Question 15 Marks
The demand function of a watch is $p=6000-2 x$. Find the demand which maximizes the revenue and also find the corresponding price.
AnswerDemand function $p=6000-2 x$
Now, revenue function $ R=p \cdot x$
$ =(6000-2 x) x$
$\therefore R=6000 x-2 x^{2}$
$\therefore \frac{d R}{d x}= 6000-4 x $
Putting $\frac{d R}{d x}=0$,
$ 6000-4 x=0$
$\therefore 6000=4 x$
$\therefore x=1500 $
Now $ \frac{d^{2} R}{d x^{2}} =0-4$
$ =-4 $
Here, putting $x=1500$ in $\frac{d^{2} R}{d x^{2}}$,
$ \frac{d^{2} R}{d x^{2}}=-4<0 $
$\therefore $ Revenue is maximum at $x=1500$.
Now we shall find the corresponding price.
Putting $x=1500$ in demand function $p=6000-2 x$,
$ \text { Price } p =6000-2(1500)$
$ =6000-3000$
$p =3000 $
View full question & answer→Question 25 Marks
Find the maximum and minimum values of $y=x^{3}-2 x^{2}-4 x-1$.
AnswerHere, $y=x^{3}-2 x^{2}-4 x-1$
$ \therefore \frac{d y}{d x}=3 x^{2}-4 x-4 $
For stationary values, $\frac{d y}{d x}=0$
$ \therefore 3 x^{2}-4 x-4=0$
$\therefore 3 x^{2}-6 x+2 x-4=0$
$\therefore 3 x(x^2)+2(x^2)=0$
$\therefore (x-2)(3 x+2)=0$
$\therefore x=2 \text { or } x=-\frac{2}{3} $
Now, $\frac{d^{2} y}{d x^{2}}=6 x-4$
$ \text { At } \boldsymbol{x}=\mathbf{2} $
$\frac{d^{2} y}{d x^{2}} =6(2)-4$
$ =8>0 $
$\therefore$ Function is minimum at $x=2$.
$ \text { At } \boldsymbol{x}= -\frac{2}{3}$
$\frac{d^{2} y}{d x^{2}} =6\left(\frac{-2}{3}\right)-4$
$ =-4-4$
$ =-8<0 $
$\therefore$ Function is maximum at $x=-\frac{2}{3}$.
Minimum value of function $y$
Putting $x=2$ in the function $y$,
$ y =(2)^{3}-2(2)^{2}-4(2)-1$
$ =8-8-8-1$
$ =-9 $
Maximum value of function $y$
Putting $x=-\frac{2}{3}$ in the function $y$,
$ y =\left(-\frac{2}{3}\right)^{3}-2\left(-\frac{2}{3}\right)^{2}-4\left(-\frac{2}{3}\right)-1$
$ =\frac{-8}{27}-\frac{8}{9}+\frac{8}{3}-1$
$ =\frac{13}{27} $
Thus, the maximum value of $y$ is $\frac{13}{27}$ and the minimum value is $-9$.
View full question & answer→Question 35 Marks
Find the maximum and minimum values of $f(x)=2 x^{3}+3 x^{2}-12 x-4$
AnswerHere, $f(x)=2 x^{3}+3 x^{2}-12 x-4$
$ \therefore f^{\prime}(x)=6 x^{2}+6 x-12 $
For stationary values, $f^{\prime}(x)=0$
$ \therefore 6 x^{2}+6 x-12=0$
$\therefore x^{2}+x-2=0$
$\therefore (x+2)(x-1)=0$
$\therefore x=-2 \text { or } x=1 $
Now, $f^{\prime \prime}(x)=12 x+6$
$ \text { At } \boldsymbol{x}=-\mathbf{2} $
$f^{\prime \prime}(-2) =12(-2)+6$
$ =-18<0 $
$\therefore $ We get the maximum value of the function at $x=-2$.
At $x=1$
$ f^{\prime \prime}(1) =12(1)+6$
$ =18>0 $
$\therefore$ We get the minimum value of the function at $x=1$.
Minimum value of $f(x)$
Putting $x=1$ in the function $f(x)$,
$ f(1) =2(1)^{3}+3(1)^{2}-12(1)-4$
$ =2+3-12-4$
$ =-11 $
Maximum value of $f(x)$
Putting $x=-2$ in the function $f(x)$,
$ f(-2) =2(-2)^{3}+3(-2)^{2}-12(-2)-4$
$ =-16+12+24-4$
$ =16 $
Thus, the maximum value of $f(x)$ is $16$ and the minimum value is $-11$.
View full question & answer→Question 45 Marks
A toy is sold at $Rs. 20 $ Total cost of producing $x$ such toys is $C=1000+16.5 x+$ $0.001 x^{2}$ rupees. How many toys should be produced for maximum profit ?
