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

If $F(x)=2x^3-21\,x^2+36x-20$, then 

  1. f has maxima at x=1

  2. f has minima at x=1

  3. f has maximum value -128

  4. f has minimum value -3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider given the function,

$F\left( x \right)=2{{x}^{3}}-21{{x}^{2}}+36x-20$      ……(1)

Differentiate with respect to x,

${{F}^{'}}\left( x \right)=6{{x}^{2}}-42x+36$          ……..(2)


For maxima and minima,

$ F\left( x \right)=0 $

$ 6{{x}^{2}}-42x+36=0 $

$ {{x}^{2}}-7x+6=0 $

$ {{x}^{2}}-6x-x+6=0 $

$ x\left( x-6 \right)-1\left( x-6 \right)=0 $

$ \left( x-6 \right)\left( x-1 \right)=0 $

$ x=1,6 $


Differentiate equation 2nd with respect to x,

${{F}^{''}}\left( x \right)=12x-42$

At $x=1\Rightarrow {{F}^{''}}\left( x \right)<0$

Hence, F(x) Is maximum.


At $x=6\Rightarrow F\left( x \right)>0$

Hence, function F(x) is minimum.

 

Hence, this is the answer.

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

If $\displaystyle xy=a^{2}$ and $\displaystyle S=b^{2}x+c^{2}y$ where a,b and c are constants then the minimum value of S is 

  1. $abc$
  2. $\displaystyle bc\sqrt{a}$
  3. $2abc$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $x y = a^2$ and $S = b^2x + c^2y$
$\Rightarrow S = b^2 x + c^2a^2/x$
$\Rightarrow \dfrac{dS}{dx} = b^2 - c^2a^2/x^2$
For maximum or minimum value of $S$
$ \dfrac{dS}{dx} = 0 = b^2 - c^2a^2/x^2 \Rightarrow x =\pm  ac/b$
Now $\dfrac{dS}{dx} = 2 c^2a^2/x^3$
Clearly at $x =  ac/b$,  $\dfrac{dS}{dx} = 2 b^3/ac > 0 $ (Assuming that $ b^3/ac>0$)
Hence minimum value of $S$ is $= b^2(ac/b)+c^2(b/ac)= 2abc$

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

If $\displaystyle \theta +\phi =\frac{\pi }{3}$ then $\displaystyle  \sin \theta \cdot\sin \phi$ has a maximum value at $\displaystyle \theta$ =

  1. $\displaystyle \dfrac{\pi }{6}$
  2. $\displaystyle \dfrac{2\pi }{3}$
  3. $\displaystyle \dfrac{\pi }{4}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $y = \sin\theta.\sin\phi = \sin\theta.\sin(\dfrac{\pi}{3}-\theta)$
For maximum value of $y$ 
$\dfrac{dy}{dx} = 0 = \cos\theta.\sin(\dfrac{\pi}{3}-\theta) - \sin\theta.\cos(\dfrac{\pi}{3}-\theta) = \sin(2\theta -\dfrac{\pi}{3})$
$\Rightarrow \theta = \dfrac{\pi}{6}$

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

The sum of two nonzero numbers is $8$. The minimum value of the sum of their reciprocals is

  1. $\displaystyle \frac{1}{4}$
  2. $\displaystyle \frac{1}{2}$
  3. $\displaystyle \frac{1}{8}$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $x$ and $y$ be two numbers 
$\Rightarrow x+y = 8$
Assume $z$ be be sum of their inverse
$z = 1/x+1/y =\dfrac{x+y}{xy} = \dfrac{8}{xy} = \dfrac{8}{x(8-x)}$
For minimum value of $z $
$\dfrac{dz}{dx} = 0 =\dfrac{16(4-x)}{(x(8-x))^2}\Rightarrow x = 4$
Hence minimum value of $z$ is $=1/4+1/4 = \dfrac{1}{2}$ 

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

Find the two positive numbers $x$ & $y$ such that their sum is $60$ and $\displaystyle xy^{3}$ is maximum

  1. $15$ & $45$
  2. $30$ & $30$
  3. $20$ & $40$
  4. $10$ & $50$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let one number be $x$
Hence the other number will be $(60-x)$.
Let 
$K=x^{3}.(60-x)$
Differentiating $K$ with respect to $x$, we get 
$\dfrac{dK}{dx}$
$=3x^{2}(60-x)-x^{3}=0$
Or 
$x^{2}[180-3x-x]=0$
Or 
$x=0$ and $x=\dfrac{180}{4}=45$.
Now its given that the numbers are positive.
Hence $x=0$ is ruled out.
Thus we get $x=45$.
Hence
$y=15$.
Therefore the numbers are $45,15$.

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

Two parts of $64$ such that the sum of their cubes is minimum will be-

  1. $44, 20$
  2. $16, 48$
  3. $32, 32 $
  4. $50, 14$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let one part be $x$.
Hence another part be $(64-x)$
Thus 
Their cubes will be 
$x^{3}+(64-x)^{3}=y$
Thus 
$y'=3x^{2}-3(64-x)^{2}=0$
Or 
$x^{2}=(64-x)^{2}$
Or 
$x=64-x$ and $x=-64+x$
Hence
$x=32$.
Hence both the parts are 
$32,32$.

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

The sum of two numbers is 6. The minimum value of the sum of their reciprocals is

  1. $\displaystyle \frac{3}{4}$
  2. $\displaystyle \frac{6}{5}$
  3. $\displaystyle \frac{2}{3}$
  4. $\displaystyle \frac{2}{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x+y=6$
$Sum =\dfrac {1}{x}+\cfrac {1}{y}$
$\dfrac {d(sum)}{dx}=\dfrac {-1}{x^2}+\dfrac {1}{(6-x)^2}=0$
$x^2=(6-x)^2$
$x=\pm (6-x)$
$x=3$,  $y=3$
$Sum =\dfrac {2}{3}$

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 $x=-\frac{5}{2}$
$\displaystyle f(-\frac{5}{2})=\frac{125}{4}$