Grade Select GradeDifferential Calculus
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.
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.
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.
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.
