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Maxima and Minima

Problems on 2 Variables

Solved Problems

Q. 1 : Examine the maximum and minimum on the surface : z = x2 + y2 – 10x – 2y + 36 and find the minimum value of z. Soln.: Tips : z = x2 + y2 – 10x – 2y + 36 For maxima–minima, on the surface, we have z y

z x

= 0,

z x

= 2x – 10 = 0

… (1)

z y

= 2y – 2 = 0

… (2)

=0

Solving equation (1) and equation (2), 2x – 10 = 0 or x = 5 2y – 2 = 0 or y = 1 Thus the point to be examined is (5, 1) Now let us find r, s, t from equations (1) and (2) r=

2z x 2

s=

2z x y

t=

2z y 2

=2 =0 =2

 rt – s2 = 2(2) – 02 = 4 > 0 Further r = 2 > 0 Hence there is a minimum at (5, 1). 1

Maxima and Minima

Problems on 2 Variables

Solved Problems

To find this minimum value put x = 5, y = 1 in the given expression for z. z = x2 + y2 – 10x – 2y + 36

i.e.

Tip 6

zmin= 25 + 1 – 50 – 2 + 36 = 10 Ans.:

Minimum value at (5, 1) is 10 units.

Q. 2) : Examine the maximum and minimum on the surface : z = x3 + 3xy2 – 15y2 – 15y2 + 72x Soln.: z = x3 + 3xy1 – 15x – 15y2 + 72x For maximum-minimum on the surface



z y

z x

= 0,

z x

= 3x2 + 3y2 – 30x + 72 = 0 z y

=0

= 6xy – 30y = 0

From equation (2) we get 6y(x – 5) = 0 or y(x – 5) = 0 

y = 0 and x = 5

From equation (1), when y = 0 3x2 – 30 x + 72 = 0 

3(x – 4) (x – 6) = 0



x = 4, x = 6

Thus the points are (4, 0), (6, 0)

2

…

(1) …

(2)

Maxima and Minima

Problems on 2 Variables

Solved Problems

Again for x = 5, from equation (1) 75 + 3y2 – 150 + 72 = 0 or y2 = 1 

y=1



The points are (5, 1), (5, –1)

Thus the points to be examined are (4, 0), (6, 0), (5, 1), (5, –1) rt – s2 = (6x – 30)2 – 36 y2 = 36 {(x – 5)2 – y2} (a) at (4, 0) rt – s2 = 36 {1} > 0 and r = 24 – 30 = –6 < 0 

There is a minimum.

(b) at (6, 0) rt – s2 = 36 (1) > 0 and r = 36 – 30 > 0 

There is a minimum.

(c) at (5, 1) rt – s2 = 36 (–1) < 0 

no maximum and no minimum.

(d) at (–5, –1) rt – s2 = 36 (–1) < 0  Ans.:

No maximum and no minimum. (a) Maximum at (4, 0),

(b) Minimum at (6, 0), (c) No maximum and no minimum at (5, 1), (d) No maximum and no minimum at (5, –1). Q. 3) : Examine the maximum and minimum on the surface : z = x3 + y3 – 3axy. (Dec. 2003) Slon.: 3

Maxima and Minima

Problems on 2 Variables

Solved Problems

z = x3 + y3 – 3axy For maxima-minima,

z y

= 0,

z y

=0

z x

= 3x2 – 3ay = 0 or x2 – ay = 0

z y

= 3y2 – 3ax = 0 or y2 – ax = 0…

… (1) (2)

Multiply equation (1) by x and equation (2) by y and subtract to get x3 – axy = 0 y3 – axy = 0 or

x3 – y3 = 0 or x = y



From equation (1), x2 – ax = 0

or

x (x – a) = 0

So that x = 0, x = a 

from equation (1), y = 0, y = a

The points to be examined are (0, 0) and (a, a) Now r =

2z x 2

s=

2z x y

t=

2z y 2

= 6x = –3a = 6y

 rt – s2 = 6s.6y – (–3a)2 = 36xy – 9a2 (a) at (0, 0), rt – s2 = –9a2 < 0 which means there are no maximum and no minimum at the point (0, 0) 4

Maxima and Minima

Problems on 2 Variables

Solved Problems

(b) at (a, a) rt – s2 = 36a2 – 9a2 > 0 and r = 6a If a > 0, r > 0 and there is a minimum at (a, a) a < 0, r < 0 and there is a maximum at (a, a) Tip 6 Ans.:

(a) No maximum and no minimum at (0, 0).

