A cylindrical hole of radius xis bored through a shere of radius R in such a way that the axis of th

Ryker Stein

Ryker Stein

Answered question

2022-05-26

A cylindrical hole of radius xis bored through a shere of radius R in such a way that the axis of the hole passes throiugh the centerof the sphere. Find the value of x that maximizes the complete surface area of the remaining solid. Hint: The area of a segment of height h on a shere of radius R is 2 π R h

Answer & Explanation

Leonard Mahoney

Leonard Mahoney

Beginner2022-05-27Added 10 answers

Step 1
Please use the given hint. The surface area of a segment of height h on a sphere of radius R is 2 π R h . Here h = R R 2 x 2 . There are two such segments removed by drilling the hole.
So surface area removed from the sphere:
4 π R 2 4 π R R 2 x 2
Surface area of the sphere remaining:
4 π R R 2 x 2
Surface area of the cylindrical surface through the sphere:
2 π x 2 R 2 x 2
Total surface S = 4 π R R 2 x 2 + 4 π x R 2 x 2
S = 4 π ( R + x ) R 2 x 2
d S d x = 4 π ( R 2 R x 2 x 2 ) R 2 x 2
Solving R 2 R x 2 x 2 = 0 , we find that S is maximized when x = R / 2

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