12th NCERT Application of Derivatives Exercise 6.5 Questions 13 to 29
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Question (13)

Find two numbers whose sum is 24 and whose product is as large as possible

Solution

Let two number be x and y from the information x + y = 24
⇒ ∴ y = 24 -x
product required max
so function is product
f(x) = xy
f(x) = x(24-x)
f(x) = 24x - x2
f'(x) = 24 - 2x
f"(x) = -2
Now f'(x) = 0
24 - 2x = 0
⇒ x = 12
f"(12) = -2 < 0
So f is maximum at x = 12
∴ y = 24 - 12 = 12
so the number are 12, 12

Question (14)

Find two poistive numbers x and y such that x + y = 60 and xy3 is maximum.

Solution

Here we can find y= 60- x as well but in function it is xy3
If we calculate y then we has to make cube of it or use product rule for derivative, which will become lengthy.
So we have calculate x = 60 - y and function will be in terms of y instead of x
x + y = 60
x = 60 -y
f = x y3
f = ( 60 - y) y3
f(y) = 60 y3 - y4
f'(y) = 180y2 - 4y3
f"(y) = 360y - 12y2
Now f'(y) = 0
180y2 - 4y3 = 0
4y2 ( 45 - y) = 0
y2 = 0 OR y - 45 = 0
y = 0 OR y = 45
f"(0) = 0 ⇒ No maxima or minima
f"(45) = 360(45) - 12(45)2
f"(45) = 12(45) (30 -45) = 12(45) (-15) < 0
⇒ f is maximum at 45
∴ y = 45
∴ x = 60 - y = 60 - 45 = 15
So the numbers are 15 and 45

Question (15)

Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum

Solution

Let number be x and y
∴ x + y = 35
x = 35 - y
f: = x2 y5
f = (35 - y)2 y5
$f'\left( y \right) = {\left( {35 - y} \right)^2}\frac{d}{{dy}}{y^5} + {y^5}\frac{d}{{dy}}{\left( {35 - y} \right)^2}$ $= {\left( {35 - y} \right)^2}5{y^4} + {y^5}\left( 2 \right)\left( {35 - y} \right)\left( { - 1} \right)$ $= 5{y^4}{\left( {35 - y} \right)^2} - 2{y^5}\left( {35 - y} \right)$ $= {y^4}\left( {35 - y} \right)\left[ {5\left( {35 - y} \right) - 2y} \right]$ $= {y^4}\left( {35 - y} \right)\left[ {175 - 5y - 2y} \right]$ $= {y^4}\left( {35 - y} \right)\left[ {175 - 7y} \right]$ $= 7{y^4}\left( {35 - y} \right)\left( {25 - y} \right)$ $f"\left( y \right) = 7\left[ {{y^4}\left( {35 - y} \right)\frac{d}{{dy}}\left( {25 - y} \right) + {y^4}\left( {25 - y} \right)\frac{d}{{dy}}\left( {35 - y} \right) + \left( {35 - y} \right)\left( {25 - y} \right)\frac{d}{{dy}}{y^4}} \right]$ $f"\left( y \right) = 7\left[ {{y^4}\left( {35 - y} \right)\left( { - 1} \right) + {y^4}\left( {25 - y} \right)\left( { - 1} \right) + 4{y^3}\left( {35 - y} \right)\left( {25 - y} \right)} \right]$ $f"\left( y \right) = - 7{y^3}\left[ {y\left( {35 - y} \right) + y\left( {25 - y} \right) - 4\left( {35 - y} \right)\left( {25 - y} \right)} \right]$ Now f'(y) = 0
7y4 ( 35 - y) ( 25 - y) = 0
y4 = 0, 35-y = 0, 25 - y = 0
y = 0 OR y = 35 OR y = 25
f"(0) = 0
f"(35) = -7(35)3 [ y(0) +35(-10) - 4(0)]
f"(35) = -7(35)3(-10) > 0
⇒ f is minimum at y = 35
$f"\left( {25} \right) = - 7{\left( {25} \right)^3}\left[ {25\left( {10} \right) + 25\left( 0 \right) - 4\left( 0 \right)} \right]$ $f"\left( {25} \right) = - 7{\left( {25} \right)^3} \times 250$ $f"\left( {25} \right) < 0$ ⇒ f(x) is maximum at y=25
∴ y =25
x = 35 - y = 35 - 25 = 10
∴ number are 10 and 25

Question (16)

