# Engineering Mathematics Questions and Answers – Maxima and Minima of Two Variables – 3

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Maxima and Minima of Two Variables – 3”.

1. What is the saddle point?
a) Point where function has maximum value
b) Point where function has minimum value
c) Point where function has zero value
d) Point where function neither have maximum value nor minimum value

Explanation: Saddle point is a point where function have neither maximum nor minimum value.

2. Stationary point is a point where, function f(x,y) have?
a) ∂f∂x = 0
b) ∂f∂y = 0
c) ∂f∂x = 0 & ∂f∂y = 0
d) ∂f∂x < 0 and ∂f∂y > 0

Explanation: Point where function f(x,y) either have maximum or minimum value is called saddle point. i.e, ∂f∂x = 0 & ∂f∂y = 0.

3. For function f(x,y) to have minimum value at (a,b) value is?
a) rt – s2>0 and r<0
b) rt – s2>0 and r>0
c) rt – s2<0 and r<0
d) rt – s2>0 and r>0

Explanation: For the function f(x,y) to have minimum value at (a,b)
rt – s2>0 and r>0
where, r = 2f∂x2, t=2f∂y2, s=2f∂x∂y, at (x,y) => (a,b).

4. For function f(x,y) to have maximum value at (a,b) is?
a) rt – s2>0 and r<0
b) rt – s2>0 and r>0
c) rt – s2<0 and r<0
d) rt – s2>0 and r>0

Explanation: For the function f(x,y) to have maximum value at (a,b)
rt – s2>0 and r<0
where, r = 2f∂x2, t=2f∂y2, s=(2f∂x∂y, at (x,y) => (a,b).

5. For function f(x,y) to have no extremum value at (a,b) is?
a) rt – s2>0
b) rt – s2<0
c) rt – s2 = 0
d) rt – s2 ≠ 0

Explanation: For the function f(x,y) to have no extremum value at (a,b)
rt – s2 < 0 where, r = 2f∂x2, t=2f∂y2, s=2f∂x∂y, at (x,y) => (a,b).
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6. Discuss minimum value of f(x,y)=x2 + y2 + 6x + 12.
a) 3
b) 3
c) -9
d) 9

Explanation: Given, f(x, y) = x2 + y2 + 6x + 12
Now, ∂f∂x = 2x + 6 and ∂f∂y = 2
Putting, ∂f∂x and ∂f∂y = 0 we get,
(x,y) = (-3,0)
Now, r= 2f∂x2 = 2>0 and t= 2f∂y2 = 2 and s= 2f∂x∂y = 0
hence, rt – s2 = 4>0 and r>0
hence. f(x,y) has minimum value at (-3,0), which is f(x,y) = 12 + 9 – 18 = 3.

7. Discuss maximum or minimum value of f(x,y) = y2 + 4xy + 3x2 + x3.
a) minimum at (0,0)
b) maximum at (0,0)
c) minimum at (2/3, -4/3)
d) maximum at (2/3, -4/3)

Explanation: Given,f(x,y) = y2 + 4xy + 3x2 + x3
Now,∂f∂x = 4y + 6x + 3x2 and ∂f∂y = 2y + 4x
Putting,∂f∂x and ∂f∂y = 0,and solving two equations,we get,
(x,y) = (0,0) or (2/3, -4/3)
Now,at (0,0) r= 2f∂x2=6+6x=6>0 and t= 2f∂y2 =2>0 and s= 2f∂x∂y=4
hence, rt – s2 = 12 – 16<0,hence it has no extremum at this point.
Now at (23,-43) r=2f∂x2= 6 + 6x = 10>0 and t= 2f∂y2 =2>0 and s= 2f∂x∂y=4
hence, rt – s2 = 20 – 16 > 0 and r>0, hence it has minimum at this point.(23, –43).

