2. Maxima and Minima in Case of Two or More Variables

(26 Solved problems and 10 Home assignments)

Maxima and Minima of functions of two variables:

Let ( )y,xfz = be a function of two independent variables x and y.

Relative maximum: ( )y,xf is said to have a relative maximum at a point ( )b,a if

( ) ( )kb,hafb,af ++> ( ) ( ) 0b,afkb,haf <−++⇒

for small positive or negative values of ‘h’ and ‘k’

or

If the value of the function f (x, y) at ( )b,a , i.e., ( )b,af is greater than the value

of the function f at all points in some small neighbourhood of ( )b,a , then ( )y,xf is said

to have a relative maximum at a point ( )b,a .

Relative minimum: ( )y,xf is said to have a relative minimum at a point ( )b,a if

( ) ( )kb,hafb,af ++< ( ) ( ) 0b,afkb,haf >−++⇒

for small positive or negative values of ‘h’ and ‘k’.

or

Differential Calculus

Maxima and Minima

(Maxima and minima in case of two or more variables)

Prepared by:

Dr. Sunil

NIT Hamirpur (HP)

(Last updated on 04-09-2008)

Latest update available at: http://www.freewebs.com/sunilnit/

Maxima and minima: Maxima and minima in case of two or more variables

Prepared by: Dr. Sunil, NIT Hamirpur (HP)

2

If the value of the function f at ( )b,a , i.e., ( )b,af is smaller than the value of the

function f at all points in some small neighbourhood of ( )b,a , then ( )y,xf is said to have

a relative minimum at a point ( )b,a .

Notation: Denoting ( ) ( )[ ]b,afkb,haf −++ by ( )b,af∆ or simply by ∆

i.e. ( ) ( )[ ]b,afkb,haf −++=∆ .

Then, ( )y,xf has the maximum at ( )b,a if ∆ has the negative sign for all small values

of h, k; i.e. 0<∆ .

Similarly, ( )y,xf has the minimum at ( )b,a if ∆ has the positive sign for all small

values of h, k; i.e. 0>∆ .

Extremum:

Extremum is a point, which is either a maximum or a minimum. The value of the

function f at an extremum (maximum or minimum) point is known as extremum

(maximum or minimum) value of the function f.

Geometrically, ( )y,xfz = represents a surface. The maximum is a point on the

surface (hill top) from which the surface descends (comes down) in every direction

towards the xy-plane. The minimum is the bottom of depression from which the surface

ascends (climbs up) in every direction towards the xy-plane. In either case, tangent planes

to the surface at a maximum or minimum point is horizontal (parallel to xy-plane) and

perpendicular to z-axis.

x-axis

y-axis

z-axis

Minimum

z = f(x, y)

x-axis

y-axis

z-axis

Maximum

z = f(x, y)



3

Saddle point:

Saddle point or minimax is a point where function is neither maximum nor

minimum. At such point, f is maximum in one direction while minimum in other

direction.

Geometrically, such a surface (looks like the leather seat on back of a horse)

forms a ridge rising in one direction and falling in another direction.

Example: z = xy, hyperbolic parabolic has a saddle point at the origin.

Necessary and sufficient conditions for extrema of a function f of two

variables:

Since ( ) ( )[ ]b,afkb,haf −++=∆ .

Expanding ( )kb,haf ++ by Taylor’s theorem, we obtain

( ) ( ) ( ) ( )[ ]

( ) ( ) ( )[ ] ......b,afkb,ahkf2b,afh!2

1

b,akfb,ahfb,afkb,haf

yy2

xyxx2

yx

++++

++=++

(i)

Since h and k are small, so neglecting higher order terms of 22 k ,hk ,h , etc. Then the

above expression reduces to

( ) ( ) ( ) ( )[ ]b,akfb,ahfb,afkb,haf yx ++=++

Thus ( ) ( )[ ]b,afkb,haf −++=∆ ( ) ( )b,akfb,ahf yx += .

(ii)

Necessary conditions:

z-axis

y-axis

x-axis

Saddle point



4

The necessary condition for ( )y,xf has the maximum or minimum at ( )b,a

i.e. ∆ has the negative or positive sign for all small values of h, k is

( ) ( ) 0b,af and 0b,af yx == ,

even though ‘h’ and ‘k’ can take both positive and negative values.

Method for developing the sufficient conditions:

With ( ) ( ) 0b,af and 0b,af yx == ,

( ) ( ) ( ) ( )[ ]

( ) ( ) ( )[ ]b,afkb,ahkf2b,afh!2

1

b,akfb,ahfb,afkb,haf

yy2

xyxx2

yx

+++

+=−++=∆

reduces to

( ) ( ) ( ) ( ) ( )[ ]b,afkb,ahkf2b,afh!2

1b,afkb,haf yy

2xyxx

2 ++=−++=∆

Denote ( ) ( ) ( ) tb,af ,sb,af ,rb,af yyxyxx === , we get

( ) ( ) [ ]tkhks2rh!2

1b,afkb,haf 22 ++=−++=∆ . (iii)

From this expression, we observe that the nature of the sign of ∆ depends on the nature

of sign of tkhks2rh 22 ++ .

Thus [ ]

++=++=∆

r

rtkhkrs2rh of signtkhks2rh of sign ofsign

22222

( ) ( )

−++=

r

srtkkshr of sign

222

. (iv)

If 0srt 2 >− , then the numerator of RHS of (iv) is positive.

In this case, sign of ∆ = sign of r.

Thus 0<∆ if 0srt 2 >− and 0r <

and 0>∆ if 0srt 2 >− and 0r > .

Therefore, the sufficient (Lagrange’s) conditions for extrema are:

I. f attains (has) a maximum at (a, b) if 0srt 2 >− and 0r <



5

II. f attains (has) a minimum at (a, b) if 0srt 2 >− and 0r >

III. Saddle point: If 0srt 2 <− , then 0<∆ or 0>∆ depending on ‘h’ and ‘k’.

Therefore, f has a saddle point (minimax) at (a, b) if 0srt 2 <− .

IV. Failure case: If 0srt 2 =− , then further investigation is needed to determine

the nature of the function f.

FINAL CONCLUSIONS:

A function of two variables ( )y,xfu = is said to have a maximum or minimum if

(1) 0x

f=

∂

∂, 0

y

f=

∂

∂,

which gives the values of x say x = a and y say y = b, for which the function is

maximum or minimum.

(2)

22

2

2

2

2

yx

f

y

f.

x

f

∂∂

∂>

∂

∂

∂

∂ at these values i.e. x = a and y = b.

i.e. ( )[ ] 0ff.fbyax

2xyyyxx >−

== i.e. 0srt 2 >− .

(3a) Then the function ( )y,xf will have a maximum value if apart from the above

conditions (1) and (2), also 0x

f2

2

<∂

∂ i.e. 0r < at x = a and y = b.

(3b) Similarly, the function ( )y,xf will have a minimum value if apart from the above

conditions (1) and (2), also 0x

f2

2

>∂

∂ i.e. 0r > at x = a and y = b.

Remarks:

(a) If

22

2

2

2

2

yx

f

y

f.

x

f

∂∂

∂=

∂

∂

∂

∂i.e. 0srt 2 =− at x = a and y = b,

Then the function ( )y,xf is neither a maximum nor a minimum.

(b) Again if

22

2

2

2

2

yx

f

y

f.

x

f

∂∂

∂<

∂

∂

∂

∂ i.e. 0srt 2 <− at x = a and y = b,



6

Then such a point where

22

2

2

2

2

yx

f

y

f.

x

f

∂∂

∂<

∂

∂

∂

∂ holds, is called a “Saddle point”.