Find two positive number whose sum is 16 and the sum of whose cubes is minimum

Solution

x + y = 16
y = 16 - y
f: sum of its cube
f: x3 + y3
f(x) = x3 + (16 - x)3
Now f'(x) = 3x2 + 3(16-x)2(-1)
f'(x) = 3x2 - 3(16-x)2
f"(x) = 6x - 6( 16 - x) (-1)
f"(x) = 6x + 6(16-x)
Now f'(x) = 0
⇒ 3x2 - 3(16 - x)2 = 0
⇒ 3[ x2 - (16 - x)2] = 0
⇒ 3[ (x+16-x)( x -16 +x)] = 0
48(2x - 16) = 0 2x - 16 = 0
x = 8
f"(8) = 6(8) + 6(16-8)
f"(8) = 48 +48 = 96 > 0
∴ f is minimum at x = 8
y = 16 - 8 = 8
so the numbers ares 8 and 8

Question (17)

A square piece of tin of side 18cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box iis the maximum possible

Solution

Let square of side x is cut from each corner to prepare a box
as square of x each cut from both corners
length of box = 18 -x - x = 18-2x
length of box = 18 -2x
breadth of box = 18 -2x
height of box = x
Volume of box = lbh
Volume of box = (18-2x)(18-2x) (x)
Volume of box = 2(9-x) 2( 9 -x) x
Volume of box = 4x (9-x)2
so f(x) = 4x (9-x)2
$f'\left( x \right) = 4\left[ {x\frac{d}{{dx}}{{\left( {9 - x} \right)}^2} + {{\left( {9 - x} \right)}^2}\frac{d}{{dx}}x} \right]$ $f'\left( x \right) = 4\left[ {x\left( 2 \right)\left( {9 - x} \right)\left( { - 1} \right) + {{\left( {9 - x} \right)}^2} \cdot 1} \right]$ $f'\left( x \right) = 4\left[ { - 2x\left( {9 - x} \right) + {{\left( {9 - x} \right)}^2}} \right]$ $f'\left( x \right) = 4\left[ {\left( {9 - x} \right)\left( { - 2x + 9 - x} \right)} \right]$ $f'\left( x \right) = 4\left[ {\left( {9 - x} \right)\left( {9 - 3x} \right)} \right]$ $f'\left( x \right) = 12\left( {9 - x} \right)\left( {3 - x} \right)$ $f"\left( x \right) = 12\left[ {\left( {9 - x} \right)\frac{d}{{dx}}\left( {3 - x} \right) + \left( {3 - x} \right)\frac{d}{{dx}}\left( {9 - x} \right)} \right]$ $f"\left( x \right) = 12\left[ {\left( {9 - x} \right)\left( { - 1} \right) + \left( {3 - x} \right)\left( { - 1} \right)} \right]$ $f"\left( x \right) = - 12\left( {9 - x + 3 - x} \right)$ $f"\left( x \right) = - 12\left( {12 - 2x} \right)$ $f"\left( x \right) = - 24\left( {6 - x} \right)$ Now f'(x) = 0
12(9-x)(3-x) 0
9-x = 0 OR 3 -x = 0
x = 9 OR x = 3
f"(9) = -24( 6 -9) = -24(-3) = 72 > 0
⇒ f is minimum at x = 9
f"(3) = -24( 6 -3) = -24(3) = -72 < 0
⇒ f is maximum at x = 3
So square of 3cm is to be cut from each corner of square sheet to get box of more volume

Question (18)

A rectangular sheet of tin 45cm by 24cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Solution

Let square of side x to be cut from each corner of rectangular sheet to prepare box Length of box = 45-2x
breadth of box = 24 -2x
height = x
volume of box = lbh
volume = (45 - 2x ) (24 - 2x) x
volume = (45 - 2x ) (12 - x) 2x
f(x) = 2x (45 - 2x ) (12 - x)
$f'\left( x \right) = 2\left[ {x\left( {45 - 2x} \right)\frac{d}{{dx}}\left( {12 - x} \right) + x\left( {12 - x} \right)\frac{d}{{dx}}\left( {45 - 2x} \right) + \left( {45 - 2x} \right)\left( {12 - x} \right)\frac{d}{{dx}}x} \right]$ $f'\left( x \right) = 2\left[ {x\left( {45 - 2x} \right)\left( { - 1} \right) + x\left( {12 - x} \right)\left( { - 2} \right) + \left( {45 - 2x} \right)\left( {12 - x} \right)} \right]$ $f'\left( x \right) = 2\left[ { - x\left( {45 - 2x} \right) - 2x\left( {12 - x} \right) + \left( {45 - 2x} \right)\left( {12 - x} \right)} \right]$ $f'\left( x \right) = 2\left[ { - 45x + 2{x^2} - 24x + 2{x^2} + 540 - 45x - 24x + 2{x^2}} \right]$ $f'\left( x \right) = 2\left[ {6{x^2} - 138x + 540} \right]$ $f'\left( x \right) = 2 \times 6\left[ {{x^2} - 23x + 90} \right]$ $f'\left( x \right) = 12\left[ {{x^2} - 23x + 90} \right]$ $f"\left( x \right) = 12\left[ {2x - 23} \right]$ Now f'(x) = 0
⇒ 12(x2 - 23x + 90) = 0
(x-5) (x-18) = 0
x - 5 = 0 OR x - 18 = 0
x = 5 OR x = 18
f"(18) = 12(36-13) = 12(13) > 0
⇒ f is minimum at x = 18
$f"\left( 5 \right) = 12\left( {10 - 23} \right) = 12\left( { - 13} \right) < 0$ ⇒ f is maximum at x =5
∴ x =5 ∴ length of square = 5 cm
So square of sides 5cm is to be cut from each corner of rectangular sheet of 45 × 24 cm to prepare box having maximum volume

