Engineering Mathematics Questions and Answers – Maxima and Minima of Two Variables – 1

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Maxima and Minima of Two Variables – 1”.

1. Consider the f(x, y) = x2 + y2 – a. For what values of a do we have critical points for the function.
a) independent of a
b) for any real number except zero
c) a ∊ (0, +∞)
d) a ∊ (-1, 1)
View Answer

Answer: a
Explanation: Consider
fx = 2x
fy = 2y
There is no a here. Thus, independent of a.

2. The critical point exist for the function f(x, y) = xn + xn-1 y +……+yn at (0,0).
a) True
b) False
View Answer

Answer: b
Explanation: Counter example is with n=1
f(x, y) = x + y.

3. f(x, y) = sin(x).cos(y) Which of the following is a critical point?
a) (Π4, Π4)
b) (- Π4, Π4)
c) (0, Π2)
d) (0, 0)
View Answer

Answer: c
Explanation: fx = cos(x).cos(y) = 0
fy = – sin(x).sin(y)
→(x, y) = (0, Π2).

4. The point (0,0) in the domain of f(x, y) = sin(xy) is a point of ___________
a) Saddle
b) Minima
c) Maxima
d) Constant
View Answer

Answer: d
Explanation: Differentiating fxx = -y2.sin(xy)
fyy = -x2.sin(xy)
fxy = -yx.sin(xy)
Observe that fxx. fyy – (fxy)2
Hence, it is a saddle point.

5. A man travelling onf(x, y) = sin(xy). His shadow passing through the origin in a straight line (sun travels with him overhead).
What is the slope of the line travelling on which would lead him to the lowest elevation.
a) There isn’t such a line
b) 1
d) 0
View Answer

Answer: a
Explanation: Differentiating yields
fxx = -y2.sin(xy)
fyy = -x2.sin(xy)
fxy = -yx.sin(xy)
Observe that (0,0) is an inconclusive point
Hence, he will never reach the lowest elevation(because there isn’t such point.
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6. let s(1) be the set of all critical points of f1(x, y) = g1(x).g2(y) and s(2) be the set of critical points of f2(g1(x), g2(y)) Which of the following is the right relation between s(1) and s(2), given that minimum number of elements in s(1) is 2.
a) s(1) = s(2)
b) s(1) ≠ s(2)
c) s(1) ∩ s(2) ≠ 0
d) depends on the functions
View Answer

Answer: b
Explanation: Differentiating f1(g1(x), g2(y)) with respect to x and y separately we get
dx = f1x g1x (x)
dy = f1y g1y (y)
This implies
g1x = 0
g1y = 0
Which are also the set of critical points of f1(x, y)
Thus we have the relation as s(1) ∩ s(2) ≠ 0.

7. f(x, y)=\(10y(10y-1)+(\frac{x^3sin(x^2)tan(x^3)}{(x-1)^3})-100y^2\). Find the critical points
a) (0,0)
b) (1,1)
c) (2,22)
d) None exist
View Answer

Answer: d
Explanation: Rewriting the function
f(x, y)=\((\frac{x^3sin(x^2)tan(x^3)}{(x-1)^3})-10y\)
Differentiating with respect to y we get
fy = -10
-10 ≠ 0
There exist no critical point.

8. Consider the vertical cone. The minimum value of the function in the region f(x,y) = c is?
a) constant
b) 1
c) 0
d) -1
View Answer

Answer: a
Explanation: f(x,y) = c is a level curve over which the function has constant value
Hence, we have the answer as a constant.

Sanfoundry Global Education & Learning Series – Engineering Mathematics.


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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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