# If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?I assumed the side to be a and the diameter to be 2r. I took the sum of there surface areas to be k and that of their volumes to be v.So,k = 6a2 + 4πr2 …..........(1)And,v = a3 + (4/3)πr3 …...........(2)Using (1) I replaced the value of a in (2) and then got v in terms of r. Then I differentiated it and equated v’ to 0. I got a value for r. With this value of r I am able to get the required result. The problem is that I am unable to prove that the value of r I get, is a point of local minima. I know since there is only one available point, it must be that of local minima since that is what the question requires. But can I not prove it somehow? I get a really complex expression for v’. Thus, I’m unable to apply 1st derivative test and the results the same when I apply the 2nd derivative test. Thanks for any help.

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## 1 Answers

Venkat
273 Points
5 years ago
I am using x for a
first equation after diff w.r.t r  gives → dx/dr = -2pi*r/3x
second after diff and simplification gives →dv/dr =  -2pi(xr-2r2)
again diff and simplifying d2v/dr2 = – 2pi(x-2pi*r2/3x-4r)

now dv/dr = 0 → -pi*r(x-2r) = 0 → x = 2r
at x = 2r , d2v/dr2 = 2pi(2r+pi*r/3) > 0
→ v is minium when x = 2r , that is x/2r = 1

Conclusion: when the ratio of an edge of the cube to the diameter of the sphere is 1:1, the volume is minimum...

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