Tag: application of derivatives - iii

Questions Related to application of derivatives - iii

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:

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

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) $0$
maths application of derivatives - iii second derivative test maxima and minima application of derivatives

  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.

Show answer
Correct Option: A
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 $x=y$
$f(y)=1$

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

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

  1. $4^{3}. 24^{4}$

  2. $12^{3}.16^{4}$

  3. $4321$

  4. $1234$

Show answer
Correct Option: B
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 $12^3.16^4$

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

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

  1. $\displaystyle \frac{125}{4}$

  2. $\displaystyle \frac{4}{125}$

  3. $\displaystyle \frac{625}{4}$

  4. $\displaystyle \frac{4}{625}$

Show answer
Correct Option: A
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 $x=-\frac{5}{2}$
$\displaystyle f(-\frac{5}{2})=\frac{125}{4}$

lf x, y are two real numbers such that $x^{2}+y^{2}=1$, then the maximum value of x+y is

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

  1. $\sqrt{2}$

  2. $\sqrt{5}$

  3. 2

  4. 6

Show answer
Correct Option: A
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 $\theta =\frac{\pi}{4}$
$\displaystyle f(\frac{\pi}{4})=\sqrt{2}$


if xy(y-x) = 16 then y has a minimum value when x=

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

  1. 1

  2. 3

  3. 2

  4. 4

Show answer
Correct Option: D
Explanation:

$xy(y-x)=16$
$xy^2-x^2y=16$
$y^2-xy-\dfrac {16}{x}=0$
$(y-\dfrac {x}{2})^2-\dfrac {x^2}{4}-\dfrac {16}{x}=0$
$y=\dfrac {x}{2}\pm \sqrt{\dfrac {x^2}{4}+\dfrac {16}{x}}$
$y'=\dfrac {1}{2}\pm \dfrac {1}{2}(\dfrac {\dfrac {2x}{4}-\dfrac {16}{x^2}}{\sqrt {\dfrac {x^2}{4}+\dfrac {16}{x}}})$
$-1=\pm (\dfrac {\dfrac {x}{2}-\dfrac {16}{x^2}}{\sqrt {\dfrac {x^2}{4}+\dfrac {16}{x}}})$
$\dfrac {16}{x}=\dfrac {256}{x^4}-\dfrac {16}{x}$
$x^3=8$
$x=2$
& $y=4$

The sum of two +ve numbers is 100. If the product of the square of one number and the cube of the other is maximum then the numbers are

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

  1. 60, 40

  2. 20, 80

  3. 80, 20

  4. 40, 60

Show answer
Correct Option: D
Explanation:

$x+y=100$
$f(x)=x^2(100-x)^3$
$f'(x)=2x(100-x)^3-3(100-x)^2x^2$
$3x=2(100-x)$
$x=40$  $y=60$

The positive number x that exceeds its square by largest amount is

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

  1. $\displaystyle \frac{1}{2}$

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

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

  4. 1

Show answer
Correct Option: A
Explanation:

$f(x)=x-x^2$
$f'(x)=1-2x$
$x=\dfrac {1}{2}$ when $f'(x)=0$

$f(x)=2{x}^{3}-9{x}^{2}+12x+4$ is decreasing when

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

  1. $-\infty< x<1$ and $2< \infty< \infty$

  2. $-1< x< 2$

  3. $1< x< 2$

  4. $0< x< 2$

Show answer
Correct Option: C
Explanation:

$f'(x)=6{x}^{2}-15x+12$
$=6({x}^{2}-3x+2)$
$=6(x-1)(x-2)$
$f'(x)< 0$ when $x$ lies between $1$ and $2$
$\therefore$ $1< x< 2$
$\therefore$ (3) is correct

According to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at $t$ hours past $2:00$ in the morning is given by $\displaystyle N\left( t \right) =-20{ \left( t-5 \right)  }^{ 2 }+500for\quad 0\le t\le 10$ . According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum?

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

  1. $5:30$

  2. $7:00$

  3. $7:30$

  4. $8:00$

  5. $9:00$

Show answer
Correct Option: B
Explanation:
Given
$N(t)=-20{ (t-5) }^{ 2 }+500\quad 0\le t\le 10$
where $N(t)$ is the depth in cm in time t for maximum depth,
$\cfrac { dN }{ dt } =-20\times 2\left( t-5 \right) =0$
$t=5$
$\cfrac { { d }^{ 2 }N }{ d{ t }^{ 2 } } =-40$ (negative)
Thus at $t=5$ hours , depth will be maximum.
The water tank starts filling at $2:00$ in morning.
Therefore maximum depth$=2:00+5$ hours
$=7:00$am (hours)