Method of finding Extrema of f(x, y):

1. Solving 0fx = and 0fy = yields critical or stationary point P of f.

2. Calculate xxfr = , xyfs = , yyft = at the critical point.

3. a. f attains (has) a maximum at (a, b) if 0srt 2 >− and 0r <

b. f attains (has) a minimum at (a, b) if 0srt 2 >− and 0r >

c. Saddle point: If 0srt 2 <− , then 0<∆ or 0>∆ depending on ‘h’ and ‘k’.

Therefore, f has a saddle point (minimax) at (a, b) if 0srt 2 <− .

d. Failure case: If 0srt 2 =− , then further investigation is needed to determine

the nature of the function f.

Remarks: Extrema occur only at stationary points. However, stationary points need not

be extrema.

Now let us solve some problems, where we have to evaluate the maxima or

minima of the given function:

Q.No.1.: Find the values of x and y for which 12x6yxu 22 +++= has a minimum

value and find this minimum value.

Sol.: Since ( ) 12x6yxy,xu 22 +++= .

Then ∂

∂

u

xx= +2 6 and

∂

∂

u

yy= 2 .

For u (x, y) to be minimum or maximum, we have ∂

∂

u

x= 0 and

∂

∂

u

y= 0 .

⇒ + =2 6 0x and 2 0y = .

⇒ = −x 3 and y = 0.

Also

[ ]

2x

u

0,3

2

2

=

∂

∂

−

,

[ ]

2y

u

0,3

2

2

=

∂

∂

−

and

[ ]

0yx

u

0,3

2

=

∂∂

∂

−

.



7

04022yx

u

y

u.

x

u2

2

2

2

2

2

>=−×=

∂∂

∂−

∂

∂

∂

∂∴ at x = −3 and y = 0.

And also

[ ]

02x

u

0,3

2

2

>=

∂

∂

−

.

Thus the function is minimum at x = −3 and y = 0.

And this minimum value ( )[ ] [ ][ ] 31218912x6yxy,xu 0,322

min =+−=+++= − . Ans.

Q.No.2.: If ( ) 9xy6y,xf += , find the values of x and y for which ( )y,xf has a stationary

value.

Sol.: Since ( ) 9xy6y,xf += .

Then y6x

f=

∂

∂ and x6

y

f=

∂

∂.

For f (x, y) to be minimum or maximum, we have 0x

f=

∂

∂ and 0

y

f=

∂

∂.

0y6 =⇒ and 0x6 = .

⇒ =y 0 and x = 0 .

Also

[ ]

0x

f

0,0

2

2

=

∂

∂ ,

[ ]

0y

f

0,0

2

2

=

∂

∂

−

and

[ ]

6yx

f

0,0

2

=

∂∂

∂.

03660yx

u

y

u.

x

u 2

22

2

2

2

2

<−=−=

∂∂

∂−

∂

∂

∂

∂∴ at 0x = and 0y = .

Thus x = 0 and y = 0 does not stand for maxima and minima.

Thus the given function has a stationary value at x = 0 and y = 0.

Q.No.3.: Find the maximum and minimum values of

(a) axy3yx 33 −+ .

(b) ( )yxsin.ysin.xsin + .

Sol.: (a) Let axy3yxu 33 −+= .

Partially differentiating u w. r. t. x and y, we get



8

ay3x3x

u 2 −=∂

∂ and ax3y3

y

u 2 −=∂

∂

For u (x, y) to be minimum or maximum, we have ∂

∂

u

x= 0 and

∂

∂

u

y= 0 .

0ay3x30x

u 2 =−⇒=∂

∂ay3x3 2 =⇒ ayx2 =⇒ . (i)

0ax3y30y

u 2 =−⇒=∂

∂ax3y3 2 =⇒ axy2 =⇒ . (ii)

Squaring (i), we get 224 yax = (iii)

Substituting the value of 2y from (ii) in (iii), we get axaxaxax 3324 =⇒=⇒=

ayx2 =∴ ayaya2 =⇒=⇒ .

x6x

u2

2

=∂

∂, y6

y

u2

2

=∂

∂ , a3

yx

u2

−=∂∂

∂

22

2

2

2

2

yx

u

y

u.

x

u

∂∂

∂−

∂

∂

∂

∂∴ = 22 a9xy36a9y6 . x6 −=−

Substituting xy as 2a , we get

22

2

2

2

2

yx

u

y

u.

x

u

∂∂

∂−

∂

∂

∂

∂= 0a27a9a36 222 >=− .

To find max. and min. value which u will attain.

Put 0x

u2

2

=∂

∂, we get

a6x6x

u2

2

==∂

∂

0a6 < if 0a < and 0a6 > if 0a > . Ans.

(b) Let ( )yxsin.ysin.xsin)y,x(P +=

( ) ( ){ }yxcos.xsinyxsin.xcosysinx

FFx +++=

∂

∂= )yx2sin(.ysin += (i)

( ) ( ){ }yxcos.ysinyxsin.ycosxsinx

FFy +++=

∂

∂= )y2xsin(.xsin += (ii)



9

For max. or min. values of F(x, y), we have (i) and (ii) equal to zero.

0Fx = 0)yx2sin(.ysin =+⇒ (iii)

0Fy = 0)y2xsin(.xsin =+⇒ (iv)

From (iii) and (iv), we have

0ysin = and 0xsin =

0x =⇒ and y = 0, which is not possible.

We have also

π=+ y2x or π2 (v)

π=+ yx2 or π2 (vi)

Taking π=+ 2y2x

π=+ 2yx2

Solving the above equation we have 3

2x

π= ,

3

2y

π=

Now taking π=+ y2x

π=+ yx2

Solving the above equations, we get 3

xπ

= , 3

yπ

= .

So the points of extremum are

ππ

3 ,

3 and

ππ

3

2 ,

3

2.

Let xxFr = , yyFt = and xyFs = .

So 2).yx2sin(.ysinFxx += ( )yx2cos.ysin2 +=

( )y2xcos.ysin2Fyy +=

( ) ( ) ysin.yx2cosyx2sin.ycosFxy +++= ( )y2x2sin += .

Now at points

ππ

3 ,

3 , we have to calculate the

2xyyyxx FF.F −

So 312

32Fxx −=−××=



10

312

32Fyy −=−××=

2

3Fxy

−= .

Now ( )( ) 04

9

4

33

2

33.3

2

>=−⇒

−−−−

So that 0FF.F 2xyyyxx >− and 3Fxx −= .

Hence F(x, y) has maximum at

π=

π=

3y ,

3x .

Maximum value of F(x, y) 8

33

3sin.

3sin.

3sin =

πππ.

Now taking points

ππ

3

2 ,

3

2, we have to calculate

2xyyyxx FF.F −

So 12

32Fxx ××=

12

32Fyy ××=

2

3Fxy = .

Now 04

9

4

33

2

3.1

2

321

2

32

2

>=−⇒

−

××

××

So that 0FF.F 2xyyyxx >− and also 03Fxx >= .

Hence the point

ππ

3

2 ,

3

2 is the point of minimum.

Minimum value of F(x, y) 8

33

3

4sin.

3

2sin.

3

2sin

−=

πππ.



11

So the maximum and minimum values of function F(x, y) is 8

33 and

8

33−

respectively.

Q.No.4.: Show that the minimum value of y

a

x

axy)y,x(f

33

++= is 2a3 .

Sol.: Given function is y

a

x

axy)y,x(f

33

++= .

2

3

xx

ay

x

ff −=

∂

∂=∴ ,

2

3

yy

ax

y

ff −=

∂

∂= .

For f (x, y) to be minimum or maximum, we have 0fx = and 0fy = .

If 0fx = 2

3

2

3

x

ay0

x

ay =⇒=−⇒ (i)

If 0fy = 2

3

2

3

y

ax0

y

ax =⇒=−⇒ (ii)

From (i) and (ii), we get

0a

x1x0

a.a

x.ax0

x

a

ax

3

3

33

23

2

2

3

3

=

−⇒=−⇒=

−

As 0a

x10x

3

3

=−⇒≠ , because when x = 0, y can not be defined.

axax 33 =⇒=⇒

∴From (i), we get aa

ay

2

3

== .

