Mathematics

Maxima and Minima

129 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 maths application of derivatives - iii second derivative test maxima and minima application of derivatives

lf x, y are two real numbers such that $x^{2}+y^{2}=1$, then the maximum value of x+y is

  1. $\sqrt{2}$
  2. $\sqrt{5}$

  3. 2

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $x=cos{\theta}$ and $y=sin{\theta}$
Then, $f(\theta)= cos{\theta}+sin{\theta}$
$f'(\theta)=-sin{\theta}+cos{\theta}$
For maxima or minima,
$f'(\theta)=0$
$\Rightarrow \theta =\frac{\pi}{4}$
$f''(\theta)=-(cos{\theta}+sin{\theta})$
$\Rightarrow f''(\frac{\pi}{4})<0$
Hence, f has a maximum value at $\theta =\frac{\pi}{4}$
$\displaystyle f(\frac{\pi}{4})=\sqrt{2}$


Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

if xy(y-x) = 16 then y has a minimum value when x=

  1. 1

  2. 3

  3. 2

  4. 4
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$xy(y-x)=16$
$xy^2-x^2y=16$
$y^2-xy-\dfrac {16}{x}=0$
$(y-\dfrac {x}{2})^2-\dfrac {x^2}{4}-\dfrac {16}{x}=0$
$y=\dfrac {x}{2}\pm \sqrt{\dfrac {x^2}{4}+\dfrac {16}{x}}$
$y'=\dfrac {1}{2}\pm \dfrac {1}{2}(\dfrac {\dfrac {2x}{4}-\dfrac {16}{x^2}}{\sqrt {\dfrac {x^2}{4}+\dfrac {16}{x}}})$
$-1=\pm (\dfrac {\dfrac {x}{2}-\dfrac {16}{x^2}}{\sqrt {\dfrac {x^2}{4}+\dfrac {16}{x}}})$
$\dfrac {16}{x}=\dfrac {256}{x^4}-\dfrac {16}{x}$
$x^3=8$
$x=2$
& $y=4$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at $t$ hours past $2:00$ in the morning is given by $\displaystyle N\left( t \right) =-20{ \left( t-5 \right)  }^{ 2 }+500for\quad 0\le t\le 10$ . According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

  1. $5:30$
  2. $7:00$
  3. $7:30$
  4. $8:00$
  5. $9:00$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given
$N(t)=-20{ (t-5) }^{ 2 }+500\quad 0\le t\le 10$
where $N(t)$ is the depth in cm in time t for maximum depth,
$\cfrac { dN }{ dt } =-20\times 2\left( t-5 \right) =0$
$t=5$
$\cfrac { { d }^{ 2 }N }{ d{ t }^{ 2 } } =-40$ (negative)
Thus at $t=5$ hours , depth will be maximum.
The water tank starts filling at $2:00$ in morning.
Therefore maximum depth$=2:00+5$ hours
$=7:00$am (hours)
Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

For what value of $x,x^{2} \ln (1/x)$ is maximum-

  1. $e^{-1/2}$
  2. $e^{1/2}$
  3. $e$
  4. $e^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $y=x^2\ln \dfrac{1}{x}$

$=x^2\ln (x^-)$
$=-x^2\ln (x)$
$\Rightarrow \dfrac{dy}{dx}=-2x\ln x-x^2.\dfrac{1}{x}$
$=-2x\ln x-x$
$=-x[2\ln x+1]=0$
$\Rightarrow x=0$ or $x=e^{-1/2}$
None $\dfrac{d^2y}{dx^2}=-2\ln x-2x\dfrac{1}{x}-1$
$=-2\ln x-3$
at $x=e^{-1/2}$     $\dfrac{d^2y}{dx^2}=-2<0$
$\Rightarrow $ maximum value is at $e^{-1/2}$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let '$a$' and '$b$' are positive number. If $(x, y)$ is a point on the curve $\displaystyle ax^2 + by^2 = ab$ then the largest possible value of $xy$ is