(b) Maximum at (a, a) if a < 0 and Minimum at (a, a) if a > 0. Q. 4) : Examine the maximum and minimum on the surface. z = x3 + 3xy – 3x2 – 3y2 + 4 Soln. : Tips. : z = x3 + 3xy2 – 3x2 – 3y2 + 4 For maxima-minima, z = 0, x

z x

Now,

z y

=0

= 3x2 + 3y2 – 6x = 0 = x2 + y2 – 2x = 0 … (1)

z y

= 6xy – 6y = 0 = y(x – 1) = 0



… (2)

y = 0 or x = 1

For y = 0 from x2 + y2 – 2x = 0, we get x2 – 2x = 0 or x(x – 2) = 0 or x = 0, x = 21  The points are (0, 0) or (2, 0) and for x = 1 from x2 + y2 – 2x = 0, we get y2 – 1 = 0 or y =  1 

The points are (1, 1), (1, –1) 5

Maxima and Minima

Problems on 2 Variables

Solved Problems

Thus the points to be examined are (0, 0), (2, 0), (1, 1), (1, –1) Now let us find r, s, t. r=

2z x 2

= 6x – 6, s =

2z x y

= 6y, t =

2z y 2

= 6x – 6

Thus rt – s2 = (6x – 6)2 – 36y2 (a) at (0, 0), rt – s2 = 36 > 0, and r = –6 < 0. Hence there is a maximum at (0, 0) (b) at (2, 0) rt – s2 = 6. 6 – 02 = 36 > 0, r = 6 < 0. Hence there is minimum at (2, 0) (c) at (1, 1), rt – s2 = 0 – 36 < 0. There is no maximum and no minimum (d) at (1, –1), rt – s2 = –36< 0. There is no maximum and no minimum at (1, –1). Ans.:

(a) Maximum at (0, 0).

(b) a minimum at (2, 0). (c) No maximum and no minimum at (1, 1). (d) No maximum and no minimum at (1, –1).

6

Maxima and Minima

Problems on 2 Variables

Solved Problems

Q. 5) : Examine the maximum and minimum on the surface : u = xy + a2

1 1    . x y

(Dec., 2005)

Soln.: 1 1    x y

u = xy + a2

For maxima, minima, we should have 

u x

= y + a2   x1  = 0

… (1)

u y

= x + a2   y1  = 0

… (2)



2



2





u x

= 0,

u y

=0

From equations (1) and (2), x2y = a2 = xy2 

x2y = xy2

or

x2y – xy2 = 0

or x = y, so that from x2y = a2 

 x = 0, y = 0

xy(x – y) = 0

x3 = a2 or x = a2/3

we get 

y = a2/3



The points to be examined are (0, 0), (a2/3, a2/3). Now, let us find r, s, t.



r=

 2u x 2

s=

 2u x y

t=

 2u y 2

=

2a 2 x3

=1 =

rt – s2 =

2a 2 y3 2a 2 x3

.

2a 2 y3

–1 7

Maxima and Minima

Problems on 2 Variables

Solved Problems

(a) at (0, 0), rt – s2 does not exist as it becomes , (b) at (a2/3, a2/3) 2a 3 x3

rt – s2 = r= 

2a 2 x3

=

.

2a 2 a2

2a 2 y3

–1 =

4 a4 a4

–1=4–13>0

= 2 > 0.