AnswerPrice of one toy $=₹ 20$
$\therefore$ Revenue from selling $x$ toys $R=20 x$
Total cost $C=1000+16.5 x+0.001 x^2$
Profit function:
$P=R-C$
$ =20 x-\left(1000+16.5 x+0.001 x^2\right)$
$ =20 x-1000-16.5 x-0.001 x^2$
$ P=3.5 x-1000-0.001 x^2$
$ \therefore \frac{d P}{d x}=3.5-0-0.001(2 x)$
$ =3.5-0.002 x$
$ \text { and } \frac{d^2 \mathrm{P}}{d x^2}=-0.002$
For maximum profit:
$\text { (1) } \frac{d P}{d x}=0$
$ \therefore 3.5-0.002 x=0$
$ \therefore 0.002 x=3.5$
$ \therefore x=\frac{3.5}{0.002}=1750$
$(2) \frac{d^2 \mathrm{P}}{d x^2}<0$
$\frac{d^2 \mathrm{P}}{d x^2}=-0.002<0$
Hence, $1750$ toys should be produced for maximum profit.
View full question & answer→Question 55 Marks
The selling price of a refrigerator as determined by the company is $Rs.10000 .$ The total cost of the production for $x$ refrigerator is $C=0.1 x^{2}+9000 x+100$ rupees. How many refrigerators should be manufactured for maximum profit ?
AnswerTotal cost function: $C=0.1 x^2+9000 x+100$
Selling price of one refrigerator $=₹ 10000$
$\therefore$ Revenue from selling $x$ refrigerator $R=10000 x$
Profit function:
$P=R-C$
$ =10000 x-\left(0.1 x^2+9000 x+100\right)$
$ =10000 x-0.1 x^2-9000 x-100$
$ \therefore P=1000 x-0.1 x^2-100$
$ \therefore \frac{d P}{d x}=1000-0.1(2 x)-0$
$ =1000-0.2 x$
$ \text { and } \frac{d^2 P}{d x^2}=-0.2$
For maximum profit:
$\text { (1) } \frac{d P}{d x}=0$
$ \therefore 1000-0.2 x=0$
$ \therefore 0.2 x=1000$
$ \therefore x=\frac{1000}{0.2}=5000$
$(2) \frac{d^2 \mathrm{P}}{d x^2}<0$
$\frac{d^2 \mathrm{P}}{d x^2}=-0.2<0$
Hence, $5000$ refrigerators should be manufactured for maximum profit.
View full question & answer→Question 65 Marks
A producer produces $x$ units at cost $200 x+15 x^{2}$ The demand function is $p=1200$ $10 x .$ Find the profit function and how many units should be produced for maximum profit ?
AnswerProduction cost function: $C=200 x+15 x^2$
Demand function: $p=1200-10 x$
Revenue function: $R=x \cdot p$
Putting, $p=1200-10 x$
$R=x(1200-10 x)$
$=1200 x-10 x^2$
Profit function:
Profit $=$ Revenue $-$ Cost
$P=R-C$
$\therefore$ Profit function $P=1200 x-10 x^2-\left(200 x+15 x^2\right)$
$\therefore P=1200 x-10 x^2-200 x-15 x^2$
$\therefore \mathrm{P}=1000 \mathrm{x}-25 \mathrm{x}^2$
Now, $\frac{d P}{d x}=1000-50 \mathrm{x}$ and $\frac{d^2 P}{d x^2}=-50$
For maximum profit:
$(1) \frac{d \mathrm{P}}{d x}=0$
$ \therefore 1000-50 x=0$
$ \therefore 50 x=1000$
$ \therefore x=\frac{1000}{50}=20$
$(2) \frac{d^2 \mathrm{P}}{d x^2}<0$ $\frac{d^2 \mathrm{P}}{d x^2}=-50<0$
Hence, $20$ units should be produced for maximum profit.
View full question & answer→Question 75 Marks
Find the maximum and minimum values of $f(x)=x^{3}-x^{2}-x+2$.
Answer$ f(x)=x^3-x^2-x+2$
$ \therefore f^{\prime}(x)=3 x^2-2 x-1 \text { and } f^{\prime \prime}(x)=6 x-2 $ For the maximum and minimum values of $f(x), f^{\prime}(x)=0$
$\therefore 3 x^2-2 x-1=0$
$ \therefore 3 x^2-3 x+x-1=0$
$ \therefore 3 x(x-1)+1(x-1)=0$
$ \therefore(x-1)(3 x+1)=0$
$ \therefore x-1=0 \text { OR } 3 x+1=0$
$ \therefore x=1 \text { OR } x=-1$ For the maximum value of $f(x)$ we must have $f^{\prime \prime}(x)<0$ and for the minimum value of $f(x)$ we must have $f^{\prime \prime}(x)>0$.
