Optimization of the sum of the surfaces of a sphere and cube

[leprofece]

If the sum of the surfaces of a cube and a sphere as is constant, deierminar the minion of the diameter of the sphere to the edge of the cube in cases in which:
272) The sum of the volumes is minimal
273) The sum of the volumes is maximum
And the answer are 272 = 1 and 273 = infinit

Ok
Vs = 4pir³/3 and Vc = l³
since diameter = R/2
V= 4piL/8³/3
THen
V= PiL³/6
I sum and Got V= PiL³/6 +L³

surfaces
cube = L² and Sphere Pi(L/2)²

When derive I got 0 and i can not get the answers

 

[soroban]

Re: Max and min 5

Hello, leprofece!

If the sum of the surfaces of a cube and a sphere is constant,
determine the minion of the diameter of the sphere
to the edge of the cube
in cases in which:

(272) The sum of the volumes is minimal

(273) The sum of the volumes is maximum

And the answer are: (272) 1 and (273) infinity.
What does this mean?

[tex]\begin{array}{c|c|c|} & \text{Area} & \text{Volume} \\ \hline \text{Sphere} & 4\pi r^2 & \tfrac{4}{3}\pi r^3 \\ \hline \text{Cube} & 6x^2 & x^3 \\ \hline\end{array}[/tex]

We have: .[tex]4\pi r^2 + 6x^2 \:=\:S\;\;[1][/tex]

. . . and: .[tex]V \;=\;\tfrac{4}{3}\pi r^3 + x^3 \;\;[2][/tex]From [1]: .[tex]x \;=\;\sqrt{\frac{S-4\pi r^2}{6}}[/tex]

Substitute into [2]: .[tex]V \;=\;\tfrac{4}{3}\pi r^3 + \left(\frac{S-4\pi r^2}{6}\right)^{\frac{3}{2}} [/tex]

You know the rest, don't you?

Set [tex]V' = 0[/tex] and solve for [tex]r[/tex].
Then [tex]d = 2r[/tex].
Then solve for [tex]x.[/tex]

 

[leprofece]

Re: Max and min 5

Hello, leprofece!


[tex]\begin{array}{c|c|c|} & \text{Area} & \text{Volume} \\ \hline \text{Sphere} & 4\pi r^2 & \tfrac{4}{3}\pi r^3 \\ \hline \text{Cube} & 6x^2 & x^3 \\ \hline\end{array}[/tex]

We have: .[tex]4\pi r^2 + 6x^2 \:=\:S\;\;[1][/tex]

. . . and: .[tex]V \;=\;\tfrac{4}{3}\pi r^3 + x^3 \;\;[2][/tex]From [1]: .[tex]x \;=\;\sqrt{\frac{S-4\pi r^2}{6}}[/tex]

Substitute into [2]: .[tex]V \;=\;\tfrac{4}{3}\pi r^3 + \left(\frac{S-4\pi r^2}{6}\right)^{\frac{3}{2}} [/tex]

You know the rest, don't you?

Set [tex]V' = 0[/tex] and solve for [tex]r[/tex].
Then [tex]d = 2r[/tex].
Then solve for [tex]x.[/tex]

HELO BUT I GOT sqrt of 6

 

[leprofece]

Re: Max and min 5

May you Check to see if the book or me is wrong?

 

[MarkFL]

Re: Max and min 5

May you Check to see if the book or me is wrong?

Posting your work will make it easier for our helpers to see if you are correct or not, otherwise we have to work the problem.

 

[leprofece]

Re: Max and min 5

Substitute into [2]: .V=4/3πr³+(S−4πr<sup>2[/SUP/6])3/2
Ok I derive respect to R

4pir2+2pir[S−4πr2]1/2 = 0

2r = S−4πr2]1/2

4r2= S−4πr2

and r =1/2 [sqrt[s/[1+pi]]
and Diameter =sqrt[s/[1+pi]

And now I solve for x
substituting r on x
I got sqrt[s/6[1+pi]]

solving d/x¿area? I got sqrt of 6
the answer sail this must be = 1 and minimun debe ser infinity

I really appreciatte your help
I posted IAM WAITING YOUR CHECKING</sup>