Question
Question: A given quantity of metal sheet is to be cast into a solid half circular cylinder with rectangular b...
A given quantity of metal sheet is to be cast into a solid half circular cylinder with rectangular base and semicircular ends. Show that in order that the total surface area may be minimum, the ratio of length of cylinder of diameter of its circular ends is π:π+2
Solution
To check if surface area is minimum or not, you first need to find critical points. And we check if the point is positive or negative.
Complete Step by Step solution:
It is given that a metal sheet is casted into a solid half circular cylinder.
That means, the area of the sheet is equal to the volume of the cylinder.
Let the area of the sheet be A (constant), as it is given and the volume of cylinders is V.
Then, V=21×Volume of cylinder.
=21×πr2h
Where, r is the radius of the circular part of the cylinder and his the height of the cylinder.
Total surface area of cylinder will be the sum of 21×curved surface area of cylinder, area of two semicircles, and the area of the rectangular base . . . . . (1)
Curved surface area of cylinder=2πrh . . . . . . (2)
Area of semicircle=2πr2 . . . . . . (3)
Area of rectangle=2r×h . . . . . . . (4)
Since in this case the height of the cylinder is the length of the rectangular base and the diameter of the semicircle is the breadth of the rectangular base.
Putting the values from equations (2), (3) and (4) into equation (1) we get
Total surface area of the cylinder,
S=21×2πrh+2×2πr2+2r×h
S=πrh+πr2+2rh . . . . . . (5)
Since, area of sheet, B equal to the volume of cylinder, we have 21×πr2h=A
h=πr22A . . . . . (6)
Put this value in (5), to get
S=πr×πr22A+πr2+2r×πr22A
=r2A+πr2+πr4A . . . . . . (7)
For the surface area to be minimum or maximumdrds=0.
By differentiating equations (7) we get
drds=2Ar2(−1)+2πr+πr24A(−1)
drds=r22A+2πr−πr24A (∵dxdxn=nxn−1)
∴drds=0
r2−2A+2πr−πr24A=0
Multiplying both the sides by πr2,we get
⇒−2πA+2π2r3−4A=0
⇒2π2r3=4A+2πA
=(2+π)2A
⇒r3=2π2(π+2)2A . . . . . (8)
Now, we need to check if the total surface area will be minimum for this value of r.
For that differentiatedrds
dr2d2s=r34A+2π+πr38A
(dr2d2s)r3=π2(π+2)A=(π+2)4Aπ2+2π+π(π+2)A8Aπ2
=π+24π2+2π+π+28π
Clearly, all the values are positive.
∴(dr2d2s)r>0.
⇒S is minimum.
∵for minima at point a, (dr2d2s)a>0
Now, we need to find the ratio of length of the cylinder to the diameter of its circular ends i.e.2rh.//
⇒2rh=2r1×πr22A
=πr3A
=π×π(π+2)AA
⇒2rh=(π+2)π
Hence, it is proved that the ratio of the length of the cylinder to the diameter of its circular ends isπ:(π+2)
Note: Equation of the question will most probably be in two variables. You have to make sure to find a relation between the variables. So that you could simplify equations in a single variable. Just like in this example, we found the relation between hand r, and converted the equation of S in terms of
r.