  1. $\displaystyle \frac {\sqrt {ab}}{2}$
  2. $\displaystyle \sqrt {ab}$
  3. $\displaystyle \frac {ab}{a + b}$
  4. $\displaystyle \frac {2ab}{a + b}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The point (x,y) on the curve can be written in polar coordinates as 
$x=\sqrt{b} $cos$\theta$ and $y=\sqrt{a} $sin$\theta$

Thus, 
$(xy) _{max}= (\sqrt{ab}$sin$\theta $cos$\theta) _{max}$

$ = (\sqrt{ab}\dfrac{sin2\theta}{2}) _{max}$
$ = \dfrac{\sqrt{ab}}{2}        \because ($sin$2\theta) _{max}= 1 $

$\therefore$ Ans. is option A.
Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $g(x)=a _{0}+a _{1}x+a _{2}x^{2}+a _{3}x^{3}$ and $ f(x)=\sqrt{g(x)}$.
$f(x)$ has its non-zero local minimum and maximum values at $-3$ and $3$ respectively. If $a _{3}\in $ the domain of the function $ \displaystyle h(x)=\sin ^{-1}\left(\dfrac{1+x^{2}}{2x}\right)$. The value of $a _{0}$ is

  1. equal to $50$
  2. greater than $54$
  3. less than $54$
  4. less than $50$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


$\displaystyle D _{h}=\left { -1, 1 \right }$, as only possible values in the domain of $h(x)$ is $1$ and $-1$
$\therefore  a _{3}=-1$
Now, $ g(x)=a _{0}+a _{1}x+a _{2}x^{2}-x^{3}$
$ {g}'(x)=a _{1}+2a _{2}x-3x^{2}$
$=-3(x-3)(x+3)$
$=-3x^{2}+27$
$\therefore  a _{1}=27, a _{2}=0$
$\therefore a _{1}+a _{2}=27$
Also, $g(-3)> 0$ and $g(3)> 0$
$\Rightarrow  a _{0}> 54$ and $a _{0}< -54$
$\therefore   a _{0}> 54$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $f(x) = ax^2+bx+c, a, b, c \in R.$ It is given $|f(x)| \le 1, \, |x| \le 1$ then the possible value of $|a+b|$, if $\dfrac{8}{3}a^2+2b^2$ is maximum, is given by

  1. $1$
  2. $0$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given |ax^2+bx+c| <= 1 for |x| <= 1, this is a classic problem related to Chebyshev polynomials. The maximum value of the expression 8/3*a^2 + 2*b^2 under these constraints occurs at specific coefficients, leading to |a+b| = 1.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $x$ and $y$ be two positive real numbers such that $xy = 1.$ The minimum value of $x + y$ is

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

Given $xy=1$ and $f(x,y)=x+y$
$\Rightarrow f(x)=x+\dfrac{1}{x}$
$f'(x)=1-\dfrac{1}{x^2}$
For maxima or minima,
$f'(x)=0$
$\Rightarrow x=\pm1$
$f''(x)=\dfrac{2}{x^3}$
$f''(x)>0$ at $x=1$
Hence f(x) has minimum at $x=1$
$f(1)=2$
So, minimum value of $x+y  \ is  \  2$.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

If $f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}}  + \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}}  + \sqrt {{x^2} + {{\left( {y - 1} \right)}^2}}  + \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 4} \right)}^2}} $ where $x,y \in R$, then the minimum value of $f\left( {x,y} \right)$ is

  1. $2 + \sqrt 5 $
  2. $5 + \sqrt 2 $
  3. $5 - \sqrt 2 $
  4. $\sqrt 5 - 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The function represents the sum of distances from (x, y) to four points: (0, 0), (1, 0), (0, 1), and (3, 4). This is a Fermat point problem. The minimum distance sum for these points is found by connecting the diagonals, resulting in 2 + sqrt(5).

