Applied problem using absolute extrema

[John O' Meara]

Hi, I have a problem solving this question, I don't know if I am on the correct track for solving this. Suppose that the sum of the surfaces of a sphere and a cube is constant. Show that the sum of their volumes is smallest when the diameter of the sphere is equal to the length of an edge of the cube.
Let x = length of an edge of the cube. r= radius of the sphere and V = total volume. S= total surface area. Then, V=\frac{4}{3} \pi r^3 + x^3
S=4\pi r^2 + 6x^2 => x^2=\frac{S-4\pi r^2}{6} =>
x^6=(\frac{S}{6}-\frac{2}{3} \pi r^2)^3
=> x^3=\sqrt{(\frac{S}{6}-\frac{2}{3} \pi r^2)^3}. It follows from this that V=\frac{4}{3} \pi r^3 + \sqrt{(\frac{S}{6}-\frac{2}{3} \pi r^2)^3}. This looks like an unyieldy equation. I have yet to differentiate V w.r.t. r, and put V'(r)=0. Do I expand it using the Binomial Theorem or am I totally wrong in my approach to the question? Thanks for the help.

 

[Eynstone]

Use implicit differentiation & don't bother to find x in terms of r.

 

[John O' Meara]

Ok, I used implicit differentiation to get \frac{dV}{dr}=4 \pi r^2 + 3x^2 \frac{dx}{dr} = 0 I cannot see anything in the question that tells me the value of dx/dr. So can someone give a hint as to where to go from here. Thanks very much.

 

[Redbelly98]

I cannot see anything in the question that tells me the value of dx/dr. So can someone give a hint as to where to go from here. Thanks very much.

Hint:

S=4\pi r^2 + 6x^2 => x^2=\frac{S-4\pi r^2}{6}