Hence extreme point is (a, a)

Now, ( )

3

3

3

3

xxx

a2

x

a.20f =

−−= ,

( )

3

3

3

3

2

2

yyy

a2

y

a.20

y

Ff =

−−=

∂

∂= ,



12

101yx

Ff

2

xy =−=∂∂

∂= .

At point (a, a) 2a

a2f

3

3

xx == , 2a

a2f

3

3

yy == .

Now, ( ) ( ) 0314122fff22

xyyyxx >=−=−×=−× .

Also 02fxx >= .

∴Minimum value of f(x, y) is 222233

a3aaaa

a

a

aa.a =++=++ . Ans.

Q.No.5.: Find co-ordinates of a point P(x, y) such that sum of squares of its distances

from rectangular axis of reference and line 8yx =+ is minimum.

Sol.: We know that distance of a point P(x, y) from any line 0cbyax =++ is given by

22 ba

cbyax

+

++

( )2

8yxyxu

222 −+

+=∴

For maxima or minima, we have 0x

u=

∂

∂ and 0

y

u=

∂

∂.

( )0

2

8yx2x2 =

−++∴ and

( )0

2

8yx2y2 =

−++

8yx3 =+ (i) 8y3x =+ (ii)

Solving (i) and (ii), we get x = 2 and y = 2.

∴ At x = 2, y = 2, the given function have max. or min. value.

Also we know that for minima, we have

0x

u2

2

>∂

∂ and 0

y

u2

2

>∂

∂.

03x

u

2y2x

2

2

>=

∂

∂

==

and 03y

u

2y2x

2

2

>=

∂

∂

==

.

∴Thus the function have minimum value at x = 2, y = 2.



13

Hence (2, 2) be the required coordinates of the point such that sum of squares of its

distances from rectangular axis of reference and line 8yx =+ is minimum. Ans.

Q.No.6.: Divide ‘a’ into three parts such that their products be a maximum.

Sol.: Let ‘a’ be the sum of three no. x, y, z.

Therefore ( )yxazzyxa +−=⇒++=

Their product can be written as

( )[ ] xyyxxayPyxaxyPxyzP 22 −−=⇒+−=⇒= .

For maximum or minimum values yx F ,F are zero.

0yxy2ayx

PF 2

x =−−=∂

∂=∴ 0yx2a =−−⇒ x2ayyx2a −=⇒+=⇒ (i)

0yx2xxay

PF 2

y =−−=∂

∂= 0y2xa =−−⇒ y2axxy2a −=⇒+=⇒ (ii)


( )y2a2ay −−=3

ayay3y4a2ay =⇒=⇒+−=⇒

Putting this value in (ii), we get

3

ax

3

a2ax =⇒−= .

3

az

3

a2a

3

a

3

aaz =⇒−=

+−=∴ .

Differentiating yx F ,F again

a3

2y2

x

PF

2

2

xx −=−=∂

∂=∴

a3

2x2

y

PF

2

2

xx −=−=∂

∂=

also

∂

∂

∂

∂=

∂∂

∂=

x

P

yyx

PF

2

xy ( )2yxy2yay

−−∂

∂= y2x2a −−= .

Putting the value of x and y.

3

a

3

a4a3

3

a2

3

a2a

yx

P2 −=

−=−−=

∂∂

∂



14

03

a

9

a3

9

aa4

9

a

3

a2.

3

a2

yx

P

y

P.

x

P 222222

2

2

2

2

2

>==−

=−−−

=

∂∂

∂−

∂

∂

∂

∂∴

Therefore the product is maximum.

Hence ‘a’ can be divided into three parts as

3

a ,

3

a ,

3

a.

Q.No.7.: Divide 120 into three parts so that the sum of the products taken two at a time

shall be maximum.

Sol.: Let zyx120 ++= . ...(i)

and zxyzxyP ++= . ...(ii)

From (i), we get yx120z −−= .

( ) ( )yx120xyx120yxyP −−+−−+=∴ .

Then ∂

∂

P

xy y x y x y= + − + − − = − −( )1 120 2 120 2

and y2x120xy2x120xy

P−−=−−−+=

∂

∂.

For maximum or minimum vales both 0x

P=

∂

∂ and 0

y

P=

∂

∂.

0yx2120 =−−⇒ and 0y2x120 =−− .

Solving these two equations, we get

40x = , 40y = and 40z = .

Also

[ ]

2x

P

40,40,40

2

2

−=

∂

∂ ,

[ ]

2y

P

40,40,40

2

2

−=

∂

∂ and

[ ]

1yx

P

40,40,40

2

−=

∂∂

∂.

03)1()2()2(yx

P

y

P.

x

P 2

22

2

2

2

2

>=−−−×−=

∂∂

∂−

∂

∂

∂

∂∴ at 40x = , 40y = and 40z = .

And also

[ ]

02x

P

40,40,40

2

2

<−=

∂

∂

Thus the function is maximum at x = 40 , y = 40 and z = 40 .

Q.No.8.: Prove that the rectangular solid of maximum volume which can be inscribed in

a sphere is a cube.



15

Sol.: Let x , y and z be the dimensions of the rectangular solid and D be the diameter of

the given sphere.

The diagonal of the solid will be diameter of the sphere.

2222 zyxD ++=∴ 222 zyxD ++=⇒ . ...(i)

Let V be the volume of the solid

222 yxDxyxyzV −−==∴ ( ) 4224222222222 yxyxyxDyxDyxV −−=−−=⇒ .

Let uV2 = . 4224222 yxyxyxDu −−=∴ . ...(ii)

Differentiate (ii) partially w. r. t. x and y separately, we get

( )222242322 yx2Dxy2xy2yx4xyD2x

u−−=−−=

∂

∂ ...(iii)

( )222232422 y2xDyx2yx4yx2yxD2y

u−−=−−=

∂

∂ ..(iv)

For max. and min. of u, 0x

u=

∂

∂, 0

y

u=

∂

∂

( ) 0yx2Dxy2x

u 2222 =−−=∂

∂∴

and ( ) 0y2xDyx2y

u 2222 =−−=∂

∂

But 0y ,0x ≠≠ being lengths

( ) 0yx2D 222 =−− ...(v)

( ) 0y2xD 222 =−− ...(vi)

From (v) 222 x2Dy −= and put in (vi), we get

0D2x4xD 2322 =−+− 3

Dx

3

DxDx3

2222 =⇒=⇒=⇒ .

Now 3

Dy

3

D

3

D2Dx2Dy

222222 =⇒=−=−= ; (y cannot be < 0)

Again differentiate (iii) partially w. r. t. x and y separately and (iv) w. r. t. y, we get

42222

2

2

y2yx12yD2x

u−−=

∂

∂ ,



16

3322

xy8yx8xyD4yx

u−−=

∂∂

∂,

22422

2

2

yx12x2xD2y

u−−=

∂

∂.

9

D8

9

D2

3

D.

3

D12

3

D.D2

x

uA

442222

3

D,

3

D2

2

−=−−=

∂

∂=

9

D4

33

D.

3

D8

3

D.

33

D8

3

D.

3

D.D4

yx

uB

4332

3

D,

3

D

2

−=−−=

∂∂

∂=

9

D8

3

D.

3

D12

9

D2

3

D.D2

y

uC

422422

3

D,

3

D2

2

−=−−=

∂

∂=

Now 081

D16

81

D64

9

D4

9

D8

9

D8BAC

884442 >−=

−

−

−=− (+ ve)

Also A is negative for 3

Dyx == .

∴ u is maximum when 3

Dyx == .

∴ 2V or V is maximum value for 3

Dyx == .