Question (19)

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution

Given radius of circle be R
f: area is to be maximum
Let rectangle ABCD is inscribed in circle of radius 'R', with centre 'O'
as ∠B = 90° , AC is diameter of circle
As $\overset{ \huge\frown}{AC}$ is semicircular arc
AC = 2R
Let length of rectangle be x and breadth be y
AB = x and B = y
Now in ΔABC
AB2 + BC2 = AC2 (by Pythagoras' theorem)
x2 + y2 = (2R)2
y2 = 4R2 - x2
Now area = lb = xy
As area of rectangle is maximum, the square of area will also be maximum
We have taken square of area to avoid the square root, otherwise y=√(4R2-x2) then we has to take derivative of square root
A = x y
A2 = x2 y2
A2 = x2 ( 4R2 - x2)
f = 4R2 x2 - x4
f(x) = 4R2 x2 - x4
f'(x) = 8R2x - 4x3
f"(x) = 8R2 - 12x2
Now f'(x) = 0
8R2x - 4x3 = 0
4x(2R2 - x2) = 0
x = 0, x2 = 2R2 ⇒ x = √2 R
Now f"(0) = 8R2 - 0 = 8R2 > 0
⇒ f is minimum at x = 0
f"(R√2) = 8R2 - 12(√2 R)2
f"(R√2) = 8R2 - 24 R2 = -16R2 < 0
⇒ f is maximum at x = √2 R
x = √2 R
y2 = 4R2 - x2 = 4R2 - 2R2 = 2R2
y = √2 R
so rectangle length x = breadth y , so it is square

Question (20)

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base

Solution

the function is volume which we required max.
volume of cylinder = π r2 h
which is in two variables we have to bring the one variable either in 'r' or in 'h'
we get value of 'r' or 'h' we have to use given that total surface area is constant
In a given sum, surface are of cylinder is given, it means surface area of cylinder is constant.
Let radius of cylinder be 'r' and height be 'h'
Here surface area it mean
Total surface area is constant 'k'
∴ TSA = k
2πr h + 2π r2 = k
2πrh = k - 2πr2
$h = \frac{{k - 2\pi {r^2}}}{{2\pi r}}$ $\text{Now} \quad V = \pi {r^2}h$ $V = \require{cancel}\cancel{\pi} {r^\cancel{2}}\left( {\frac{{k - 2\pi {r^2}}}{{2\cancel{\pi} \cancel{ r}}}} \right)$ $V = r\left( {\frac{{k - 2\pi {r^2}}}{2}} \right)$ $f\left( r \right) = \frac{{kr}}{2} - \pi {r^3}$ $f'\left( r \right) = \frac{k}{2} - 3\pi {r^2}$ $f"\left( r \right) = - 6\pi r$ Now f'(r) = 0 $\therefore \frac{k}{2} - 3\pi {r^2} = 0 \quad \text{k is constant}$ $k - 6\pi {r^2} = 0$ ∴ Now replace the value of 'k'
2πrh + 2π r2 - 6π r2 = 0
2π r h - 4 π r2 0
2πr ( h - 2r) = 0
r = 0 or r = h/2
Now f"(0) = -6π(0) = 0
so no maxi or minimum at r = 0
$f"\left( {\frac{h}{2}} \right) = - 6\pi \left( {\frac{h}{2}} \right) = - 3\pi h < 0 \quad\text{ as}\quad h > 0$ ⇒ so f is maximum at r= h/2
∴ so volume of cylinder of given surface area is maximum at r = h/2
⇒ h = 2r ( 2r = diameter d)
⇒ h = d
∴ so volume is maximum when height is equal to diameter of cylinder of given surface area

Question (21)

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area?