There is a minimum at (a2/3 , a2/3)

Ans.:(a)Maximum and minimum does not exist at (0, 0). (b) Minimum at (a2/3, a2/3). Q. 6) : A rectangular box, open at the top is to have a volume 108 m3. Find the dimensions of the box so that the surface area is minimum. (May 2003) Soln.: Tips : Let x, y, z be the length, breadth and height of the box. Then the volume of the box is given by v = xyz or 108 = xyz. Tip 1 

z=

108 xy

… (1)

If S is the surface area of the box, then S = xy + 2yz + 2zx, (since the top with sides x and y are open.) We want the minimum of S, for which S x

= 0,

S y

=0

Now from equation (1), eliminating the third variable z, we get 8

Maxima and Minima

Problems on 2 Variables

S = xy + 2(y + x) or

S = xy +





216 y

+

Solved Problems

108 xy

216 x

S x

= y–

216 x2

= 0 or x2y = 216

S y

= x–

216 y2

= 0 or xy2 = 216

x2y = xy2

or x2y – xy2 = 0 or xy (x – y) = 0 or x = 0, y = 0, x = y x = 0, y = 0 is not possible as the box can not be constructed 

x = y, which gives

From xy2 = 216, y3 = 216 or y = 6  x = 6 Thus the point to be examined is (6, 6) Now

r=

2S x 2

t= s= t= 

rt – s2 =

at (6, 6),

= (216x )(2)

2S y 2

3

= (216y )(2) 3

2S = x y

1

(216 )(2) x3

=

(216 )(2) y3

(216 )(2) (216 )(2) . y3 x3

rt – s2 =

(216 )(2) 63

–1 (216 )(2) 63

–1

= 2.2 – 1 = 4 – 1 = 3 > 0 and

r=

(216 )(2) 63

=

(216 )(2) 63

=2>0

Hence there is a minimum at (6, 6). 9

Maxima and Minima

Problems on 2 Variables

Solved Problems

Therefore the dimensions of the box are x = 6, y = 6, z =

108 x y

=

108 6 6

=3

Thus the sides of the box are 6 m, 6 m, 3 m with minimum surface area xy + 2yz + 2zx = 6.6 + 2.3 + 2.6.3 S = 36 + 36 + 36 = 108 sq.m. Ans.: m

Dimensions of the box are : 6, 6, 3 and S = 108 sq.

Q. 7) : Find the minimum distance from the origin of any point (x, y, z) on the plane z2 = 1 + xy. Soln.: Tips : Let P (x, y, z) be any point 

d(OP) =

(x  0)2  (y  0)2  (z  0)2

=

x2  y2  z2

If the distance is maximum or minimum, then its square or square root is maximum or minimum. Let the distance of any point from the origin be given by

x2  y2  z2

.

Let u = x2 + y2 + z2

… (1)

The point of intersection of this distance (line) with the plane can be obtained by eliminating z between the two equations. 

u = x2 + y2 + 1 + xy

For maxima-minima, we have 10

Maxima and Minima

Problems on 2 Variables

u y

u x

= 0,

u y

= 2x + y = 0

… (2)

u x

= 2y + x = 0

… (3)

Solved Problems

=0

Solving equation (2) and equation (3), we get x = 0, y = 0 



The point to be examined is (0, 0) r=

 2u x 2

=2

t=

 2u y 2

=2

s=

 2u x y

=1

rt – s2 = 2.2 – 1 = 3 > 0 and r = 2 > 0.

Therefore there is a minimum at (0, 0). 

x = 0, y = 0



from z2 = 1 + xy, z2 = 1

So that z =  1. Thus the points of minimum are (0, 0,  1)  The minimum distance from the origin of any point on the given plane is  2u x y

Ans.:

=

x2  y2  z2

=

0 2  0 2  12

Minimum distance 1 unit.

11

1 unit

Maxima and Minima

Problems on 2 Variables

Q. Find all stationary value of :F(x,y) = 𝒙𝟑 + 𝟑𝒙𝒚𝟐 − 𝟏𝟓𝒙𝟐 − 𝟏𝟓𝒚𝟐 + 𝟕𝟐𝒙 Fy = 6xy -30y Fy = 𝟑𝒙𝟐 + 𝟑𝒚𝟐 − 𝟑𝟎𝒙 + 𝟕𝟐 Now fx = fy = 0   

𝟑𝒙𝟐 + 𝟑𝒚𝟐 − 𝟑𝟎𝒙 + 𝟕𝟐 = 𝟎 6xy – 30y = 0 6y (x – 5) = 0 x=5,y=0

If x = 5 , y = ±1 Points are (5 , ±1) , (5 , -1) If y =0 (1)  𝟑𝒙𝟐 − 𝟑𝟎𝒙 + 72 = 0  x = 4, 6  Points are ( 4,0 ) , ( 6,0 ) r = fxx = 6x- 30 s = fxy = 6y t = fyy = 6x – 30 pts

r

(4,0) 6 0

0

36 > 0

{ f has max at (4,0) } (6,0)