$ f^{\prime \prime}(x)=6 x-2 $
At $x=1$, $ f^{\prime \prime}(x)=6(1)-2=4>0$
So, at $x=1, f(x)$ is minimum. At $x=-\frac{1}{3}$
$f^{\prime \prime}(x)=6\left(-\frac{1}{3}\right)-2=-2-2=-4<0$
So, at $x=-\frac{1}{3}, f(x)$ is maximum.
Maximum value of $f(x)$ :
Putting, $x=-\frac{1}{3}$ in $f(x)=x^3-x^2-x+2$
Minimum value of $f(x)$ : Putting, $x=1$ in $f(x)=x^3-x^2-x+2$
$f(x)_{\min .}=(1)^3-(1)^2-1+2$
$ =1-1-1+2=1$
Hence, at $x=-\frac{1}{3}$, the maximum value of $f(x)$ obtained is $\frac{59}{27}$ and at $x=1$, the minimum value of $f(x)$ obtained is $1.$ View full question & answer→Question 85 Marks
Find the values of $x$ which maximize or minimize $f(x)=2 x^{3}+3 x^{2}-36 x+10$. Also find -the maximum and minimum values of $f(x)$.
Answer$ f(x)=2 x^3+3 x^2-36 x+10$
$ \therefore f^{\prime}(x)=2\left(3 x^2\right)+3(2 x)-36(1)+0$
$ =6 x^2+6 x-36$
$ \text { and } f^{\prime \prime}(x)=6(2 x)-6(1)-0$
$ =12 x-6 $ For maximum and minimum value of $f(x), f^{\prime}(x)=0$
$ \therefore 6 x^2+6 x-36=0$
$ \therefore x^2+x-6=0$
$ \therefore x^2+3 x-2 x-6=0$
$ \therefore x(x+3)-2(x+3)=0$
$ \therefore(x+3)(x-2)=0$
$ x+3=0 \text { OR } x-2=0$
$ x=-3 \text { OR } x=2 $ For maximum value of $f(x)$ we must have $f^{\prime \prime}(x)<0$ and for minimum value of $f(x)$
we must have $f^{\prime \prime}(x)>0$. $f^{\prime \prime}(x)=12 x-6$ At $x=-3$, $f^{\prime \prime}(x)=12(-3)-6=-36-6=-42<0$
So, at $x=-3, f(x)$ is maximum. At $x=$
$2, f^{\prime \prime}(x)=12(2)+6=24+6=30>0$
So, at $x=2, f(x)$ is minimum
Maximum value of $f(x)$ :
Putting, $x=-3$ in $f(x)=2 x^3+3 x^2-36 x+10$
$f(x)_{\max }=2(-3)^3+3(-3)^2-36\{-3)+10$
$ =2(-27)+3(9)+108+10$
$ =-54+27+118=91$
Minimum value of $f(x)$ :
Putting, $x=2$ in $f(x)=2 x^3+3 x^2-36 x+10$
$f(x)_{\min .}=2(2)^3+3(2)^2-36(2)+10$
$ =2(8)+3(4)-72+10$
$ =16+12-62=-34$
Hence, at $x=-3$, maximum value of $f(x)$ obtained is $91$ and $x=2$, minimum value of $f(x)$ obtained is $-34 .$
View full question & answer→Question 95 Marks
Find the values of $x$ which maximize or minimize $y=2 x^{3}-15 x^{2}+36 x+12$. Also find the maximum and minimum values of $y$.
Answer$ y=2 x^3-15 x^2+36 x+12$
$ \therefore \frac{d y}{d x}=2\left(3 x^2\right)-15^{\prime}(2 x)+36(1)+0$
$ =6 x^2-30 x+36$
$ \text { and } \frac{d^2 y}{d x^2}=6(2 x)-30(1)+0$
$ =12 x-30 $ For maximum or minimum value of $y$, $ \frac{d y}{d x}=0$
$ \therefore 6 x^2-30 \mathrm{x}+36=0$
$ \therefore x^2-5 \mathrm{x}+6=0$
$ \therefore x^2-3 \mathrm{x}-2 \mathrm{x}+6=0$
$ \therefore x(x-3)-2(x-3)=0$
$ \therefore(x-3)(x-2)=0$
$ \therefore x-3=0 \text { OR } x-2=0$
$ \therefore x=3 \text { OR } x=2 $
For maximum value of $y$ we must have $\frac{d^2 y}{d x^2}<0$ and for minimum value of $y$,
We must have $\frac{d^2 y}{d x^2}>0$ Now, at $x=3$
$ \frac{d^2 y}{d x^2}=12(3)-30=36-30=6>0 $
So, at $x=3$, $y$ will be minimum.