Solution

Closed cylinder tank , volume (V) = 100
f: surface area min
Let radius of tank be r and height be 'h'
Now volume is given 100
V = 100
π r2 h = 100
$h = \frac{{100}}{{\pi {r^2}}}$ The function is : surface area
f = TSA of cylinder
f = 2πrh + 2 πr2
$f = 2\cancel{\pi} \cancel{r}\left( {\frac{{100}}{{\cancel{\pi} {r^\cancel{2}}}}} \right) + 2\pi {r^2}$ $f\left( r \right) = \frac{{200}}{r} + 2\pi {r^2}$ $f'\left( r \right) = \frac{{ - 200}}{{{r^2}}} + 4\pi r$ $f"\left( r \right) = \frac{{400}}{{{r^3}}} + 4\pi$ Now f'(r) = 0
$\Rightarrow \frac{{ - 200}}{{{r^2}}} + 4\pi r = 0$ $\Rightarrow - 200 + 4\pi {r^3} = 0$ $4\pi {r^3} = 200$ ${r^3} = \frac{{50}}{\pi }$ $r = {\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}$ $f"{\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}} = \frac{{400}}{{\frac{{50}}{\pi }}} + 4\pi$ $f"{\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}} = 8\pi + 4\pi = 12\pi > 0$ $\text{f is minimum at r} = {\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}$ $\text{Now}\quad h = \frac{{100}}{{\pi {r^2}}}$ $h = \frac{{100}}{{\pi {{\left( {\frac{{50}}{\pi }} \right)}^{\frac{2}{3}}}}}$ $h = \frac{{100{\pi ^{\frac{2}{3}}}}}{{\pi {{\left( {50} \right)}^{\frac{2}{3}}}}} = \frac{{2 \times 50 \times {\pi ^{\frac{2}{3}}}}}{{\pi {{\left( {50} \right)}^{\frac{2}{3}}}}}$ $h = \frac{{2 \times {{\left( {50} \right)}^{\frac{1}{3}}}}}{{{{\left( \pi \right)}^{\frac{1}{3}}}}} = 2{\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}$ $\text{so}\quad r = {\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}\quad \text{and} \quad h = 2{\left( {\frac{{50}}{\pi }} \right)^{\frac{1}{3}}}$

Question (22)

A wire of length 28m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Solution

length of wire = 28 m
f: area of circle + area of square
Let from piece of wire we prepare the circle of radius 'r'
C = 2π r
∴ length of wire = 2πr
length of remaining wire for Square = 28 - 2πr
From remaining wire square is prepared so
perimeter of square = 28 - 2 π r
4(side) = 28 - 2 π r
$side = \left( {\frac{{14 - \pi r}}{2}} \right)$ f: sum of area
f = Area of circle + Area of square
f = πr2 + (side)2
$f\left( r \right) = \pi {r^2} + {\left( {\frac{{14 - \pi r}}{2}} \right)^2}$ $f'\left( r \right) = 2\pi r + \cancel{2}\left( {\frac{{14 - \pi r}}{2}} \right)\left( {\frac{{ - \pi }}{\cancel{2}}} \right)$ $f'\left( r \right) = 2\pi r - \frac{\pi }{2}\left( {14 - \pi r} \right)$ $f"\left( r \right) = 2\pi - \frac{\pi }{2}\left( { - \pi } \right)$ $f"\left( r \right) = 2\pi + \frac{{{\pi ^2}}}{2}$ Now f'(r) = 0
$\Rightarrow 2\pi r - \frac{\pi }{2}\left( {14 - \pi r} \right) = 0$ $\therefore \quad 2\pi r = \frac{\pi }{2}\left( {14 - \pi r} \right)$ $4\cancel{\pi} r = \cancel{\pi} \left( {14 - \pi r} \right)$ 4r = 14 - πr
4r + πr = 14
r (π + 4) = 14
$r = \frac{{14}}{{\pi + 4}}$ $f"\left( {\frac{{14}}{{\pi + 4}}} \right) = 2\pi + \frac{{{\pi ^2}}}{2} > 0$ ⇒ f is minimum at
$r = \frac{{14}}{{\pi + 4}}$ C = 2πr
$C = 2\pi \left( {\frac{{14}}{{\pi + 4}}} \right) = \frac{{28\pi }}{{\pi + 4}}$ length of wire for square (l) = 28 -2πr
$l = 28 - \frac{{28\pi }}{{\pi + 4}}$ $l = \frac{{28\pi + 112 - 28\pi }}{{\pi + 4}} = \frac{{112}}{{\pi + 4}}$ length of wire are $l = \frac{{28\pi }}{{\pi + 4}},\frac{{112}}{{\pi + 4}}$

Question (23)

Prove that the volume of the largest come that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere