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If $P ( \alpha , \beta )$ moves on $x ^ { 2 } + y ^ { 2 } - 2 x + 6 y + 1 = 0$ then minimum value of $a ^ { 2 } + \beta ^ { 2 } - 2 a - 4 \beta$ is 

  1. -3

  2. -1

  3. 1

  4. 3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The circle is x^2 - 2x + 1 + y^2 + 6y + 9 = -1 + 1 + 9, which is (x-1)^2 + (y+3)^2 = 9. The center is (1, -3) and radius is 3. We want the minimum of a^2 + b^2 - 2a - 4b. This is (a-1)^2 + (b-2)^2 - 5. The point (a, b) is on the circle. The minimum distance from (1, 2) to the circle (center (1, -3), radius 3) is |distance between (1, 2) and (1, -3)| - radius = |2 - (-3)| - 3 = 5 - 3 = 2. The minimum value of the squared distance (a-1)^2 + (b-2)^2 is 2^2 = 4. Thus, 4 - 5 = -1.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

Suppose A is a complex number and $ n \in N, $ such that $A^{n} = (A + 1)^{n} =1, $ then the least value of $n$ is

  1. $3$
  2. $6$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Since\quad { z }^{ n }=1\ \Rightarrow \quad { \left| z \right|  }^{ n }=1\ \quad \quad \quad \left| z \right| =1\ similarly,\ { \left( z+1 \right)  }^{ n }=1\ \Rightarrow \quad { \left| z+1 \right|  }^{ n }=1\ \left| z+1 \right| =1\ Let\quad z=a+ib\ \left| z \right| =\left| z+1 \right| \quad \Rightarrow \quad { a }^{ 2 }+{ b }^{ 2 }={ \left( a+1 \right)  }^{ 2 }+{ b }^{ 2 }\ { a }^{ 2 }+{ b }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2a+1\ \Rightarrow \quad 2a+1=0\ \therefore \quad a=\frac { -1 }{ 2 } \ putting\quad the\quad value\quad of\quad a\quad in\quad eq.\ \Rightarrow \quad { \left( \frac { -1 }{ 2 }  \right)  }^{ 2 }+{ b }^{ 2 }=1\ \Rightarrow \quad { b }^{ 2 }=\frac { 3 }{ 4 } \ \Rightarrow \quad b=\pm \frac { \sqrt { 3 }  }{ 2 } \ Now,\quad z+1=\quad \frac { 1 }{ 2 } \pm \frac { \sqrt { 3 }  }{ 2 } \ \Rightarrow \quad z+1={ e }^{ \pm \frac { zni }{ 3 }  }\ { \left( z+1 \right)  }^{ n }=\quad { e }^{ \pm \frac { zni }{ 3 }  }\ For\quad { \left( z+1 \right)  }^{ n }\quad to\quad be\quad 1\quad cos\quad \pm \frac { zn }{ 3 } =1\quad and\quad sin\quad \pm \frac { zn }{ 3 } =0\ This\quad can\quad only\quad happen\quad if\quad \pm \frac { zn }{ 3 } =2ak\quad for\quad integer\quad k.\ Solving\quad for\quad n,\quad we\quad get:\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \pm \frac { zn }{ 3 } =2ak\quad \Rightarrow \quad n=6k\ \quad \quad \quad \quad \quad \quad \quad k=\frac { 6 }{ n } \ least\quad value=\quad 6\ $

Multiple choice maths decimal numbers adding and subtracting decimals addition and subtraction of decimals operations on decimals

If k is an integer and $\displaystyle \left( 0.0025 \right) \left( 0.025 \right) \left( 0.00025 \right) \times { 10 }^{ k }$ is an integer, what is the least possible value of k ?

  1. -12

  2. -6

  3. 0

  4. 6

  5. 12
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Given expression:

 $(25 \times 10^{-4}) $ $(25 \times 10^{-3}) $$(25 \times 10^{-5}) $
$\rightarrow$ $15625 \times 10^{-12}$.
So, to make the result an integer, we must multiply by $10^{12}$
Least possible value of k should be 12. (option E)