From (i), we get 3

D

3

D

3

DDyxDz

222222 =−−=−−=

3

Dzyx ===∴ i.e. length = breadth = height.

Hence for maximum volume solid is a cube.

Q.No.9: The temperature T at any point (x, y, z) in space is defined by 2xyz400T = .

Find the maximum temperature on the surface of a unit sphere 1zyx 222 =++ .

Sol.: Given equation of the sphere is 1zyx 222 =++ 222 yx1z −−=⇒ (i)

Given 2xyz400T = ( )22 yx1xy400T −−=⇒ 33 xy400yx400xy400T −−=⇒ (ii)



17

Now, 0y400yx1200y4000x

T 32 =−−⇒=∂

∂ 01x3y 22 =−+⇒ 22 x31y −=⇒ (iii)

0xy1200x400y4000y

T 23 =−−⇒=∂

∂01y3x 22 =−+⇒

( ) 01x313x 22 =−−+⇒ [using (ii)]

4

1x01x93x 222 =⇒=−−+⇒

2

1x =⇒

Putting in (iii), we get2

1y = .

Putting in (i), we get 2

1z2 = .

Now to check Maxima and Minima

x2400x

T2

2

−=∂

∂ and y2400

y

T2

2

−=∂

∂

1200x

T

2

1x

2

2

−=∂

∂

=

and 1200y

T

2

1y

2

2

−=

∂

∂

=

Also 800yx

Tx2400400

yx

T

2

1x

22

−=∂∂

∂⇒−=

∂∂

∂

=

Clearly

22

2

2

2

2

yx

T

y

T

x

T

∂∂

∂>

∂

∂

∂

∂,

Now 012002

12400

x

T

2

1y ,

2

1x

2

2

<=×−=

∂

∂

==

.

∴ The function T has maxima at the calculated values of x, y and 2z .

Put the values of x, y and 2z in 2xyz400T = , we get

2

1

2

1

2

1400T ×××= 50T =⇒ units. Ans.

which is the maximum temperature on the surface of a unit sphere 1zyx 222 =++ .

Q.No.10.: A rectangular tank open at the top and is to hold a given volume . Find the



18

dimensions of the box requiring least material, for its construction.

Sol.: Let x , y and z ft. be the dimensions of the rectangular tank open at the top so that

material for construction will be least if surface area is least.

Let surface area, S = F(x, y, z) = xy + 2yz + 2zx . ....(i)

Also given volume = xyz = V. ...(ii)

Eliminating z from (i) with the help of (ii), we get

( )

++=++=

y

1

x

1V2xy

xy

Vxy2xyS .

∴ 0x

V2y

x

S2

=−=∂

∂ and 0

y

V2x

y

S2

=−=∂

∂. ...(iii)

Solving these, we get x = y = 2z. ...(iv)

Now 32

2

x

V4

x

Sr =

∂

∂= , 1

yx

Ss

2

=∂∂

∂= ,

32

2

y

V4

y

St =

∂

∂= .

At x = y , ( )

6

6

6

62

33

2

x

x164

x

xV161

x

V4.

x

V4srt

−=

−=−=− = always + ve and r is also + ve.

Henve S is minimum for x = y = 2z.

Also from (ii) , x = y = (2V)1/3

. ...(v)

Thus from (iv) and (v), we get

length = breadth = twice height = (2 volume)1/3

.

Q.No.11.: Find the point on the surface 1xyz2 += , nearest to the origin.

Sol.: Let ( )x y z, , be the point lies on the surface 1xyz2 += ,which is nearest to the

origin.

∴Distance between origin and this point 222 zyx ++= . Squaring both sides, we get

(Distance)2 222 zyx)say(u ++== subject to the condition 1xyz2 += . ...(i)

1xyyxu 22 +++=∴ .

Then yx2x

u+=

∂

∂ and xy2

y

u+=

∂

∂.

For maximum or minimum vales both 0x

u=

∂

∂ and 0

y

u=

∂

∂.



19

0yx2 =+⇒ and 0xy2 =+ .

⇒ = −xy

2 and y

x= −

2. ⇒ =x 0 and y = 0 .

Also

[ ]

2x

u

0,0

2

2

=

∂

∂ ,

[ ]

2y

u

0,0

2

2

=

∂

∂ and

[ ]

1yx

u

0,0

2

=

∂∂

∂.

03122yx

u

y

u.

x

u2

2

2

2

2

2

>=−×=

∂∂

∂−

∂

∂

∂

∂∴ at x = 0 and y = 0.

And also

[ ]

02x

u

0,0

2

2

>=

∂

∂. Also from (i) , 1z ±= .

Thus the function is minimum at x = 0 , y = 0 and 1z ±= .

Thus ( )1 ,0 ,0 and ( )1 ,0 ,0 − are the points on the surface 1xyz2 += ,which are nearest to

the origin.

Q.No.12.: A tent having the form of a cylinder, surmounted by a cone is to contain a

given volume. Prove that for the canvas to be minimum, height of the cone is

twice that of the cylinder.

Sol.: Let the height of the cone is = y

Let the height of the cylinder is = x

Then, we have to prove , x2y =

Let v be the total volume of the tent.

Then, yr3

1xrv 22 π+π= (i)

If S is the total surface area of the tent, then

� r x r 2S π+π= 22 yrr r x 2S +π+π=⇒ (ii)

Putting the value of x from (i) in (ii), we get

22

2yrr

3

y

r

vr 2S +π+

−

ππ=

y2yr2

r

3

r 2

y

S

22×

+

π+

π−=

∂

∂∴ .



20

For maxima and minima, 0y

S=

∂

∂.

0yr

yr

3

r 2

22=

+

π+

π−∴

2

22

2

yr

y

3

2

+=

⇒ 222 y9y4r4 =+⇒

r5

2yr4y5

22 =⇒=⇒ (iii)

Differentiating (ii) partially w. r. t. r, we get

+π

∂

∂+

−

π×π

∂

∂=

∂

∂ 22

2yrr

r3

y

r

vr 2

rr

S

22

222

2yr

r yr

3

y2

r

v2

+

π++π+

π−

−= .

For max. or min., 0r

S=

∂

∂

0yr

yr

3

y 2

r

v2

22

222

2=

+

π++π+

π−

−⇒

0yr

ryr

3

y2

3

yx

r

r 2

22

222

2=

++++−

+

π

π−⇒

Putting r5

2y = in above equation, we get

0r3

5r

53

9r

53

4r

53

4x2 =++−−−⇒ r

53

6r

53

5

53

1x2 =

+=⇒

2

yr

5

1x ==⇒ x2y =⇒ , which is the required proof.

Now, ( ) 2222

22

2

2

yr

yyr

yr

r yr

y

S

+×π−

+

π+=

∂

∂ ( )( ) 2/322

222

yr

yyrr

+

−+π=

( )0

yr

r 2/322

3

>+

π= (iv)



21

22

22

222

2242

2

yr

yr

r yrr 2

yr

r 2r2

r

v2

r

S

+

+

π−+π

++

π+×=

∂

∂

( )0

yr

yr 2r

yr

r 2

r

v42/322

23

223>

+

π+π+

+

π+= (v)

∂

∂

∂

∂=

∂∂

∂

y

S

r

S

ry

S2

+

π+

π−

∂

∂=

22 yr

yr

3

r 2

r

S

22

22

22

yr

yr

ryr y yr

3

r2

+

+×π−π×+

+−

=( )( ) 2/322

222

yr

yr yry

3

2

+

π−+π+

π−=

( ) 2/322

3

yr

y

3

2

+

π+

π−= .