Solution

Radius of sphere is R which is constant
Let radius of cone be 'r' and let it be 'h'
From diagram CM = h
CO + OM = h
R + OM = h
OM = h - R
In ΔOMB, ∠90°
By Pythagoras' theorem OM2 + BM2 = OB2
(h - R)2 + r2 = R2
r2 = R2 -(h-R)2
r2 = R2 -h2 + 2Rh - R2
r2 = 2Rh - h2
f: volume of cone V
$V = \frac{1}{3}\pi {r^2}h$ $V = \frac{\pi }{3}\left( {2Rh - {h^2}} \right) \cdot h$ $f\left( h \right) = \frac{\pi }{3}\left( {2R{h^2} - {h^3}} \right)$ $f'\left( h \right) = \frac{\pi }{3}\left( {4Rh - 3{h^2}} \right)$ $f"\left( h \right) = \frac{\pi }{3}\left( {4R - 6h} \right)$ Now f'(h) = 0
$\frac{\pi }{3}\left( {4Rh - 3{h^2}} \right) = 0$ $4R\cancel{{h}} = 3{h^\cancel{2}}$ $h = \frac{{4R}}{3}$ $f"\left( h \right) = \frac{\pi }{3}\left[ {4R - \cancel{6}^2\left( {\frac{{4R}}{\cancel{3}}} \right)} \right]$ $f"\left( h \right) = \frac{\pi }{3}\left[ {4R - 8R} \right] = - \frac{{4\pi R}}{3} < 0$ ⇒ f is maximum at h = 4R/3
r2 = 2Rh - h2
${r^2} = 2R\left( {\frac{{4R}}{3}} \right) - {\left( {\frac{{4R}}{3}} \right)^2}$ ${r^2} = \frac{8}{3}{R^2} - \frac{{16}}{9}{R^2}$ ${r^2} = \frac{{24{R^2} - 9{R^2}}}{9} = \frac{8}{9}{R^2}$ $\text{vol. of cone =} \frac{1}{3}\pi {r^2}h$ $\text{vol. of cone =} \frac{1}{3}\pi \left( {\frac{{8{R^2}}}{9}} \right)\left( {\frac{{4R}}{3}} \right)$ $\text{vol. of cone =} \frac{8}{{27}}\left( {\frac{4}{3}\pi {R^3}} \right)$ $\text{vol. of cone =} \frac{8}{{27}}\text{(volume of sphere)}$

Question (24)

Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base

Solution

Volume of cone is constant let V = k
Let r be radius and h be height of cone, l be slant length
From the formula for volume of cone
$V = \frac{1}{3}\pi {r^2}h = k$ $h = \frac{{3k}}{{\pi {r^2}}}$ For the cone l2 = r2 + h2
f: Curvered surface area of cone
f = π r l
As CSA of cone is least, the square of area will also be least
$A = \pi rl$ ${A^2} = {\pi ^2}{r^2}{l^2}$ ${A^2} = {\pi ^2}{r^2}\left( {{r^2} + {h^2}} \right)$ ${A^2} = {\pi ^2}{r^4} + {\pi ^2}{r^2}{h^2}$ ${A^2} = {\pi ^2}{r^4} + {\pi ^2}{r^2}{\left( {\frac{{3k}}{{\pi {r^2}}}} \right)^2}$ ${A^2} = {\pi ^2}{r^4} + \frac{{9{k^2}{\cancel{\pi ^2}}{r^2}}}{{{\cancel{\pi ^2}}{r^4}}}$ $f\left( r \right) = {\pi ^2}{r^4} + \frac{{9{k^2}}}{{{r^2}}}$ $f'\left( r \right) = 4{\pi ^2}{r^3} + 9{k^2}\left( {\frac{{ - 2}}{{{r^3}}}} \right)$ $f'\left( r \right) = 4{\pi ^2}{r^3} - \frac{{18{k^2}}}{{{r^3}}}$ $f"\left( r \right) = 12{\pi ^2}{r^2} + \frac{{54{k^2}}}{{{r^4}}}$ Now f'(r) = 0
$\Rightarrow 4{\pi ^2}{r^3} - \frac{{18{k^2}}}{{{r^3}}} = 0$ $4{\pi ^2}{r^6} = 18{k^2}$ $4{\pi ^2}{r^6} = 18{\left( {\frac{\pi }{3}{r^2}h} \right)^2}$ $4{\pi ^2}{r^6} = \cancel{18}^2\frac{{{\pi ^2}}}{\cancel{9}}{r^4}{h^2}$ $2{r^2} = {h^2} \Rightarrow r = \frac{h}{{\sqrt 2 }}$ $f"\left( {\frac{h}{{\sqrt 2 }}} \right) = 12{\pi ^2}{r^2} + \frac{{54{k^2}}}{{{r^4}}} > 0 \quad \text{as}\quad r> 0$ $\Rightarrow \text{f is minimum at}\quad r = \frac{h}{{\sqrt 2 }}$ $\therefore h = \sqrt 2 r$ Altitude of cone is √2 times radius of base

Question (25)

