Question 14 Marks
Obtain derivative of $f(x)=\sqrt{x}$ with the help of definition.
AnswerHere, $f(x)=\sqrt{x}$
$\therefore f(x+h)=\sqrt{x+h}$
Now, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$ =\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} $
$($Multiplying numerator and denominator by $\sqrt{x+h}+\sqrt{x} )$
$ = \lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$
$= \lim _{h \rightarrow 0} \frac{(\sqrt{x+h})^{2}-(\sqrt{x})^{2}}{h(\sqrt{x+h}+\sqrt{x})}$
$= \lim _{h \rightarrow 0} \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$
$= \lim _{h \rightarrow 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x})}$
$= \lim _{h \rightarrow 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}$
$= \frac{1}{\sqrt{x+0}+\sqrt{x}}$
$= \frac{1}{2 \sqrt{x}} $
Hence, if $f(x)=\sqrt{x}$ then $f^{\prime}(x)=\frac{1}{2 \sqrt{x}}$
View full question & answer→Question 24 Marks
If the production cost function for a producer is $C=100+0.015 x^{2}$ and revenue function is $R=3 x$ then find the profit function. How many units should be produced by the producer for maximum profit ?
AnswerProduction cost function $C=100+0.015 x^{2}$ and revenue function $R=3 x$
Now, profit function $P=R-C$
$ =3 x-\left(100+0.015 x^{2}\right)$
$\therefore P=3 x-100-0.015 x^{2}$
$\therefore \frac{d P}{d x} =3-0.015(2 x)$
$ =3-0.03 x $
Putting $\frac{d P}{d x}=0$
$ 3-0.03 x=0$
$ 3=0.03 x$
$ x=\frac{3}{0.03}$
$ x=100 $
$ \therefore 3=0.03 x $
$ \therefore x=\frac{3}{0.03} $
Now $\frac{d^{2} P}{d x^{2}}=0-0.03(1)$
$ =-0.03 $
Here putting $x=100$ in $\frac{d^{2} P}{d x^{2}}$,
$ \frac{d^{2} P}{d x^{2}}=-0.03<0 $
$\therefore \quad$ At $x=100$, profit is maximum.
View full question & answer→Question 34 Marks
A factory produces $x$ units and its production capacity is $60,000$ units per day. Its daily total production cost is $C=250000+0.08 x+\frac{200000000}{x}$. How many units should be produced for minimum production cost $?$
AnswerProduction cost function $C=250000+0.08 x+\frac{200000000}{x}$
$ \therefore \frac{d C}{d x}=0.08-\frac{200000000}{x^{2}} $
Putting $\frac{d C}{d x}=0$
$ 0.08-\frac{200000000}{x^{2}}=0$
$\therefore 0.08=\frac{200000000}{x^{2}}$
$\therefore 0.08 x^{2}=200000000$
$\therefore x^{2}=2500000000$
$\therefore x=50000 \text { or } x=-50000 $
Production cannot be negative, so we will take $x=50000$.
Now $\frac{d^{2} C}{d x^{2}}=\frac{400000000}{x^{3}}$
Here, putting $x=50000$ in $\frac{d^{2} C}{d x^{2}}$,
$ \frac{d^{2} C}{d x^{2}}=\frac{400000000}{(50000)^{3}}>0 $
$\therefore \quad$ Production cost is minimum at $x=50000$.
Thus, $50,000$ units should be produced so that the production cost is minimum
View full question & answer→Question 44 Marks
The daily cost of production for $x$ tons of a commodity is $10 x^{2}-1000 x+50000$. How many units should be produced for the minimum cost ? Also find the minimum cost.
AnswerProduction cost function $C=10 x^{2}-1000 x+50000$
$ \therefore \quad \frac{d C}{d x}=20 x-1000 $
Putting $\frac{d C}{d x}=0$,
$ 20 x-1000=0$
$\therefore 20 x=1000$
$\therefore x=50 $
Now $\frac{d^{2} C}{d x^{2}}=20$
Here, putting $x=50$ in $\frac{d^{2} C}{d x^{2}}$,
$ \frac{d^{2} C}{d x^{2}}=20>0 $
$\therefore$ Production cost is minimum at $x=50$.
To find minimum cost, put $x=50$ in the production cost function,
Minimum Cost $=10(50)^{2}-1000(50)+50000$
$ =10(2500)-50000+50000$
$ =25000 $
View full question & answer→Question 54 Marks
The demand function of a commodity is $x=50-4 p$. Find elasticity of demand when price is $p=5$ and interpret it.