Let 2

2

y

Sr

∂

∂= ,

2

2

r

St

∂

∂= and

ry

Sf

2

∂∂

∂=

2ft r −( ) ( ) ( )

2

2/322

2

2/322

22

2232/322

3

yr

y

3

2

yr

yr 2r

yr

r 2

r

v4

yr

r

+

π+

π−−

+

π+π+

+

π+×

+

π=

at r5

2y =

2ft r −( )

+

π+π

++

π+π+

+

π+×

+

π=

3

22

22

2/322

22

2232/3

22

3

r5

4r

r5

42rr

yr

yr 2r

r5

4r

r 2

r

v4

r5

4r

r

+

×π

+π

−2/3

22

3

r5

4r

r55

4

3

2

v at r5

2y = and 32 r

3

5r

53

2

5

1r

5

rx π=

+π== .



22

2ft r −

π+

π−−

π+π+π=

27

4

3

2

3

513

3

52

3

54

27

55

027

14

3

519

27

55>

π+

π= (vi)

From (iv), (v) and (vi), we find that the conditions satisfied from these equations is for

minima.

∴For minimum canvas to be used x2y = . Ans.

Q.No.13.: Find the stationary value of 222 zyxu ++= . If 2a3zxyzxy =++ .

{Without Lagrange Theorem}

Sol.: 222 zyxu ++= (i)

2a3zxyzxy =++ (ii)

From (ii), we get

yx

xya3z

2

+

−= (iii)

Putting above value of z in (i), we get

22

22

yx

xya3yxu

+

−++= .

( )( ) ( )( )( )

+

−−−+

+

−+=

∂

∂2

22

yx

xya31y.yx

yx

xya32x2

x

u

( )( )

( )32222

yx

xya3yxyya32x2

+

+−−−−+=

( )( )( )3

2222

yx

a3yya32x2

+

+−−=

( )( )3

44

yx

ya92x2

+

−−= (iv)

and ( ) ( )

( )

+

−−+−

+

−+=

∂

∂2

222

yx

xya3yxx

yx

ya32y2

y

u

( )( )3

44

yx

xa92y2

+

−−= (v)



23

Now putting y

u0

x

u

∂

∂==

∂

∂

( )( )

0yx

ya92x2

3

44

=+

−−

( ) ( ) 0ya9yx3xy3yxx 442233 =−−+++⇒

0a9xy3yx3xyyx 4322344 =−++++⇒ (vi)

Similarly 0a9yx3yx3yxyx 4322344 =−++++ (vii)

Subtracting (vi) and (vii), we get

0xy3yx3yxxy 3333 =−+− ( ) 0yxyx 22 =−⇒

( )( ) 0yxyxyx =+−⇒

Now, Case 1. 0x = . From (vi), we get

a3ya9y 44 =⇒= .

Now putting above value in (iii), we get

a3a3

a3z

2

== .

Now, 22222 a3a30zyx ++=++ 2222 a6zyx =++⇒ (viii)

Case 2. y = 0. From (vi), we get

a3ya9y 44 =⇒= .

and from (iii), we get a3z =

Now from (i), we get

Now, 22222 a30a3zyx ++=++ 2222 a6zyx =++⇒ (ix)

Case 3. Either yx −= .

It is not possible as it does not satisfy (iii)

Case 4. If x = y.

From (vi), we get

0a9x3x3xxx 444444 =−++++ 44 a9x9 =⇒ .

Using (iii), we get aa2

aa3z

22

=−

= .



24

azyx ===∴ (x)

Now, according to definition of stationary value f(a, b) is said to be stationary value of

f(x, y).

If [ ] 0b ,afx = and [ ] 0b ,afy =

i. e. the function is stationary at [a, b].

Stationary value of function will be

222 zyxu ++= 2222 a3aaau =++=⇒ (xi)

From (viii), (ix) and (xi) stationary value of u are 2a6 and 2a3 . Ans.

Q.No.14.: A rectangular box open at the top is to have volume equal to 32 cubic ft. Find

the dimensions of the box requiring least material for its construction.

Sol.: Let x , y and z ft. be the dimensions of the rectangular box open at the top so that

material for construction will be least if surface area is least.

Let surface area, S = F(x, y, z) = xy + 2yz + 2zx . ....(i)

Also given volume = xyz = 32. ....(ii)

Eliminating z from (i) with the help of (ii), we get

( )S xy y xxy

xyx y

= + + = + +

2

3264

1 1 .

∴ 0x

64y

x

S2

=−=∂

∂ and 0

y

64x

y

S2

=−=∂

∂.

Solving these, we get x = y = 4.

Now 32

2

x

128

x

Sr =

∂

∂= , 1

yx

Ss

2

=∂∂

∂= ,

32

2

y

128

y

St =

∂

∂= .

At x = y = 4, =−×=− 1)22(srt 2 + ve and r is also + ve.

Hence S is minimum for x = y = 4. Then from (ii), z = 2.

Hence the dimensions of the box are x = 4ft.,y = 4ft. and z = 2ft.Ans.

Q.No.15.: Prove that maximum value of 8

1CcosBcosAcos = .

Sol.: CcosBcosAcos ( )[ ]BAWcosBcosAcos +−=

)BAcos(AcoaBcos +−= [ ]WCBA =++∵



25

( )BAcosBcoscoaA)B,A(f +−=

[ ])BAsin(.Acos)BAcos(.AsinBcosfA +−+−−=

[ ])BAsin(.Acos)BAcos(.AsinBcos +++= )BA2sin(Bcos +=

[ ])BAsin(.Bcos)BAcos(.BsinAcosfB +−+−−=

[ ])BAsin(.Bcos)BAcos(.BsinAcos +++= )B2Asin(Acos +=

)BA2cos(.Bcos2fr AA +==

)B2Acos(Acos)B2Asin(Asinfs AB +++−== ( ) )B2A2cos(B2AAcos +=++=

)B2Acos(Acos2ft BB +== .

For maximum and minimum values 0fA = and 0fB =

0)BA2sin(Bcos =+

0)B2Asin(Acos =+

From above equations, we get the following four pairs of equations

(i) 0Bcos = , 0Acos = , 2

WB = ,

2

WA = .

It is not possible as two angles in same triangle cannot be °90 .

(ii) 0Bcos = , 0)B2Asin( =+ , 2

WB = , 0AWB2A =⇒=+ .

Not possible as in same triangle no angle can be zero.

(iii) Similarly, we reject pair

( ) 0BA2sin =+ , 0Acos = ,

( ) 0BA2sin =+

0BA2 =+ (i)

0B2A =+ (ii)

Solving (i) and (ii), we get

3

WBA == ,

3

WBA ==

1Wcos3

Wcos2r −== ,

2

1

3

WWcos

3

W4coss

−=

+== ,



26

1Wcos3

Wcos2t −== .

04

3srt 2 >>− and 0r < .

Hence f(A, B) is maximum at 3

WBA == .

and maximum value 8

1

3

W2cos

3

Wcos

3

Wcos =−= . Ans.

Q.No.17.: If x, y, z be the perpendicular from any point within the triangle on the sides

a, b, c of a triangle ABC of area ∆ . Show that the minimum value of

222

2222

cba

4zyx

++

∆=++ .

Sol.: ∆ = Area of the triangle ABC = area of three small triangles with in the ABC∆ .

cz2

1by

2

1ax

2

1++=∆

∆=++∴ 2czbyax (i)

Let 222 zyxu ++= (ii)

From (i) ( )byax2c

1zbyax2cz −−∆=⇒−−∆=

Put the value of z in (ii), we get

( )22

22 byax2c

1yxu −−∆++= (iii)

Now differentiate (iii) w. r. t. x, we get

( )( )abyax2c

2x2

x

u2

−−−∆+=∂

∂.