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height tan-1√2

Solution

A cone has fixed slant height l
Let radius be 'r' and height be h and semi vertical angle be θ
So ∠DAC = θ ( 0 < θ <90° )
$\sin \theta = \frac{{DC}}{{AC}} = \frac{r}{l} \Rightarrow r = l\sin \theta$ $\cos \theta = \frac{{AD}}{{AC}} = \frac{h}{l} \Rightarrow h = l\cos \theta$ f: volume of cone
$f = \frac{1}{3}\pi {r^2}h$ $f = \frac{1}{3}\pi {\left( {l\sin \theta } \right)^2}\left( {l\cos \theta } \right)$ $f\left( \theta \right) = \frac{{\pi {l^3}}}{3}{\left( {\sin \theta } \right)^2}\left( {\cos \theta } \right)$ $f'\left( \theta \right) = \frac{{\pi {l^3}}}{3}\left[ {{{\sin }^2}\theta \left( { - \sin \theta } \right) + \cos \theta \left( {2\sin \theta \cos \theta } \right)} \right]$ $f'\left( \theta \right) = \frac{{\pi {l^3}}}{3}\left[ { - {{\sin }^3}\theta + 2\sin \theta {{\cos }^2}\theta } \right]$ $f"\left( \theta \right) = \frac{{\pi {l^3}}}{3}\left[ { - 3{{\sin }^2}\theta \cos \theta + 2\left( {\sin \theta \left( { - 2\cos \theta \sin \theta } \right) + {{\cos }^2}\theta \cos \theta } \right)} \right]$ $f"\left( \theta \right) = \frac{{\pi {l^3}}}{3}\left[ { - 3{{\sin }^2}\theta \cos \theta - 4{{\sin }^2}\theta \cos \theta + 2{{\cos }^3}\theta } \right]$ $f"\left( \theta \right) = \frac{{\pi {l^3}}}{3}\left[ {2{{\cos }^3}\theta - 7{{\sin }^2}\theta \cos \theta } \right]$ $f"\left( \theta \right) = \frac{{\pi {l^3}}}{3}{\cos ^3}\theta \left[ {2 - 7\frac{{{{\sin }^2}\theta \cos \theta }}{{{{\cos }^3}\theta }}} \right]$ $f"\left( \theta \right) = \frac{{\pi {l^3}}}{3}{\cos ^3}\theta \left[ {2 - 7{{\tan }^2}\theta } \right]$ Now f'(θ) = 0 $\frac{{\pi {l^3}}}{3}\left( { - {{\sin }^3}\theta + 2\sin \theta {{\cos }^2}\theta } \right) = 0$ $\Rightarrow {\sin ^3}\theta = 2\sin \theta {\cos ^2}\theta$ $\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 2 \Rightarrow {\tan ^2}\theta = 2$ $\tan \theta = \sqrt 2$ $\theta = {\tan ^{ - 1}}\sqrt 2$ $f"\left( {{{\tan }^{ - 1}}\sqrt 2 } \right) = \frac{{\pi {l^3}}}{3}{\cos ^3}\theta \left[ {2 - 7\left( 2 \right)} \right]$ $f"\left( {{{\tan }^{ - 1}}\sqrt 2 } \right) = \frac{{\pi {l^3}}}{3}{\cos ^3}\theta \left( {2 - 14} \right)$ $f"\left( {{{\tan }^{ - 1}}\sqrt 2 } \right) = - 12\frac{{\pi {l^3}}}{3}{\cos ^3}\theta$ $\text{as}\quad \theta \in \left( {0,\frac{\pi }{2}} \right)$ ${\cos ^3}\theta > 0$ $f\left( {{{\tan }^{ - 1}}\sqrt 2 } \right) < 0$ ⇒ f is maximum at tan2θ = 2
θ = tan-1√2
∴ semi vertical angle is tan-1√2 for having maximum volume of given slant height

Question (26)

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is ${\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$

