Tag: maxima and minima

Questions Related to maxima and minima

Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

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

  1. 30, 34

  2. 31, 33

  3. 32, 32.

  4. 35, 29

Show answer
Correct Option: C
Explanation:

Let one part be x.
Hence another part will be 64-x.
Let 
$f(x)=x^{3}+(64-x)^{3}$.
$f'(x)$
$=3x^{2}-3(64-x)^{2}$
$=0$
Or 
$x^{2}=(64-x)^{2}$
Or 
$x=64-x$ or $x=-64+x$
Considering equation $x=64-x$, we get 
$x=32$.
Hence another part will also be 32.

Divide 20 into two parts such that the product of one part and the cube of the other is maximum.

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

  1. 13 and 7

  2. 14 and 6

  3. 15 and 5

  4. 16 and 4

Show answer
Correct Option: C
Explanation:

Let the two parts be $x$ and $20-x$
Let $y=(20-x)x^3$
$\Rightarrow y=20x^3-x^4$

For maximum or minimum,
$\dfrac{dy}{dx}=0$
$\Rightarrow 60x^2-4x^3=0$
$\Rightarrow 4x^2(15-x)=0$
$\Rightarrow x=0, x=15$

$\dfrac{d^2y}{dx^2}=120x-12x^2$
At $x=15$, $\dfrac{d^2y}{dx^2}<0$
Hence, $y$ has a maximum at $x=15$

So, the two numbers are 15, 5

Let x and y be two real numbers such that x > 0 and xy$=1.$ The minimum value of x+y is

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

  1. 1

  2. 1/2

  3. 2

  4. 1/4

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Correct Option: C
Explanation:

Let $ z= x+y = x+1/x$,   since $(xy=1)$
$\dfrac{dz}{dx} = 1-1/x^2$
For minimum value of $z$
$\dfrac{dz}{dx} =0= 1-1/x^2\Rightarrow x=1$, $since (x>0)$
Therefore minimum value of $z=2$

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

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

  1. $15$ & $45$

  2. $30$ & $30$

  3. $20$ & $40$

  4. $10$ & $50$

Show answer
Correct Option: A
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 $45,15$.

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

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

  1. $c\sqrt {ab}$

  2. $-c\sqrt {ab}$

  3. $2c \sqrt {ab}$

  4. $-2c \sqrt {ab}$

Show answer
Correct Option: C
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 } $

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

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

  1. $8$

  2. $-8$

  3. $16$

  4. $-16$

Show answer
Correct Option: D
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 $f(x)$
$f(x)=-8+(\dfrac { 1(-16) }{ 2 } )$
$f(x)=-16$

The difference between two numbers is $a$. If their product is minimum, then numbers are-

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

  1. $-a/2, a/2$

  2. $-a, 2a$

  3. $-a/3, 2a/3$

  4. $-a/3, 4a/3$

Show answer
Correct Option: A
Explanation:

x-y=a                (1)
$f(x)=xy=x(x-a)$
${ f }^{ ' }(x)=2x-a $     (2)
2x-a=0
$x=a/2$
Substitute value of $x=a/2$ in (1) to get $y=-a/2$

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

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

  1. $44, 20$

  2. $16, 48$

  3. $32, 32 $

  4. $50, 14$

Show answer
Correct Option: C
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 
$32,32$.

Consider a function $f(x) = \displaystyle \frac{sin x}{2}$. Let $g(x) = \int  f(x)dx$, where constant of integration is zero.
On the basis of above information, answer the following questions The number of local minima of $g(x)$ in (2$\pi$,12$\pi$) are

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

  1. $4$

  2. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: B
Explanation:
Given,
$f\left( x \right) =\cfrac { \sin { x }  }{ 2 } $
$g\left( x \right) =\int { f\left( x \right) dx } $
$=\int { \cfrac { \sin { x }  }{ 2 } dx } $
$=-\cfrac { 1 }{ 2 } \cos { x } +c$
$c=0$ (given in question
Now for critical points,
$g'\left( x \right) =0$
$-\cfrac { 1 }{ 2 } \times \left( -\sin { x }  \right) =0$
$\sin { x } =0$
$x=n\pi \quad \left( n=0,1,2... \right) $
For maxima/minima
$g''\left( x \right) =\cos { x } $ [ positive for $\left( 0,\cfrac { \pi  }{ 2 }  \right) ,\left( x,\cfrac { 3\pi  }{ 2 }  \right) ...$]
We have to consider positive values for minima
Between $\left( 2\pi ,12\pi  \right) $ there will be total $10$ critical points out of which $5$ points will give minima.
$(5)$ is the correct answer.

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

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

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

  2. $\displaystyle \frac{6}{5}$

  3. $\displaystyle \frac{2}{3}$

  4. $\displaystyle \frac{2}{5}$

Show answer
Correct Option: C
Explanation:

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