For max. or min. 0x

u=

∂

∂

( )( ) 0abyax2c

2x2

2=−−−∆+

byax2a

xc2

−−∆= , (iv)

Now, again differentiating w. r. t. y, we get



27

( )( ) 0bbyax2c

2y2

y

u2

=−−−∆+=∂

∂

byax2b

yc2

−−∆= (v)

From (iv) and (v), we get

xa

by

b

yc

a

xc 22

=⇒= (vi)

Put the value of y in (iv), we get

xa

bbax2

a

xc2

−−∆= xbxaa2xc 222 −−∆=⇒

222 cba

a2x

++

∆=⇒

Putting the value of x in (vi), we get

a

b

cba

a2y

222×

++

∆=

222 cba

b2y

++

∆=⇒

Similarly, 222 cba

c2z

++

∆=

Putting these values of x, y, z in (ii), we get

222 zyxu ++=

2

222

2

222

2

222 cba

c2

cba

b2

cba

a2

++

∆+

++

∆+

++

∆=

( )( )2222

2222

cba

cba4

++

++∆=

( )222

2

cba

4

++

∆= . Ans.

Q.No.18.: A rectangular strip b×� of metal is bent up at the sides to form a trough.

Without ends. Find the width of the side bases and the angle through which the

side must bent so that the trough may have a maximum capacity.

Sol.: Let the length � of the strip is bent up by an angle of θ , the length of bent part of

strip are of equal on both sides by x.

Let h be the height and y be the base.

θ=⇒=θ∴ sinxhx

hsin , θ=⇒=θ cosxh

x

ycos .



28

Let F be the capacity of the trough.

( ) byh2

12hbx2F ××××+×−= � ( ) b.cosx.sinxsinxbx2 θθ+θ××= �

( ) θ+θ−= 2sinb2

xsinx2xb

22

� .

Now, ( )2

2sinb2sinx4b

x

FFx

θ×+θ×=

∂

∂= � ( ) θ+θ×= 2sinbxsinx4b �

( ) θ+θ−=∂

∂= 2sinbsin40b

x

FF

2

2

xx θ+θ−= 2sinbsinb4

( ) θ+θ−=θ∂

∂=θ 2cos2.

2

bxcos.x2xb

FF

22

� ( ) θ+θ−= 2cosbxcos.x2xb 22�

( ) θ+θ−−=θ∂

∂=θθ 2sinbx2sin.x2xb

FF 22

2

2

�

( ) θ+θ−=θ∂∂

∂=θ 2cosbx2cos.x4b

x

FF

2

x � .

For x be max. or min.

0Fx = , 0F =θ

0Fx = ( ) θθ−=θ−⇒ cossin2bxsinx4b � .

( )bx2

x4bcos

�−=θ .

Also, 0F =θ

( ) ( ) 01cos2bxcos.x2xb 222 =−θ+θ−⇒ �

( ) ( ) ( )0bx

bx4

x4bbx2

bx2

x4b.x2xb 2

22

2222 =−

−×+

−−⇒

��

( )( ) ( )0bx

2

x4b

x2

x4x2bx 22

=−−

+−−

⇒��

( )( ) ( ) 2

2

bx2

x4b

2

x4x2b =

−+

−−⇒

��

( ) ( ) ( )[ ] 2xx4x22

x4=−+−

−⇒ ��

� ( ) 2x

2

x2.x4 =−⇒ � ( )[ ] 0x4xx2 =−−⇒ �

( )[ ] 0x4xx =−−⇒ �

As 0x ≠



29

( ) 0x4x0x4x =+−⇒=−= ��3

xx3�

� =⇒=⇒ .

When 3

x�

= .

60cos2

1

2

3

23

34

3.b2

34b

cos ===

−

=

−×

=θ�

�

�

��

�

��

.

°=θ∴ 60 .

When °=θ 60 , 3

x�

= .

Now θ+θ−= 2sinbsinb4Fxx °+°−= 120sinb60sinb42

3bx

2

3b4 ×+×−=

2

3b3b2 +−=

2

b 33

2

3b3

2

3b3b4 −=

−=

+−= .

( ) θ−θ−−=θθ 2sinbx2sinx2xbF 22� °−°

×−−= 120sin

9.b260sin

92

3.b

22��

�

2

3

9

b2

2

3

9

2

3b

222

×−×

−−=

��

2

3

9

b2

2

3

9

23b

222

×−

−−=

��

18

b3

18

b33

18

b32

18

3b 2222�� −

=−

==−

=

( ) θ+θ−=θ 2cosbx2cos.x4bFx � °+°×

−−×= 120cos

3.b260

3.4b

��

2

1

3

b2

2

1

3b ×−×

−×=

��

2

b

6

b3

6

b2b �� −=

−=

−−= .

2xxx FFF θθθ −×

22

2

b

6

b3

2

b33

−−

−×

−=

��

4

b

62

b33 2222��

−×

×=

4

b2

4

bb3 222222��

=−

=

02

b 22

>�

.



30

Also 02

b 33<

−.

Hence the length have maximum capacity when 3

x�

= , 3

60π

=°=θ .

Hence width of bases 3

�= and angle

3

π.

Q.No.19.: Divide 24 into three parts such that the continued product of the first, the

square of 2nd

and the cube of the third may be a maximum.

Sol.: Let the three parts be x, y and ( )[ ]yx24 +−

And let, ( )[ ] 32xyyx24f +−=

Then, ( )[ ] ( ) 3222 xy1yx24yx3x

f−++−=

∂

∂ 322322 yx3yx4yx72 −−= (i)

2

2

x

f

∂

∂ 3222 xy6yx12xy144 −−=

and [ ] ( ) 323 xy1yx24yx2y

f−+−−=

∂

∂ 3243 xy3yx2yx48 −−= (ii)

2

2

y

f

∂

∂ 343 yx3x2x48 −−=

From (i) we can get 22322

yx9yx8yx144yx

f−−=

∂∂

∂ (iii)

F has maximum and minimum, then 0x

f=

∂

∂ and 0

y

f=

∂

∂

From (i), we get

0yx3yx4yx72 322322 =−−

=

∂

∂0

x

f∵

72y3x40y3x472 =+⇒=−−⇒ (iv)

From (ii), we get

0xy3yx2yx48 3243 =−−

=

∂

∂0

y

f∵

48y3x20y3x248 =+⇒=−−⇒ (v)

Using (iv) and (v), we get



31

12x = and 8y =

Now f having maximum and minimum if at 12x = and 8y =

0yx

f

y

f.

x

f2

2

2

2

2

2

>

∂∂

∂−

∂

∂

∂

∂

( ) ( ) ( ) ( )[ ] ( ) ( ) ( )[ ]343322212861221248812681212812144 ×−−××−××−××⇒

( ) ( ) ( )[ ] 08)12(981288121443232 >××−××−××−

01528754688 >⇒

So, f having maximum and minimum at 12x = and 8y =

Now f has maximum value if 0x

f

8y12x

2

2

<

∂

∂

==

( ) ( ) ( ) ( )[ ] 0368648126812128121443222 <−=××−××−××⇒

So f having maximum and minimum if at 12x = and 8y = .

So the required three parts are ( )[ ] 4 8, ,12812-24 8 ,12 =+ . Ans.

Q.No.20.:. Examine ( ) xy3yxxf 33 ++= for maximum and minimum values.

Sol.: For maxima and minima values. Solve two equations

0x

f=

∂

∂ and 0

y

f=

∂

∂

i. e. 0y3x3 2 =+ and 0x3y3 2 =+

Solve these equations, we get

0yx2 =+ and 0xy2 =+

1 ,0y0yy4 −=⇒=+⇒ .

Similarly, 1 ,0y −= .

So, we get four points (0, 0), ( ),1 ,0 − 0) ,1(− and )1 ,1( −− .

For maxima or minima 2nd

condition is

22

2

2

2

2

yx

f

y

f.

x

f

∂∂

∂>

∂

∂

∂

∂

4

1xy9y6.x6 >⇒>⇒ .



32

Only point )1 ,1( −− satisfies this condition .