Solution

Let radius of cone be 'r' and height be h, slant height be 'l'
l2 = r2 + h2
Total surface area is constant let it be 'k'
π r l + πr2 = k
π r l = k - πr2
squaring on both sides
π2 r2 l2 = ( k - πr2)2
π2 r2 (r2 + h2)= k2 - 2kπr2 + π2 r4
π2 r4 + π2 r2 h2 = k2- 2kπr2 + π2 r4
π2 r2 h2 = k2 - 2kπr2
${h^2} = \frac{{{k^2} - 2k\pi {r^2}}}{{{\pi ^2}{r^2}}}$ Volume is maximum, the square of volume is also maximum $\therefore {V^2} = \frac{{{\pi ^2}}}{9}{r^4}{h^2}$ ${V^2} = \frac{{{\cancel{\pi ^2}}}}{9}{r^{\cancel{4}2}}\left( {\frac{{{k^2} - 2k\pi {r^2}}}{{{\cancel{\pi ^2}}{\cancel{r^2}}}}} \right)$ ${V^2} = \frac{{{r^2}}}{9}\left( {{k^2} - 2k\pi {r^2}} \right)$ $f\left( r \right) = \frac{{{k^2}{r^2}}}{9} - \frac{{2k\pi {r^4}}}{9}$ $f'\left( r \right) = \frac{{2{k^2}r}}{9} - \frac{{2k\pi }}{9}\left( {4{r^3}} \right)$ $f"\left( r \right) = \frac{{2{k^2}}}{9} - \frac{{8k\pi }}{9}\left( {3{r^2}} \right)$ $f"\left( r \right) = \frac{{2{k^2}}}{9} - \frac{{24k\pi {r^2}}}{9}$ Now f'(r) = 0
$\frac{{\cancel{2}{k^\cancel{2}}\cancel{r}}}{\cancel{9}} - \frac{{\cancel{2} \cancel{k}\pi }}{\cancel{9}}\left( {4{r^{\cancel{3}2}}} \right) = 0$ $k = 4\pi {r^2}$ Replace value of k
πrl + π r2 = 4 π r2
π r l = 3 π r2
l = 3r
$f"\left( r \right) = \frac{{2{k^2}}}{9} - \frac{{8k\pi {r^2}}}{3}$ $f"\left( r \right) = \frac{{2k}}{9}\left( {k - 12\pi {r^2}} \right)$ $f"\left( r \right) = \frac{{2k}}{9}\left( {\pi rl + \pi {r^2} - 12\pi {r^2}} \right)$ $f"\left( r \right) = \frac{{2k}}{9}\left[ {\pi r\left( {3r} \right) - 10\pi {r^2}} \right]$ $f"\left( r \right) = \frac{{2k}}{9}\left( {3\pi {r^2} - 10\pi {r^2}} \right)$ $f"\left( r \right) = \frac{{2k}}{9}\left( { - 7\pi {r^2}} \right) < 0$ ⇒ f is maximum at l = 3r
$\frac{r}{l} = \frac{1}{3}$ $\text{But} \quad \frac{r}{l} = \sin \theta$ $\sin \theta = \frac{1}{3}$ $\Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$ So for maximum volume of cone for given surface area
$\text{semi vertical angle is }\quad {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$

### Choose the correct answer in the Exercises 27 and 29

Question (27)

The point on the curve x2 = 2y which is nearest to the point (0, 5) is
(A) (2√2, 4)     (B) (2√ 0)
(C) (0, 0)     (D) (2, 2)

Solution

$\frac{{{x^2}}}{2} = y$ $A\left( {x,y} \right) = \left( {x,\frac{{{x^2}}}{2}} \right) \quad\text{is on the curve}$ $AB = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {\frac{{{x^2}}}{2} - 5} \right)}^2}}$ $\Rightarrow AB = \sqrt {{x^2} + {{\left( {\frac{{{x^2}}}{2} - 5} \right)}^2}}$ As distance is minimum Square of distance is also minimum
$\therefore A{B^2} = {x^2} + {\left( {\frac{{{x^2}}}{2} - 5} \right)^2}$ $f\left( x \right) = {x^2} + {\left( {\frac{{{x^2}}}{2} - 5} \right)^2}$ $f'\left( x \right) = 2x + 2\left( {\frac{{{x^2}}}{2} - 5} \right)\left( {\frac{{\cancel{2}x}}{\cancel{2}}} \right)$ $f'\left( x \right) = 2x + 2x\left( {\frac{{{x^2}}}{2} - 5} \right)$ $f'\left( x \right) = 2x + {x^3} - 10x$ $f"\left( x \right) = 2 + 3{x^2} - 10 = 3{x^2} - 8$ Now f'(x) = 0
x3 - 8x = 0
x(x2 -8) = 0
x= 0 OR x = 2√2
f"(0) = -8 < 0
f is maximum
f"(2√2) = 3(8) - 8 = 16 > 0
so f is maximum at x = 2√2
x = 2√2; y = 4
(x, y) = (2√2 , 4) So option A is correct

Question (28)

For all real values of x, the minimum values of
$\frac{{1 - x + {x^2}}}{{1 + x + {x^2}}}\quad \text{is}$

