Tag: application of derivatives - iii

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

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

  1. 30, 34

  2. 31, 33

  3. 32, 32.

  4. 35, 29
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. 13 and 7

  2. 14 and 6

  3. 15 and 5

  4. 16 and 4
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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

  1. $4$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
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

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