Now for maxima 0fxx < at point )1 ,1( −− will ⇒ that this is maxima.

06x6fxx <−== at )1 ,1( −− . So this point is a maximum value and no minimum

value of function.

Now, condition for saddle point.

i. e. ( )4

1xyff.f

2xyyyxx <⇒< .

Three points (0, 0), ( ),1 ,0 − 0) ,1(− satisfies the above equation.

Therefore these points are saddle points for the curve xy3yx 33 ++ .

Q.No.21.: Find the shortest and longest distance of the point )1 2, ,1( − to the sphere

24zyx 222 =++ .

Sol.: Let us consider a point A(x, y, z) on the surface of the sphere.

( ) ( ) ( )2221z2y1xAP ++−+−=∴

Let, ( ) ( ) ( )2221z2y1xF ++−+−= 1z2z4y4y1x2x 222 ++++−++−=

6z2y4x2z2yx 22 ++−−++= (i)

Now the given equation of sphere is 24zyx 222 =++ (ii)


6z2y4x224F ++−−= z2y4x230 +−−=

also from (ii), we get 22 yx24z −−±=

When 22 yx24z −−= i. e. positive, then

22 yx242y4x230F −−+−−=

( ) ( )2

yx24

x2

yx242

x222F

2222x −

−−

−=

−−

−+−=∴ (iii)

( ) ( )4

yx24

y24

yx242

y22F

2222y −

−−

−=−

−−

−= (iv)

For maximum or minimum.

0Fx = and 0Fy =



33

( ) 22

2222yx24x2

yx24

x202

yx24

x2−−=−⇒=

−−

−⇒=−

−−

−

Squaring both sides, we get

24yx2yx24x 22222 =+⇒−−= (v)

( ) 22

22yx242y04

yx24

y2−−=−⇒=−

−−

−


( ) 96y5x4yx244y 22222 +⇒−−= (vi)

Multiplying (v) by 2 and subtracting from (vi), we get

48y3 2 = 4y16y2 ±=⇒=⇒ .

But from (iv) y should be negative

4y −=∴

2416242

1x ==−= .

But from (iii) x should be negative. 4x −=∴

( ) ( ) 24164244224z22 ==−−=−−−−=∴ .

Hence the points are ( )2 ,4 ,2 −− .

Now

( ) 2/322

2

xx

yx24

y248F

−−

+−= .

xxF at ( )2 ,4 ,2 −−( )

28

16

16424

32482/3

−=−

=−−

+−= .

yyF at ( )2 ,4 ,2 −−( )

58

848

16x24

x2482/32

2

−=+−

=−−

+−= .

xyF at ( )2 ,4 ,2 −−( )

28

82

16x24

xy22/32

−=×−

=−−

−= .

For maxima or minima

( ) 0FF.F2

xyyyxx >−

( )( ) ( ) 06041002522 >⇒>−⇒>−−−−⇒ , which is true.



34

Also 02Fxx <−= .

( )2 ,4 ,2 −−∴ are the points of maxima for F.

( ) ( ) ( )222.max 122412F ++−−+−−= 549369 =++ .

When F is maximum then AP is also maximum.

54AP .max =∴ .

Now, when 22 yx24z −+−= i. e. negative.

Then, 22 yx242y4x230F −−−−−=

2yx24

x2F

22x −

−−=∴ (vii)

4yx24

y2F

22y −

−−= (viii)

For max. or min.

0Fx = and 0Fy =

( ) 22

2222yx24x2

yx24

x202

yx24

x2−−=−⇒=

−−

−⇒=−

−−

−


24yx2yx24x 22222 =+⇒−−=

( ) 22

22yx242y04

yx24

y2−−=−⇒=−

−−

−


( ) 96y5x4yx244y 22222 +⇒−−=

Solving above equations, we get

2x ±= and 2y ±=

But from (vii) and (viii) x and y should be positive

2x =∴ and 2y = and 2z −= .

Hence the points are ( )2 4, ,2 −



35

xxF at ( )2 ,4 ,2 −( )

−−

−=

2/322

2

yx24

y242 .

( )2

16424

16242

2/3=

−−

−=

yyF at ( )2 ,4 ,2 −( )

5

yx24

x242

2/322

2

=

−−

−= .

xyF at ( )2 ,4 ,2 − 28

16

yx24

xy222

==−−

.

Now, for max. or mini.

( ) 0FF.F2

xyyyxx >−

060410 >⇒>−∴ , which is true.

Also, 02Fxx >= .

( )2 ,4 ,2 −∴ are the point of minimum for F.

when F is minimum AP is also minimum

( ) ( ) ( ) 6122412AP222

.min =+−+−+−=∴ .

∴The longest distance of point )1 2, ,1( − from sphere 24zyx 222 =++ is 54 and the

shortest distance is 6 . Ans.

Q.No.22.: Find the extreme values of 222 zyx ++ , when Pczbyax =++ .

Sol.: A maximum or minimum value is called extreme value of the function.

Let 222 zyxu ++= (i)

And also Pczbyax =++

c

byaxPz

−−=∴

Putting in (i), we get

( )22

22 byaxPc

1yxu −−++= (ii)

Differentiate partially (ii) w. r. t. x, we get

( ) abyaxPc

2x2

x

u2

−×−−+=∂

∂

Differentiate partially (ii) w. r. t. y, we get



36

( ) bbyaxPc

2y2

y

u2

−×−−+=∂

∂

for max. and mini. Values both

0x

u=

∂

∂ and 0

y

u=

∂

∂

( )( ) 0byaxPac

2x2

2=−−−+ (iii)

( ) 0byaxPc

b2y2

2=−−− (iv)

Solving (iii) and (iv), we get

22 ac

abyaPx

+

−=

Put in (iv), we get

0ybac

abyaPabbPyc 2

22

2 =+

+

−+− ( )( ) ( ) ( )222222 acbPabyaPabybcac +=−+++⇒

222 cba

bPy

++=∴ ,

and also 222 cba

aPx

++= ,

222 cba

cPz

++= .

Now ( )ac

a22

x

u22

2

−−=∂

∂2

2

c

a22 += and

2

2

y

u

∂

∂2

2

c

b22 += ,

22

2

c

ab

yx

u=

∂∂

∂

22

2

2

2

2

y.x

u

y

u.

x

u

∂∂

∂−

∂

∂

∂

∂∴

( )0

c

ba

c

ab24

4

22

2

22

>−+

+⇒

Clearly 0y.x

u

y

u.

x

u2

2

2

2

2

2

>

∂∂

∂>

∂

∂

∂

∂

Also 0c

b22

x

u2

2

2

2

>+=∂

∂.

∴u has minimum value at 222 cba

aPx

++= ,

222 cba

bPy

++= and

222 cba

cPz

++= .

And minimum value of 222 zyx ++



37

( ) ( ) ( ) 222

2

2222

22

2222

22

2222

22

cba

P

cba

Pc

cba

Pb

cba

Pa

++=

+++

+++

++= . Ans.

Q.No.23.: Find the maximum and minimum values of

x72y15x15xy3x)y,x(f 2223 +−−+= .

Sol.: Differentiating f partially w.r.t. x and y, we get

72x30y3x3x

ff 22

x +−+=∂

∂= ,

y30xy6y

ff y −=

∂

∂= .

The stationary (critical) points are given by fx = 0 and fy = 0.

From 0y30xy6f y =−= 0)5x(y6 =−⇒ .

Thus, either 0y = or x = 5.

Since 072x30y3x3f 22x =+−+= .

For y = 0, 072x30x3 2 =+− ⇒ x = 6 or 4.

For x = 5, 072150y375 2 =+−+ 1y ±=⇒

Thus, the four stationary points are given by

(6, 0), (4, 0), (5, 1), )1 ,5( − .

