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UPSC NDA I 2019 Mathematics Solved Paper
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© examsnet.com
Question : 69
Total: 120
A given quantity of metal is to be cast into a half cylinder (i.e., with a rectangular base and semicircular ends). If the total surface area is to be minimum, then the ratio of the height of the half cylinder to the diameter of the semicircular ends is
[NDA I 2019]
π : (π+2)
( π+2) : π
1:1
None of the above
Validate
Solution:
👈: Video Solution
(A)
Let the radius of cylinder be r and height be h, then
Volume of half cylinder V =
1
2
π
r
2
h
…(i)
Surface area of half cylinder
S = 2r.h + πrh + πr
2
⇒S =
4
V
π
r
+
π
.2
V
π
r
+
π
r
2
(from (i) rh =
2
V
π
r
)
⇒
d
s
d
r
=
−
4
V
π
r
2
−
2
V
r
2
+
2 πr = 0
⇒ 2 πr =
4
V
π
r
2
+
2
V
r
2
=
h
2
r
=
π
π
+
2
⇒ 2πr =
4.
π
r
2
h
2
π
r
2
+
2
r
2
.
1
2
π
r
2
h
© examsnet.com
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