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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 = 6a 2 + 4 π r 2 …..........(1) And, v = a 3 + (4/3) π r 3 …...........(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 1 st derivative test and the results the same when I apply the 2 nd derivative test. Thanks for any help.

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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