Mathematics

Maxima and Minima

191 Questions

Maxima and minima problems involve finding the highest and lowest values of mathematical functions within given intervals. These calculus questions require differentiation and analytical logic. They are common in civil service and state level mathematics examinations.

Extreme function valuesComplex number minimizationMinimum variables calculationMaximum matrix analysisCalculus optimization

Maxima and Minima Questions

Multiple choice

Given a function: (x, y) = 4x2 + 6y2 − 8x − 4y + 8 The optimal value of f(x, y)

  1. is a minimum equal to 10/3

  2. is a maximum equal to 10/3

  3. is a minimum equal to 8/3

  4. is a maximum equal to 8/3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Find partial derivatives: df/dx = 8x - 8 = 0 => x = 1. df/dy = 12y - 4 = 0 => y = 1/3. The second derivative test (f_xx = 8, f_yy = 12, f_xy = 0) shows a minimum since f_xx > 0 and D = f_xx*f_yy - f_xy^2 = 96 > 0. Value at (1, 1/3) is 4(1)^2 + 6(1/3)^2 - 8(1) - 4(1/3) + 8 = 4 + 2/3 - 8 - 4/3 + 8 = 4 - 2/3 = 10/3.

Multiple choice

The function f(x) = 2x2 - 3x2 - 36x + 2 has its maxima at

  1. x = - 2 only

  2. x = 0 only

  3. x = 3 only

  4. both x = - 2 and x = 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

f(x) = 2x^3 - 3x^2 - 36x + 2. f'(x) = 6x^2 - 6x - 36. Setting f'(x) = 0 gives x^2 - x - 6 = 0, so (x-3)(x+2) = 0. Roots are x = 3 and x = -2. f''(x) = 12x - 6. f''(-2) = -24 - 6 = -30 (<0, maxima). f''(3) = 36 - 6 = 30 (>0, minima). Thus, maxima at x = -2.

Multiple choice maths construction circumscribing and inscribing a circle on a regular hexagon constructions related to a circle construction of polygons construction of tangent to a circle construction of tangents construction of line segment and circle of given radius construction related to lines

To construct a quadrilateral minimum of its _________ elements are required.

  1. $3$
  2. $4$
  3. $5$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To construct a unique quadrilateral, we will be need a minimum of $5$ dimensions.

If we have five dimensions, we can draw a side first then mark angle on both ends then we can construct a quadrilateral uniquely.
Or if we have three sides and two included angles then also we can construct a unique quadrilateral.
Unless we are constructing any one of the special quadrilaterals.

Multiple choice business mathematics and statistics applications of calculus revenue functions from marginal revenue functions minimization of cost function and maximization of revenue function and profit function integral calculus – ii

$y = 48x - 2x^2$
where, $y=$ Total revenue $ $ $.
$x = $ Output
At what output is total revenue a maximum?

  1. $2$
  2. $12$
  3. $48$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow$  We have given, $y=48x-2x^2$

$\Rightarrow$  Differentiate on both sides w.r.t $x$ we get,
$\Rightarrow$  $\dfrac{dy}{dx}=48-4x$          --- ( 1 )
$\Rightarrow$  Let $\dfrac{dy}{dx}=0$
$\Rightarrow$  $48-4x=0$
$\Rightarrow$  $4x=48$
$\Rightarrow$  $x=12$ is a turning point.
$\Rightarrow$  $\dfrac{d^2y}{dx^2}=-4$         [Differentiate ( 1 ) on both sides]
$\Rightarrow$  So, the turning point is a maximum 

Multiple choice maths surface area and volume of cube and cuboid finding out the diagonal of cube and cuboid length of the diagonal of cube diagonal of cube and cuboid

A Pythagorean triplet whose smallest member is $8$, is:

  1. $8, 15, 18$
  2. $8, 13, 16$
  3. $8, 14, 17$
  4. $8, 15, 17$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We can get Pythagorean triplet by using general form $2m,\ m^{2}-1,\ m^{2}+1 $
Let us first take 

$m^{2}-1=8$
So, $m^{2}=8+1=9$
Which gives $m=3$
Therefore $2m=6$ and  $ \displaystyle m^{2}+1 = 10  $
The triplet is thus $6,8,10$, but $8$ is not the smallest member of this triplet.

So let us try
$2m=8$
then $m=4$
We get $ \displaystyle m^{2}+1 = 16-1=15$
and $ \displaystyle m^{2}+1 =16+1=17$
The triplet is $8,15,17$ with $8$ as the smallest member.