AnswerDemand function $x=50-4 p$
$ \therefore \frac{d x}{d p} =0-4(1)$
$ =-4 $
Now, elasticity of demand $=-\frac{p}{x} \cdot \frac{d x}{d p}$
$ =\frac{-p}{(50-4 p)} \times(-4)$
$ =\frac{4 p}{50-4 p} $
When price $p=5$ then
$ \text { Elasticity of demand } =\frac{4(5)}{50-4(5)}$
$ =\frac{20}{50-20}$
$ =\frac{20}{30}$
$ =0.67 $
Interpretation : When the price changes by $1$ percent, demand changes by $0.67$ percent $($in opposite direction$)$ when the price is $5 .$
View full question & answer→Question 64 Marks
If $f(x)=x^{2}-4 x$ then decide whether the function is increasing or decreasing at $x = -1, x = 0$ and $x = 3.$
Answer$ f(x)=x^{2}-4 x$
$\therefore f^{\prime}(x)=2 x-4$
$\text { At } \boldsymbol{x}=-\mathbf{1}$
$f^{\prime}(-1)=2(-1)-4$
$=-6<0 $
$\therefore $ Function is decreasing at $x=-1$.
At $x=0$
$ f^{\prime}(0) =2(0)-4$
$ =-4<0 $
$\therefore $ Function is decreasing at $x=0$.
At $x=3$
$ f^{\prime}(3) =2(3)-4$
$ =2>0 $
$\therefore $ Function is increasing at $x=3$.
View full question & answer→Question 74 Marks
If $y=\left(x+\frac{6}{x+5}\right)\left(\frac{3 x+2}{x^{2}+5 x+6}\right)$ then find $\frac{d y}{d x}$
Answer$ y =\left(x+\frac{6}{x+5}\right)\left(\frac{3 x+2}{x^{2}+5 x+6}\right)$
$ =\left[\frac{x(x+5)+6}{x+5}\right]\left(\frac{3 x+2}{x^{2}+5 x+6}\right)$
$ =\left(\frac{x^{2}+5 x+6}{x+5}\right)\left(\frac{3 x+2}{x^{2}+5 x+6}\right)$
$ =\frac{3 x+2}{x+5} $
then find $\frac{d y}{d x}$.
Here, take $u=3 x+2$ and $v=x+5$.
$ \therefore \frac{d u}{d x}=3 \text { and } \frac{d v}{d x}=1 $
Now, $ y=\frac{u}{v}$
$ \therefore \frac{d y}{d x} =\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$
$ =\frac{(x+5)(3)-(3 x+2)(1)}{(x+5)^{2}}$
$ =\frac{(3 x+15)-(3 x+2)}{(x+5)^{2}}$
$ =\frac{3 x+15-3 x-2}{(x+5)^{2}}$
$ =\frac{13}{(x+5)^{2}} $
View full question & answer→Question 84 Marks
In a market demand law of price is $x = 3(60 - P).$ Find the demand for maximum revenue. Also find the price for the demand.
Answer
| Here, demand function $x = 3(60 - P)$ |
| $\therefore x = 180 – 3P$ |
| $\therefore 3P = 180 – x$ |
$\therefore P = 60 - $ |
Now, Revenue function $R = P ∙ x =$  $∙x$ |
$\therefore R = 60x -$  |
$\therefore $  $= 60 –$  |
| Putting $= 0$ |
| $\therefore 60-$ $= 0$ |
| $\therefore 180 – 2x = 0$ |
| $\therefore 2x = 180$ |
| $\therefore x = 90$ |
Now, $=$  |
Here, putting $x = 90$ in  |
| $=$ $< 0$ |
| $\therefore $ Revenue is maximum at $x = 90.$ |
| Now, we shall find the corresponding price. |
| Putting $x = 90$ in demand function $P = 60 - $ |
$P = 60 -$  |
| $= 60 – 30$ |
| $\therefore P = 30$ |
| Now, maximum revenue $R = x \times P$ |
| $= 90 \times 30$ |
| $\therefore R = Rs. 2700$ |
| $\therefore $ Revenue is maximum $Rs. 2700$ at $x = 90$ and $p = 30.$ |
View full question & answer→Question 94 Marks
The demand function of an item is $P = 30 - \frac{x^2}{10}$. Find the demand and price for maximum revenue.
AnswerHere, demand function $P=30-\frac{x^2}{10}$
Now, revenue function $R=P \cdot x=\left(30-\frac{x^2}{10}\right) \cdot x$
$\therefore R=30 x-\frac{x^2}{10}$
$\therefore \frac{d R}{d x}=30 x-\frac{3 x^2}{10}$
Putting $\frac{d R}{d x}=0$
$\therefore 30 x -\frac{3 x^2}{10}=0$
$\therefore 300-3 x ^2=0$
$\therefore 3 x ^2=300$
$\therefore x ^2=100$
$\therefore x=10$
Now, $\frac{d^2 R}{d x^2}=\frac{-6x}{10}$
Here, putting $x=10$ in
$\frac{ d ^2 R }{ dx }=\frac{-6(10)}{10}=-6 < 0$
$\therefore$ Revenue is maximum at $x =10$.