To determine the nature of these points, calculate fxx, fyy and fxy, we obtain

30x6Afxx −== , y6Bfxy == , 30x6Cf yy −== .

Thus, ( ) ( )[ ]22222 y5x36y3630x6BAC −−=−−=−

1. At the stationary point (6, 0), we have 063036A >=−= and 036BAC 2 >=− .

So (6, 0) is a minimum point of the given function f and the minimum value of f at (6,

0) is 1086.7236.15063 =+−+ .

2. At (4, 0): 063024A <−=−= and 036BAC 2 >=− . So a maximum occurs at the

point (4, 0) and the maximum value of f at (4, 0) is 112.

3. At (5, 1), A = 0, 036BAC 2 <−=− . So (5, 1) is a saddle point (It is neither

maximum nor minimum).



38

4. At )1,5( − , A = 0, AC = B2 036 <−= . So )1,5( − is neither a maximum nor a

minimum (it is a saddle point).

Q.No.24.: Find the shortest distance from origin to the surface 2xyz2 = .

Sol.: Given equation of surface is 2xyz2 = .

Let d be the distance from origin (0, 0, 0) to any point (x, y, z) of the given surface, then

( ) ( ) ( )2220z0y0xd −+−+−= 2222 zyxd ++=⇒ .

Eliminating z2 using the equation of the surface 2xyz2 = . So replace

xy

2z

2 = .

)y,x(fxy

2yxd

222 =++=∴ .

Now yx

2x2f

2x −= , 2y

xy

2y2f −= .

The stationary (critical) points are given by fx = 0 and fy = 0.

Solving fx = 0 and fy = 0, we get

0yx

1yx

2

3

=−

and 0xy

1xy

2

3

=−

33 xy1yx == ( ) 0yxxy 22 =−⇒ .

Since 0x ≠ , 0y ≠ , so 1yx =±= .

Thus, the two stationary points are (1, 1) and ( )1,1 −− .

Now yx

42f

3xx += , 3yy

xy

42f += ,

22xyyx

2f = .

At (1, 1): 06fxx >= , 03266.6ff.f 2xyyyxx >=−=− .

At ( )1,1 −− : 32fff.6f 2xyyyxxxx =−+= .

So minimum occurs at )2 ,1 ,1( and )2 ,1 ,1( −− .

Hence the shortest distance is ( ) 24211222 ==++ . Ans.

Q.No.25.: The temperature T at any point (x, y, z)in space is T(x, y, z) = kxyz2, where k

is a constant. Find the highest temperature on the surface of the sphere



39

2222 azyx =++ .

Sol.: Given the temperature T at any point (x, y, z) in space is T(x, y, z) = kxyz2.

Eliminating the variable z, using 2222 yxaz −−= , we get

)y,x(F)yxa(kxykxyz)z,y,x(T 2222 =−−== .

Now )yx3a(kyF 222x −−= , )y3xa(kxF 222

y −−= .

The stationary (critical) points are given by Fx = 0 and Fy = 0.

0)yx3a(ky 222 =−− , 0)y3xa(kx 222 =−−

⇒ 222 ayx3 =+ , 222 ay3x =+ and x =0, y = 0

Solving, we get 2

ayx ±== .

Also kxy6Fxx −= , kxy6Fyy −= , ( )222xy y3x3akF −−= .

At (0, 0), yyxx F0F == , 2xy kaF = .

0ka0.0 2 <−∴ ⇒ (0, 0) is a saddle point.

At both the points

2

a,

2

a and

−−

2

a,

2

a,

04

ka6F

2

xx <−= and 0ak24

kaak

4

9FF.F 42

24422

xyyyxx >=−=− .

T∴ attains a maximum value at both these points

2

a,

2

a and

−−

2

a,

2

a.

The maximum value of T is 8

ka

2

a

4

a.k

422

=

. Ans.

Q.No.26.: Find the shortest distance between the lines

1

7z

2

5y

1

3x −=

−

−=

−

and 1

1z

6

1y

7

1x +=

−

+=

+

Sol.: The given equations of lines are



40

1

7z

2

5y

1

3x −=

−

−=

− (i)

and 1

1z

6

1y

7

1x +=

−

+=

+ (ii)

Equating each of the fractions of (i) to λ , we get λ+= 3x , λ−= 25y , λ+= 7z .

Thus any point P on the first line (i) is given by

( )λ+λ−λ+ 7 ,25 ,3 .

Similarly, any point Q on the second line (ii) is ( )µ+−µ−−µ+− 1 ,61 ,71 .

The distance between the given

( ) ( ) ( )222176125713PQ µ−+λ++µ++λ−+µ−+λ+= .

Consider ( ) 10540866PQ),(f 222 +λµ−µ+λ==µλ .

The problem is to find minimum value of f as a function of the two variables µλ, .

µ−λ=λ 4012f , λ−µ=µ 40172f .

Solving 04012 =µ−λ and 040172 =λ−µ , we get

0=λ and 0=µ as the only stationary point.

12f =λλ , 172f =µµ , 40f −=λµ .

Now 012f >=λλ and 0fff 2 >− λµµµλλ , a minimum occurs at 0 ,0 =µ=λ .

The minimum (shortest) distance is given by

292116864PQ 222 ==++= . Ans.

Home Assignments

Q.No.1.: Test the functions for maxima, minima and saddle points:

a. 1yxyx 2244 +−−+ .

b. 2yz2xy2z3y2x 222 −−−++ .

Ans.: a. Maximum at (0, 0)

Maximum value is 1.



41

Minima at four points ( )21/ ,2/1 ±±

Minimum value at these 4 points is 2

1.

Saddle points at four points ( ) ( )0 ,21/ ,2/1 ,0 ±±

b. Maximum at (1, 1), minimum at ( )1 ,1 −− .

Q.No.2.: Find the extrema of f(x, y) : ( ) 2x2x622 eyx ++ .

Ans.: Minima at (0, 0) (minimum value 0) and at ( )0 ,1− (minimum value )e 4− .

Saddle point at

− 0 ,

2

1.

Q.No.3.: Examine the following function f(x, y) for extrema : )yxsin(ysinxsin +++ .

Ans.: Maximum at

ππ

3 ,

3, maximum value

2

33.

Q.No.4.: Find the shortest distance from the origin to the plane 3z2y2x =−− .

Ans.: Shortest distance is 1 (from 0, 0, 0) to the point

−−

3

2 ,

3

2 ,

3

1 on the plane.

Q.No.5.: Find the shortest distance between the lines

2

5z

2

6y

3

2x

−

−=

−

−=

−

and 6

8z

1

3y

2

5x −=

−=

−.

Ans.: Shortest distance is 3 between the points (5, 4, 3) and (3, 2, 2).

Q.No.6.: If the perimeter of a triangle is constant, prove that the area of this triangle is

maximum, when the triangle is equilateral.

Ans.: Maximum when 3

s2cba === .

Q.No.7.: Find the volume of the largest rectangular parallelopiped with edges parallel to

the axes, that can be inscribed in the:

a. Sphere

b. Equation of the ellipsoid is 36z9y4x4 222 =++ .



42

Ans.: a. Volume : 3

azy x,

33

a8 3

===

b. Volume: 316 (with a = 3, b = 3, c = 2).

Q.No.8.: Find the dimensions of a rectangular box, with open top, so that the total surface

area of the box is a minimum, given that the volume of the box is constant say

V.

Ans.: x = y = 2z = (2V)1/3

.

Q.No.9.: Find the dimensions of the rectangular box, with open top, of maximum

capacity whose surface area is 432 sq. cm.

Ans.: 12, 12, 6.

Q.No.10.: If the total surface area of a closed rectangular box is 108 sq. cm., find the

dimensions of the box having maximum capacity.

Ans.: 18 ,18 ,18 .

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Documents

2. Maxima and Minima in Case of Two or More Variables