Tag: maths

Questions Related to maths

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

Show answer
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:

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

If the sum of two +ve numbers is 18, then the maximum value of their product is

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

  1. 81

  2. 85

  3. 72

  4. 80

Show answer
Correct Option: A
Explanation:

$x+y=18$
$Product =xy$
$\dfrac {d(Product)}{dx}=(18-x)-x$
$0=18-2x$
$x=9$
$y=9$
$Product = 81$