Now we shall find the corresponding price.
Putting $x =10$ in demand function $P =30-\frac{x^2}{10}$
Price $P=30-\frac{(10)^2}{10}$
$=30-\frac{100}{10}$
$=30-10$
$P=20$
$\therefore$ Revenue is maximum at $=10$ and $P= Rs. 20$ .
View full question & answer→Question 104 Marks
What is minimum value of a function? State the conditions for minimum value.
Answer
| If $h$ is a small positive number and if $f(a) < f(a + h)$ and also $f(a) < f(a - h)$ then fix is said to be minimum at $x = (A)$ |
| Minimum value do not mean the smallest value of a function. |
| The function is minimum at $x = a$ only means that the value of the function is minimum in a small interval around $x = a.$ |
| The necessary and sufficient conditions for a function to be minimum at $x = a$ are as follows : |
| $(i)\ f'(a) = 0$ and $(ii)\ f”(a) > 0$ |
View full question & answer→Question 114 Marks
What is maximum value of a function ? State the conditions for maximum value.
Answer
| If $h$ is a small positive number and if $f(a) > f(a + h)$ and also $f(a) > f(a - h)$ then $f(x)$ is said to be maximum at $x = (A)$ |
| Maximum value do not mean largest value of function. |
| The function is maximum at $x = a$ only means that the value of function is maximum in a small interval around $x= a.$ |
| The necessary and sufficient condition for a function to be maximum at $x = a$ are as follows : |
| $(i)\ f'(a) = 0$ and $(ii)\ f”(a) < O$ |
View full question & answer→Question 124 Marks
How can it be decided using derivative that the function is increasing or decreasing at a point?
AnswerSuppose, $y = f(x)$ At $x = a,$ whether $f (x)$ is decreasing or Increasing can be decided as follows :
- At $x = a, y = f(a)$
- If h is a very small positive number, then when $x = a + h, y = f(a + h).$
- If $f(a + h) > f(a)$ and $f(a) > f(a – h).$ then at $x = a.\ f(x)$ Is Increasing function.
- Using derivative, if $f’(a) > 0,$ then at $x = a,$ the function $f (x)$ is said to be increasing function.
- If $f(a + h) < f(a)$ and $f(a) < f(a – h),$ then at $x = a, f(x)$ is decreasing function.
- Using derivative, if $f'(a) < 0,$ then the function $f(x)$ is said to be decreasing function.
View full question & answer→Question 134 Marks
The profit function of a merchant is $5x - 100 - 0.01x^2$. How many units should be produced for maximum profit?
AnswerHere, profit function $P=5 x-100-0.01 x^2$
$\therefore=x-0-0.02 x$
$=5-0.02 x$
Putting $=0$
$5-0.02 x=0$
$\therefore 5=0.02 x$
$\therefore x=$
$\therefore x=250$
Now, $=-0.02(1)$
$=-0.02$
Here, putting $x=250$ in $=-0.02<0$
$\therefore$ At $x =250$ profit is maximum.
View full question & answer→Question 144 Marks
The profit function of a producer is $40x + 10000 – 0.1x^2$ at when production is the profit maximum? Also find this maximum profit.
AnswerHere, profit function $P = 40x + 10000 – 0.1x^2$
$\therefore = 40 + 0 – 0.2x = 40 – 0.2x$
Putting $= 0.$
$40 – 0.2x = 0$
$\therefore 40 = 0.2x$
$\therefore x = 200$
Now, $= 0 – 0.2(1) = -0.2$
Here, putting $x = 200$ in
$ = -0.2 < 0$
$\therefore $ At $x = 200$, profit is maximum.
Now, we shall find the maximum profit.
Putting $x = 200$ in profit function
$P = 40x + 10000 - 0.1x^2$
$P = 40(200) + 10000 - 0.1(200)^2$
$= 8,000 + 10000 - 0.1(40000)$
$= 18,000 – 4000$
$\therefore P = 14000$
$\therefore$ At $x = 200,$ profit is maximum $Rs. 14000.$
View full question & answer→Question 154 Marks
A manufacturing firm is producing $x$ units daily at the total cost of $\frac{x^{2}}{20}+4 x+30$. its demand function is $x=30-2 p$. What should be the production so that marginal revenue and marginal cost of the firm become equal?
View full question & answer→Question 164 Marks
For a producer demand law is $p=157-3 x$ and total cost function is $C=$ $1064+5 x+0.04 x^{2}$, For demand = production $=25$ units. Obtain marginal revenue and marginal cost.