At $x=2$, $ \frac{d^2 y}{d x^2}=12(2)-30=24-30=-6<0 $
So, at $x=2$, $y$ will be maximum.
Maximum value of $y$ :
Putting, $x=2$ in $y=2 x^3-15 x^2+36 x+12$
$y_{\max }=2\left(2^3\right)-15\left(2^2\right)+36(2)+12$
$ =2(8)-15(4)+72+12$
$ =100-60$
$ =40$
Minimum value of $y$ :
Putting, $x=3$ in $y=2 x^3-15 x^2+36 x+12$
$y \min =2\left(3^3\right)-15\left(3^2\right)+36(3)+12$
$ =2(27)-15(9)+108+12$
$ =54-135+120$
$ =174-135$
$ =39$
Hence, at $x=2$, maximum value of $y$ obtained is $40$ and at $x=3$, minimum value of $y$ obtained is $39 .$
View full question & answer→Question 105 Marks
The demand function for an item is $x=60-3 p$ and total cost function is $C=\frac{x^{2}}{20}+50$ what should be the production for maximum profit? Find this maximum profit.
Answer$\bar{P}=20 x-\frac{23}{60} x^{2}-50, x=26 \cdot P_{\max }=211$
View full question & answer→Question 115 Marks
For a firm demand function and cost function are as follows: $x=75-3 p, C=100+3 x$ , How much will the firm produce for maximum profit? Find the maximum profit.
Answer$P =22 x -\frac{1}{3} x^{2}-100, x=33 . P_{\max }=263$
View full question & answer→Question 125 Marks
The demand function of an item is $x=50+p-p^{2}$. Find the elasticity of demand when price is $5$ units and interpret it.
AnswerElasticity of demand $1.5$ Interpretaüon $p\ 5$ then $x\ 30$. Mence, the change of $1 \%$ in price, results in $1.5 \%$ of change in demand, $($in opposite direction$).$
View full question & answer→Question 135 Marks
Obtain the maximum and minimum values of $f(x)=2 x^{3}-21 x^{2}+60 x-8$.
Answer$x=2, f(x)_{\max }=44 ; x=5, f(x)_{\min }=17$
View full question & answer→Question 145 Marks
If $f ( x )=2 x^{3}-15 x^{2}+36 x +10$, obtain the minimum and maximum values of $f ( x )$.
Answer$x=3, f(x)_{\min }=37 ; x=2, f(x)_{\max }=38$
View full question & answer→Question 155 Marks
Obtain the maximum and minimum values of $f(x)=2 x^{3}+3 x^{2}+12$
Answer$x=-1, f(x)_{\max }=13 ; x=0, f(x)_{\min }=12$
View full question & answer→Question 165 Marks
If $y=x^{3}+6 x^{2}-7$, for what value of $x$ will the value of $y$ be maximum? Find this maximum value.
Answer$x=-4, y_{\max }=25$
View full question & answer→Question 175 Marks
If $y=x^{3}-3 x^{2}+5$, for what value of $x$, will the value of $y$ be minimum? Obtain this minimum value.
View full question & answer→Question 185 Marks
Obtain the maximum and minimum values of $f(x)=x+\frac{1}{x}$
Answer$x=-1, f(x)_{\max }=-2 ; x=1, f(x)_{\min }=2$
View full question & answer→Question 195 Marks
$f(x)=4 x^{3}+19 x^{2}-14 x+3$, for what values of $x, f(x)$ will be maximum or minimum?
Answer$x=\frac{-7}{2}, f(x)_{\max }=\frac{453}{4} ; x=\frac{1}{3}, f(x)_{\min }=\frac{16}{27}$
View full question & answer→Question 205 Marks
Obtain maximum and minimum value of $y=x^{3}-9 x^{2}+24 x+2$
Answer$x=2, f(x)_{\max }=22 ; x=4, f(x)_{\min }=18$
View full question & answer→Question 215 Marks
If $f(x)=x^{3}+6 x^{2}-15 x+7$, obtain the maximum and minimum values of $f(x)$
Answer$x=-5, f(x)_{\max }=107 ; x=1, f(x)_{\min }=-1$
View full question & answer→Question 225 Marks
If $y=\frac{2 x^{3}}{3}-\frac{3 x^{2}}{2}-9 x+11$ for what values of $x, y$ will be maximum or minimum?
AnswerAt $x=-\frac{3}{2}, y$ is maximum; at $x=3, y$ is minimum
View full question & answer→Question 235 Marks
If $y=\frac{x^{3}}{3}+\frac{x^{2}}{2}-2 x+2$ for what values of $x, y$ will be maximum or minimum?
Answer$\overline{\text { At } x}=-2, y$ is maximum ; At $x=1, y$ is minimum
View full question & answer→