere or the radius of the sphere as a function of the side of the cube intuition says the volume is optimized by giving all the area to the sphere, but that probably won't be enough for your teacher

OpenStudy (anonymous):

so for a sphere that "fills" a cube 2r=x

OpenStudy (anonymous):

for a cube with 1m² surface area x=(1/6)^(3/2). For a sphere with 1m² surface area r=(1/3)*(sqrt(1/(4*pi)))

OpenStudy (anonymous):

from Ac+As=1 As=1-Ac then solve for r

OpenStudy (anonymous):

then plug what you found is equal to r into the equation for volume where you see an r. take the derivative of what you get, and see where it is zero.

OpenStudy (anonymous):

the equation of the volume of Vc+Vs?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[x³+\frac{ 4 }{ 3 }*\pi*\frac{ 1-6x² }{ 4\pi }*\sqrt{\frac{ 1-6x² }{ 4\pi }}\]

OpenStudy (anonymous):

\[x³+\frac{ 1-6x² }{ 3 }*\sqrt{\frac{ 1-6x² }{ 4*\pi }}\]

OpenStudy (anonymous):

you still need to take the derivative

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