8. Find the minimum value of xy+a3 (1x + 1y).
a) 3a2
b) a2
c) a
d) 1

Explanation:
Given,f(x,y) = $$xy+a^3(\frac{1}{x}+\frac{1}{y})$$
Now, $$\frac{∂f}{∂x}=y-\frac{a^3}{x^2}$$ and $$\frac{∂f}{∂y}=x-\frac{a^3}{y^2}$$
Putting, $$\frac{∂f}{∂x}$$ and $$\frac{∂f}{∂y}$$=0,and solving two equations,we get,
(x,y)=(a,a) or (-a,a)
Now, at (a,a) r = $$\frac{∂^2 f}{∂x^2}=\frac{2a^3}{x^3}$$=2>0 and $$t=\frac{∂^2 f}{∂y^2}=\frac{2a^3}{y^3}$$=2>0 and $$s=\frac{∂^2 f}{∂x∂y}$$=1
hence, rt-s2=3>0 and r>0,hence it has minimum value at (a,a).
Now, at (-a,a) r=$$\frac{∂^2 f}{∂x^2}=\frac{2a^3}{x^3}$$=-2<0 and $$t=\frac{∂^2 f}{∂y^2}=\frac{2a^3}{y^3}$$=2>0 and s=$$\frac{∂^2 f}{∂x∂y}$$=1
hence, rt-s2=-5<0,hence it has no extremum at this point.
Hence maximum value is, f(a,a)=a2+a3 $$(\frac{1}{a}+\frac{1}{a})=a^2+2a^2=3a^2$$

9. Divide 120 into three parts so that the sum of their products taken two at a time is maximum. If x, y, z are two parts, find value of x, y and z.
a) x=40, y=40, z=40
b) x=38, y=50, z=32
c) x=50, y=40, z=30
d) x=80, y=30, z=50

Explanation: Now, x + y + z = 120 => z = 120 – x – y
f = xy + yz + zx
f = xy + y(120-x-y) + x(120-x-y) = 120x + 120y – xy – x2 – y2
Hence, ∂f∂x = 120 – y – 2x and ∂f∂y = 120 – x – 2y
putting ∂f∂x and ∂f∂y equals to 0 we get, (x, y)=>(40, 40)
Now at (40,40), r=2f∂x2 = -2 < 0, s = 2f∂x∂y = -1, and t = 2f∂y2 = -2
hence, rt – s2 = 5 > 0
since, r<0 and rt – s2 > 0 f(x,y) has maixum value at (40,40),
Hence, maximum value of f(40,40) = 120 – 40 – 40 = 40,
Hence, x = y = z = 40.

10. Find the maximum value of Sin(A)Sin(B)Sin(C) if A, B, C are the angles of triangle.
a) 3√38
b) 3√48
c) –3√38
d) π8

Explanation: Given f(A,B,C)=Sin(A)Sin(B)Sin(c),

Since A, B, C are the angle of triangle, hence, C = 180 – (A+B),

hence, f(x,y) = Sin(x)Sin(y)Sin(x+y), where A = x and B = y

Hence, ∂f∂x = Cos(x)Sin(y)Sin(x+y) + Sin(x)Sin(y)Cos(x+y) = Sin(y)Sin(y+2x)
and, ∂f∂y = Sin(x)Cos(y)Sin(x+y) + Sin(x)Sin(y)Cos(x+y) = Sin(x)Sin(x+2y)

Hence, putting ∂f∂x and ∂f∂y = 0, we get (x,y)=(60,60), (120,120)
Hence, at (x,y) = (60,60)we get,r = -√3, s = -√3/2, t = -√3, hence, rt-s2= 94∂x>0

hence, r<0 andrt-s2>0 hence, f(x,y) or f(A,B) have maximum value at (60,60)

Hence, at (x,y)=(120,120)we get,r=√3,s=√3/2,t=√3,hence,rt-s2 = 94∂x>0
And this value is 3√38
hence, r>0 and rt-s2 >0 hence, f(x,y) or f(A,B) have minimum value at (60,60)
and this value is –3√38.

11. The drawback of Lagrange’s Method of Maxima and minima is?
a) Maxima or Minima is not fixed
b) Nature of stationary point is can not be known
c) Accuracy is not good
d) Nature of stationary point is known but can not give maxima or minima

Explanation: In lagrange’s theorem of maxima of minima one can not determine the nature of stationary points.

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