6>0

6

{ f has min at (6,0) } 12

Solved Problems

Maxima and Minima

(5,1)

0

Problems on 2 Variables

Solved Problems

6

36 < 0

6

36 < 0

0

{ Sadle Point} (5, -1)

0

0

{ Sadle Point} fmax = (𝟒)𝟑 − 𝟏𝟓(𝟒)𝟐 + 𝟕𝟐 × 𝒙 = 112 fmin = (𝟔)𝟑 − 𝟏𝟓(𝟔)𝟐 + 𝟕𝟐 × 𝟔 = 108

Q . 1) Show that minimum value of :f(x,y) = xy + 𝒂𝟑 {𝟏𝒙 + 𝟏𝒚} 𝟑

𝒂 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: − 𝒇𝒙 = y - 𝒙𝟐 𝟑

𝒇𝒚 = x - 𝒂𝒚𝟐

𝒇𝒙 = 𝒇𝒚 =0 𝟑

y = 𝒂𝒙𝟐 ∴ x=

&

𝟑

x = 𝒂𝒚𝟐

𝒂𝟑 𝒙𝟒 𝒂𝟔

∴ 𝒙𝟑 = 𝒂𝟑 13

Maxima and Minima

Problems on 2 Variables

⇒x = a ⇒y=a ∴ Stationary point is (a,a) r = 𝒇𝒙𝒙 =

𝟐𝒂𝟑 𝒙𝟑

s = 𝒇𝒙𝒚 = 1 t = 𝒇𝒚𝒚 =

𝟐𝒂𝟑 𝒚𝟑

at (a,0) r=2>0 s=1 t=2 rt - 𝒔𝟐 = 4 – 1 = 3 > 0 ∴ 𝒇 𝒊𝒔 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝒂𝒕 (𝒂, 𝒂) & 𝒇𝒎𝒊𝒏 = 𝒂𝟐 +𝒂𝟑 {𝒂𝟏 + 𝒂𝟏} = 𝟑𝒂𝟐

14

Solved Problems

Maxima and Minima

Problems on 2 Variables

Solved Problems

Q . 2) Divide 120 into three parts so that sum of their products taken two at a time shall be maximum. Solution:- Let x, y, z be three parts of 120 i.e x + y + z = 120 function to be maximized is f(x, y, z) = xy + yz + xz = xy + y (120 – x – y) + x (120 - x - y) =xy + 120y – xy - 𝒚𝟐 + 120x - 𝒙𝟐 - xy f(x, y, z) = -𝒙𝟐 - 𝒚𝟐 + 120x + 120y – xy 𝒇𝒙 = −𝟐𝒙 + 𝟏𝟐𝟎 − 𝒚 𝒇𝒚 = −𝟐𝒚 + 𝟏𝟐𝟎 − 𝒙 𝒇𝒙 = 0 = 𝒇𝒚 2x + y = 120 2y + x = 120 On solving x = 40 , y = 30 + 10 = 40 15

Maxima and Minima

Problems on 2 Variables

Solved Problems

r = 𝒇𝒙𝒙 = -20

t=2

s=1 ∴ rt - 𝒔𝟐 = 4 – 1 = 3 > 0 𝟏 𝟑

𝟏 𝟑

∴ f is minimum at ((𝟐𝒗) , (𝟐𝒗) ) 𝒚

𝒗

But z = 𝒙𝒚 = z=

𝒗

𝟏 (𝟐𝒗)𝟑

=

𝟏 (𝟐𝒗)𝟑 −𝟐 𝟑

𝟏 𝟑

𝟏 𝟑

𝟏 𝟑

𝟏 (𝟐𝒗)𝟑

𝒗 𝟐 (𝟐)𝟑

𝟏 𝟑

𝒗

z=𝟐 𝒗

𝟏 𝟑

2z = 𝟐 𝒗 = (𝟐𝒗) x=y

𝟏 𝟑

𝟏 𝟑

−𝟐 𝟑

𝟏 𝟑

∴ Dimension are (𝟐𝒗) , (𝟐𝒗) , (𝟐 𝒗 ) 20