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

If $xy={c}^{2}$ then the minimum value of $ax+by(a> 0, b> 0)$ is :

  1. $c\sqrt {ab}$
  2. $-c\sqrt {ab}$
  3. $2c \sqrt {ab}$
  4. $-2c \sqrt {ab}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$xy={ c }^{ 2 }$
$y={ c }^{ 2 }$
Put the value of $y={ c }^{ 2 }$ in $ ax+by$
$f(x)=a{ c }^{ 2\quad \quad  }y+by=0$
${ f }^{ ' }(x)=-a{ c }^{ 2\quad  }{ y }^{ 2\quad  }+b=0$
$-a{ c }^{ 2\quad  }+b{ y }^{ 2\quad  }=0$
$b{ y }^{ 2\quad  }=a{ c }^{ 2 }$
$y=+,-c\sqrt { (b/a)\quad  } $
${ f }^{ ''\quad  }(x)=2b{ c }^{ 2\quad  }/{ x }^{ 2 }$
$x=c\sqrt { b/a } $
${ f }^{ ''\quad  }(c\sqrt { (b/a } )=2b{ c }^{ 2\quad  }/{ c }^{ 2 }(b/a)=2a>0$
While $x=-c\sqrt { b/a } $will give maxima.
Put $x=c\sqrt { b/a }$ 
$a(c\sqrt { (b/a) } )+b({ c }^{ 2\quad  }\sqrt { a) } /c\sqrt { b } =2c\sqrt { ab } $

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

If $xy=4$ and $x<0$ then maximum value of $x+16y$ is-

  1. $8$
  2. $-8$
  3. $16$
  4. $-16$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$f(x)=x+16y$         (1)
$xy=4 $                         (2)
Substituting $y=\dfrac { 4 }{ x } $ in (1).
$f(x)=x+\dfrac{ 16.4 }{ x } $


${ f }^{ ' }(x)=1-\dfrac { 64 }{ { x }^{ 2 } } $

${ f }^{ ' }(x)=\dfrac { { x }^{ 2 }-64 }{ { x }^{ 2 } } $
$x=\pm 8$
Given $x<0, x=-8,y=-\dfrac 12$
Substitute this value in $f(x)$
$f(x)=-8+(\dfrac { 1(-16) }{ 2 } )$
$f(x)=-16$

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

Observe the following lists

List-I List-II
(A) Maximum value of  $xy$ subject to  ${x}+{y}=7$ is 1) $72$
(B) If  $l^{2} + m^{2} = 1$ , then the maximum value of $l + m$ is 2) $1$
(C) If $x +y = 12$, then the minimum Value of $x^{2}  +y^{2}$   is 3) $\sqrt{2}$
(D) Minimum value $x^{2} - 8x +17$ is  4) $\displaystyle \frac{49}{4}$
5) $0$

  1. A - 4, B -3, C -1, D -2.

  2. A - 4, B -3, C -2, D -1.

  3. A - 2, B -3, C -5, D -4.

  4. A - 2, B -3, C -1, D -4.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(A) use A.M. & G.M.
$\displaystyle \frac {x+y}{2}\geq (xy)^{\frac {1}{2}}$
$\displaystyle (\dfrac {7}{2})\geq (xy)^{\frac {1}{2}}$
$(xy)\leq(\dfrac {7}{2})^2$
(B) $y=l+\sqrt {1-l^2}$

$\displaystyle \frac {dy}{dl}=1-\frac {l}{\sqrt {1-l^2}}$

$\displaystyle \frac {dy}{dl}=0$ when $\displaystyle l=\frac {1}{\sqrt 2}$
So $\displaystyle m=\frac {1}{\sqrt 2}$
$\Rightarrow l=\displaystyle \frac{1}{\sqrt{2}}$

$\Rightarrow l+m=\sqrt{2}$

(C) $s=x^2+(12-x)^2$
$\displaystyle \frac {ds}{dx}=2x-2(12-x)$
$\displaystyle \frac {ds}{dx}=0$ when $x=6$ $y=6$
$s=36+36=72$
(D) $f'(x)=2x-8$
$f'(x)=0$ at $x=y$
$f(y)=1$

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

lf $\mathrm{x}+\mathrm{y}=28$ then the maximum value of $\mathrm{x}^{3}\mathrm{y}^{4}$ is

  1. $4^{3}. 24^{4}$
  2. $12^{3}.16^{4}$
  3. $4321$
  4. $1234$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $x+y=k$ then maximum value of $x^{m}y^{n}$ is at $\displaystyle x=\frac{km}{m+n}, y=\frac{km}{m+n}$ where $x,y>0$ and $m,n \ge{1} $
Here $k=28, m=3,n=4$
So,$x=12, y=16$
Hence maximum value is $12^3.16^4$

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

lf $2\mathrm{x}+\mathrm{y}=5$ then the maximum value of $\mathrm{x}^{2}+3\mathrm{x}\mathrm{y}+\mathrm{y}^{2}$ is

  1. $\displaystyle \frac{125}{4}$
  2. $\displaystyle \frac{4}{125}$
  3. $\displaystyle \frac{625}{4}$
  4. $\displaystyle \frac{4}{625}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$2x+y=5$
$\Rightarrow y=5-2x$
$f(x)=x^2+3x(5-2x)+(5-2x)^2$
$f(x)=-x^2-5x+25$
$f'(x)=-2x-5$
For maxima or minima,
$f'(x)=0$
$\Rightarrow x=-\frac{5}{2}$
$f''(x)=-2$
$f''(-\frac{5}{2})=-2<0$
So, f(x) has a maximum at $x=-\frac{5}{2}$
$\displaystyle f(-\frac{5}{2})=\frac{125}{4}$

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

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 $x+y  \ is  \  2$.