This set of Engineering Mathematics Questions and Answers for Entrance exams focuses on “Maxima and Minima of Two Variables – 2”.
1. Find the critical points of the function.
f(x, y)=\(\frac{sin^{-1}(y^2).(y^2+3y).(sin(y^6+7y))}{(y^9+y^{10})}+10x\)
a) (0,0)
b) (0,-90)
c) (90, 0)
d) None exist
View Answer
Explanation: Find fx = 10
10 ≠ 0
Hence, no points exist.
2. For function f(x, y) = sin-1(x2 + y2) critical points are found. Now a new graph g(x, y) is formed by coupling graphs f(x, y) and f(x, y) = – sin-1(x2 + y2). What are the critical points of g(x, y).
a) (0,0)
b) There are infinite such points
c) Only positive (x, y) are critical points
d) (90,-90)
View Answer
Explanation: The function takes constant values along a circle(observe the function)
But it is composed of arc sine function. Hence, we will have critical points at equal intervals
Hence, there are infinite such points.
3. Consider the circular region x2 + y2 = 81, What is the maximum value of the function?
f(x, y) = x6 + y2(3x4 + 1) + x2.(3y4 + 1) + y6
a) 90
b) 80
c) 81 + 813
d) 100
View Answer
Explanation: Rewrite the function as
f(x, y) = x2 + y2 + (x2 + y2)3
Put x2 + y2 = 81
= 81 + 813.
4. What is the maximum value of the function f(x, y) = x2(1 + 3y) + x3 + y3 + y2(1 + 3x) + 2xy over the region x=0; y=0; x + y=1.
a) 0
b) -1
c) Has no maximum value
d) 2
View Answer
Explanation: Rewrite the function as
f(x, y) = (x + y)2 + (x + y)3
Put x + y = 1
= 2.
5. If the Hessian matrix of a function is zero then the critical point is?
a) It cannot be concluded
b) Always at Origin
c) Depends on Function
d) (100,100)
View Answer
Explanation: If the Hessian matrix is zero then the second derivative test fails and nothing can be said about the critical points.
6. The maximum value of the function is?
f(x, y) = sin(x).cos(2y).cos(x + 2y) + sin(2y).cos(x + 2y).cos(x) in the region x=0; y=0; x+2y = 3
a) 90
b) cos(1)
c) sin(1).cos(1)
d) sin(3).cos(3)
View Answer
Explanation: Rewrite the function as
f(x, y) = cos(x + 2y) * (sin(x).cos(2y) + cos(x).sin(2y))
f(x, y) = cos(x + 2y).sin(x + 2y)
Put x+2y = 3
= sin(3).cos(3).
7. Find the minimum value of the function f(x, y) = x2 + y2 +199 over the real domain.
a) 12
b) 13
c) 0
d) 199
View Answer
Explanation: Find
fx = 2x
fy = 2y
The critical point is
x=0
y=0
(0,0) is the critical point
Put it back into the function we get
z = 0 + 0 + 199 = 199 is the required minimum value.
8. What is the maximum value of the function f(x, y) = 3xy + 4x2y2 in the region?
x=0; y=0; 2x + y = 2
a) 1
b) 0
c) 100
d) 10
View Answer
Explanation: Differentiating we have
fx = 3y + 8xy2 = 0
fy = 3x + 8x2y
x = 0
y = 0
(0,0) lies in the region
Substitute x=0
f(0, y) = 0
Substitute y=0
f(x, 0) = 0
Substitute y = 2 – 2x
10 is the maximum value.
Sanfoundry Global Education & Learning Series – Engineering Mathematics.
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