Chapter 6 Application of Derivatives Ex 6.5 Solutions
Question - 21 : - Of all the closed cylindrical cans (right circular), of agiven volume of 100 cubic centimetres, find the dimensions of the can which hasthe minimum surface area?
Answer - 21 : -
Let r and h bethe radius and height of the cylinder respectively.
Then, volume (V) of the cylinderis given by,
Surface area (S) of the cylinderis given by,
∴By second derivative test, the surface area is theminimum when the radius of the cylinder is
Hence, the requireddimensions of the can which has the minimum surface area is given by radius =
and height = 
Question - 22 : - A wire of length 28 m is to be cut into two pieces. One ofthe pieces is to be made into a square and the other into a circle. What shouldbe the length of the two pieces so that the combined area of the square and thecircle is minimum?
Answer - 22 : -
Let a piece of length l becut from the given wire to make a square.
Then, the other piece of wire to be madeinto a circle is of length (28 − l) m.
Now, side of square =
Let r bethe radius of the circle. Then,
The combined areas of thesquare and the circle (A) is given by,
Thus, when
By second derivative test,the area (A) is the minimum when
.
Hence, the combined area isthe minimum when the length of the wire in making the square is
cm while the length of thewire in making the circle is
Question - 23 : - Prove that the volume ofthe largest cone that can be inscribed in a sphere of radius R is 
of the volume of thesphere.
Answer - 23 : -
Let r and h bethe radius and height of the cone respectively inscribed in a sphere ofradius R.
Let V be the volume of thecone.
Then, 
Height of the cone is given by,
h = R +AB
∴ By second derivative test, the volume of the cone isthe maximum when
Question - 24 : - Show that the rightcircular cone of least curved surface and given volume has an altitude equalto 
time the radius of the base.
Answer - 24 : -
Let r and h bethe radius and the height (altitude) of the cone respectively.
Then, the volume (V) of the coneis given as:
The surface area (S) of the coneis given by,
S = πrl (where l isthe slant height)
Thus, it can be easilyverified that when
∴ By second derivative test, the surface area of thecone is the least when
Hence, for a given volume,the right circular cone of the least curved surface has an altitude equalto times the radius ofthe base.
Question - 25 : - Show that the semi-verticalangle of the cone of the maximum volume and of given slant height is
Answer - 25 : -
Let θ be the semi-verticalangle of the cone.
It is clear that 
Let r, h, and l bethe radius, height, and the slant height of the cone respectively.
The slant height of the cone is given as constant.Now, r = l sin θ and h = l cos θ
The volume (V) of the cone isgiven by,
∴By second derivative test, the volume (V) isthe maximum when
Hence, for a given slantheight, the semi-vertical angle of the cone of the maximum volume is
Answer - 26 : -
Let r bethe radius, l bethe slant height and h bethe height of the cone of given surface area, S.
Also, let α be the semi-vertical angle of the cone.
Then S = πrl + πr2
Let V bethe volume of the cone.
Differentiating (2) with respect to r, we get
For maximum or minimum, put 
Differentiating again with respect to r, weget
As S = πrl + πr2
⇒ 4πr2 = πrl + πr2
⇒ 3πr2 = πrl
⇒ l =3r
Now,in ΔCOB
Question - 27 : - The point on the curve x2 =2y which is nearest to the point (0, 5) is
(A) 
(B) 
Answer - 27 : -
The given curve is x2 =2y.
For each value of x, the position of thepoint will be
Let us denote PA2 by Z. Then,
Z = y2 – 8y +25
Differentiating both sides with respectto y, we get
For maxima or minima, wehave
is(A) 0 (B) 1
(C) 3 (D) 
Answer - 28 : - Let
∴By second derivative test, f is theminimum at x = 1 and the minimum value is given by 
Question - 29 : - The maximum value of 
is
(A) 
(B) 
(C) 1 (D) 0
Answer - 29 : - Let
Then, we evaluate the valueof f at critical point
and at the end points of the interval [0, 1]{i.e., at x = 0 and x = 1}. Hence, we can conclude that the maximumvalue of f in the interval [0, 1] is 1.
The correct answer is C.