Solution

$f\left( x \right) = \frac{{1 - x + {x^2}}}{{1 + x + {x^2}}}$ $f'\left( x \right) = \frac{{\left( {1 + x + {x^2}} \right)\left( { - 1 + 2x} \right) - \left( {1 - x + {x^2}} \right)\left( {1 + 2x} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}$ $f'\left( x \right) = \frac{{ - 1 - \cancel{x} - {x^2} + \cancel{2x} + 2{x^2} + \cancel{2{x^3}} - 1 + \cancel{x} - {x^2} - \cancel{2x} + 2{x^2} - \cancel{2{x^3}}}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}$ $f'\left( x \right) = \frac{{2{x^2} - 2}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}$ $f'\left( x \right) = \frac{{2\left( {{x^2} - 1} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}$ $f"\left( x \right) = 2\left[ {\frac{{{{\left( {1 + x + {x^2}} \right)}^2}\left( {2x} \right) - \left( {{x^2} - 1} \right)2\left( {1 + x + {x^2}} \right)\left( {1 + 2x} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^4}}}} \right]$ $f"\left( x \right) = \frac{{2 \times 2\left(\cancel{ {1 + x + {x^2}}} \right)\left[ {x\left( {1 + x + {x^2}} \right) - \left( {{x^2} - 1} \right)\left( {1 + 2x} \right)} \right]}}{{{{\left( {1 + x + {x^2}} \right)}^{\cancel{4}3}}}}$ $f"\left( x \right) = \frac{4}{{{{\left( {1 + x + {x^2}} \right)}^3}}}\left[ {x + \cancel{{x^2}} + {x^3} - \cancel{{x^2}} - 2{x^3} + 1 + 2x} \right]$ $f"\left( x \right) = \frac{{4\left( {3x - {x^3} + 1} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^3}}}$ f'(x) = 0 ⇒ x2 - 1 = 0, x = ±1
$f\left( 1 \right) = \frac{{4\left( {3 - 1 + 1} \right)}}{{{{\left( 3 \right)}^3}}} > 0 \Rightarrow\text{ f is minimum}$ $f\left( { - 1} \right) = \frac{{4\left( { - 3 + 1 + 1} \right)}}{1} = - 12 < 0 \Rightarrow \text{ f is maximum}$ ∴ x = 1 ⇒ (B) is correct option

Question (29)

The maximum value of [x(x-1) + 1]1/3, 0≤ x ≤ 1 is
$(A){\left( {\frac{1}{3}} \right)^{\frac{1}{3}}}$ $(B)\frac{1}{2}$ (C) 1
(D) 0

Solution

$f\left( x \right) = {\left[ {x\left( {x - 1} \right) + 1} \right]^{\frac{1}{3}}}$ $f\left( x \right) = {\left( {{x^2} - x + 1} \right)^{\frac{1}{3}}}$ Differentiating with respect to x $f'\left( x \right) = \frac{1}{3}{\left( {{x^2} - x + 1} \right)^{\frac{{ - 2}}{3}}} \cdot \left( {2x - 1} \right)$ $f'\left( x \right) = \frac{{2x - 1}}{{3{{\left( {{x^2} - x + 1} \right)}^{\frac{2}{3}}}}}$ $f"\left( x \right) = \frac{1}{3}\left[ {\frac{{{{\left( {{x^2} - x + 1} \right)}^{\frac{2}{3}}} \cdot 2 - \left( {2x - 1} \right) \cdot \frac{2}{3}{{\left( {{x^2} - x + 1} \right)}^{\frac{{ - 1}}{3}}} \cdot \left( {2x - 1} \right)}}{{{{\left( {{x^2} - x + 1} \right)}^{\frac{4}{3}}}}}} \right]$ $f"\left( x \right) = \frac{1}{3}\frac{2}{{{{\left( {{x^2} - x + 1} \right)}^{\frac{4}{3}}}}}\left[ {{{\left( {{x^2} - x + 1} \right)}^{\frac{2}{3}}} - \frac{{\left( {2x - 1} \right)\left( {2x - 1} \right)}}{{3{{\left( {{x^2} - x + 1} \right)}^{\frac{1}{3}}}}}} \right]$ $f"\left( x \right) = \frac{2}{{9{{\left( {{x^2} - x + 1} \right)}^{\frac{5}{3}}}}}\left[ {3\left( {{x^2} - x + 1} \right) - \left( {4{x^2} - 4x + 1} \right)} \right]$ $f"\left( x \right) = \frac{{2\left( { - {x^2} + x + 2} \right)}}{{9{{\left( {{x^2} - x + 1} \right)}^{\frac{5}{3}}}}}$ Now f'(x) = 0 ⇒ 2x - 1 = 0, x = ½
$f"\left( {\frac{1}{2}} \right) = \frac{{2\left( { - \frac{1}{4} + \frac{1}{2} + 2} \right)}}{{9{{\left( {\frac{1}{4} - \frac{1}{2} + 1} \right)}^{\frac{5}{3}}}}} > 0$ ⇒ f is minimum at x = ½
$f\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{4} - \frac{1}{2} + 1} \right)^{\frac{1}{3}}} = {\left( {\frac{3}{4}} \right)^{\frac{1}{3}}}$ $f\left( 0 \right) = {\left( {0 - 0 + 1} \right)^{\frac{1}{3}}} = 1$ $f\left( 1 \right) = {\left( {1 - 1 + 1} \right)^{\frac{1}{3}}} = 1$ $\text{max value}= \max \left\{ {1,1,{{\left( {\frac{3}{4}} \right)}^{\frac{1}{3}}}} \right\}$ max. value of f(x) = 1 ;   0 ≤ x ≤ 1
∴ C is correct option