Hence, option $D.$

Multiple choice maths decimal numbers comparing and ordering of decimals more or less comparing decimals

Of the numbers 0.16, $\displaystyle \sqrt{0.16}$, $\displaystyle (0.16)^{2}$ and $\displaystyle 0.1\overline{6}$ the least number is

  1. $\displaystyle (0.16)^{2}$
  2. $\displaystyle \sqrt{0.16}$

  3. 0.16
  4. $\displaystyle 0.1\overline{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

0.16 = 0.16
$\displaystyle \sqrt{0.16}=0.4 $  $\displaystyle \Rightarrow 0.0256=\left ( 0.16 \right )^{2}$
$\displaystyle \left ( 0.16 \right )^{2}=0.0256 $  $\displaystyle \Rightarrow 0.0256=\left ( 0.16 \right )^{2}$
$\displaystyle 0.\overline{16}=0.1666....$  $\displaystyle \Rightarrow 0.0256=\left ( 0.16 \right )^{2}$
is the least

Multiple choice range and mean deviation statistics and probability maths coefficient of variance variance and standard deviation

The highest score of a certain data exceeds in lowest score by $16$ and coefficient of range is $\cfrac{1}{3}$. The sum of the highest score and the lowest score is

  1. $36$
  2. $48$
  3. $24$
  4. $18$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the highest score be $x _{m}$ and 

the lowest score be $x _{0}$
Given that highest score exceeds lowest score by $16$
$\implies x _m=x _0+16\implies x _m-x _0=16$ ————(1)

Coefficient of range is given by $\dfrac{x _m-x _0}{x _m+x _0}$

Given that coefficient of range is $\dfrac 13$

$\implies \dfrac{x _m-x _0}{x _m+x _0}=\dfrac 13$ ———(2)

Substitute (1) in (2) we get

$\dfrac{16}{x _m+x _0}=\dfrac 13$

$\implies x _m+x _0=16\times3=48$

Therefore sum of the highest score and lowest score is $48$

Multiple choice physics turning on a pivot the turning of couple couple turning effect of force the turning effect of a force moment of force or torque

The minimum value of ${ \omega  } _{ 0 }$ below which the ring will drop down is 

  1. $\sqrt { \dfrac { g }{ 2\mu (R-r) } } $
  2. $\sqrt { \dfrac { 3g }{ 2\mu (R-r) } } $
  3. $\sqrt { \dfrac { g }{ \mu (R-r) } } $
  4. $\sqrt { \dfrac { 2g }{ \mu (R-r) } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This relates to the critical angular velocity required for a ring to maintain contact or prevent slipping in a rotating system. The derivation leads to the expression in option C.

Multiple choice maths addition of vectors vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If $a,b,c$ are unit vectors, then the maximum value of $|a+2b|^{2}+|b+3c|^{2}+|c+4a|^{2}$ is 

  1. $50$
  2. $21$
  3. $48$
  4. $58$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$\left|a+2b\right|^2+\left|b+3c\right|^2+\left|c+4a\right|^2$

$=\left(a+2b\right)^2+\left(b+3c\right)^2+\left(4a+c\right)^2$

$=a^2+4ab+4b^2+b^2+6bc+9c^2+c^2+8ac+16a^2$

$=17a^2+4ab+8ac+5b^2+10c^2+6bc$

$=50$
Multiple choice maths concepts of seven and eight digit numbers comparison of numbers comparing numbers operations on rational numbers indian place value chart largest and smallest numbers writing and expanding numbers

The ascending order of minimum values of the function  $P:\sin ^{ -1 }{ x } -\cos ^{ -1 }{ x } $, $Q=\tan ^{ -1 }{ x } -\cot ^{ -1 }{ x } $, $R=\sec ^{ -1 }{ x } -\csc ^{ -1 }{ x } $

  1. P, Q, R

  2. P, R, Q

  3. Q, P, R

  4. Q, R, P
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For P: sin^-1(x) - cos^-1(x) ranges from -pi/2 to pi/2, minimum is -pi/2. For Q: tan^-1(x) - cot^-1(x) ranges from -pi/2 to pi/2, minimum is -pi/2. For R: sec^-1(x) - csc^-1(x) ranges from -pi/2 to pi/2, minimum is -pi/2. However, evaluating the functions at their domain boundaries reveals P < Q < R.