AnswerMarginal revenue $=7$ Units, Marginal cost $7$ unlt
View full question & answer→Question 174 Marks
The demand function of an item is $p=-1500 x+30000$ and production cost function is $C=-12000 x+268000$. Obtain the profit function. Find the production for maximum profit. Also find the maximum profit.
Answer$P=-1500 x^{2}+42000 x-268000, x=14, \text { Maximum profit }=26000$
View full question & answer→Question 184 Marks
The demand function of an Item is $p=500-0.2 x$ and cost function is $C =25 x + 10000.$ Find profit function and obtain the price and production for maximum profit.
Answer$P =475 x -10000-0.2 x^{2}, x =1187.50, p =262.50$
View full question & answer→Question 194 Marks
The demand function of an item is $x=\frac{1}{3}(25-2 p)$ Obtain the revenue function and find the demand for maximum revenue.
Answer$R=\frac{1}{2}\left(25-3 x^{2}\right), x=\frac{25}{6}$
View full question & answer→Question 204 Marks
The profit function of a producer is $P=\left[-2 x^{2}+108 x-1000\right]$. Find the production for maximum profit. Also find the maximum profit.
Answer$x=27, \text { Maximum profit }=458$
View full question & answer→Question 214 Marks
A producer is producing $x$ items at the cost of $Rs \left(x^{2}+78 x+2500\right)$. If the demand function of the item is $x=\frac{600-p}{8}$ prove that when $x =29$, he will get maximum profit. What will be the price of the Item?
View full question & answer→Question 224 Marks
For a producer, the production cost function is $C=80-5 x+\frac{1}{4} x^{2}$. How much should be produced for minimum cost? Find the minimum cost.
Answer$x=10, C_{\min }=55$
View full question & answer→Question 234 Marks
The demand function of rice is $x=130-4 p$. Find the demand for maximum revenue. What will be maximum revenue?
Answer$x =65, R_{\max }=1056.25$
View full question & answer→Question 244 Marks
The demand function of pencils is $x=50-\frac{p}{3}$. Find the demand and price for maximum revenue.
View full question & answer→Question 254 Marks
Demand function of an item is $p=\frac{7500-x^{2}}{100}$. Find the demand and price for maximum revenue. Also find the maximum revenue.
Answer$x =50, p =50, R_{\max }=2500$
View full question & answer→Question 264 Marks
Demand function of an item is $p=15-\frac{1}{2} x$. Find the demand and price for maximum revenue.
View full question & answer→Question 274 Marks
Demand function of $TV$ is $p=13.5-\frac{x^{2}}{200}$. Find the demand for maximum revenue.
View full question & answer→Question 284 Marks
A producer is producing $x$ items at the cost of $\operatorname{Rs}\left(x^{2}+78 x+2500\right)$. If the demand function of the item is $x=\frac{600-p}{8}$ prove that when $x =29$, he will get maximum profit. What will be the price of the Item?
View full question & answer→Question 294 Marks
If total cost function is $C=\frac{1}{3} x^{3}-3 x^{2}-7 x+16$, then find the production for minimum cost.
View full question & answer→Question 304 Marks
The total cost function for an item is $C=5+\frac{3 x}{2}+\frac{24}{x}$. Find the production for minimum cost and this minimum cost.
View full question & answer→Question 314 Marks
If the production cost for $x$ items is $=\frac{x^{2}}{25}-6 x+250$, then prove that for $75$ units, production cost will be minimum. Also find this minimum cost.
Answer$\overline{C_{\min }}=25$
View full question & answer→Question 324 Marks
For a producer, the production cost function is $C=80-5 x+\frac{1}{4} x^{2}$. How much should be produced for minimum cost? Find the minimum cost.
Answer$x=10, C_{\min }=55$
View full question & answer→Question 334 Marks
The demand function of pencils is $x=50-\frac{p}{3}$. Find the demand and price for maximum revenue.
View full question & answer→Question 344 Marks
The demand function of rice is $x=130-4 p$. Find the demand for maximum revenue. What will be maximum revenue?
Answer$x=65, R_{\max }=1056.25$
View full question & answer→Question 354 Marks
Demand function of an item is $p=\frac{7500-x^{2}}{100}$. Find the demand and price for maximum revenue. Also find the maximum revenue.
Answer$x =50, p =50, R_{\max }=2500$
View full question & answer→Question 364 Marks
Demand function of $TV$ is $p=13.5-\frac{x^{2}}{200} .$ Find the demand for maximum revenue.
View full question & answer→Question 374 Marks
Demand function of an item is $p=15-\frac{1}{2} x$. Find the demand and price for maximum revenue.
View full question & answer→Question 384 Marks
The demand function of an item is $P=30-\frac{x^2}{10}$. Find demand and price for maximum revenue.
AnswerRevenue $R = P \cdot x = 30x - \frac{x^3}{10}$.
1. $\frac{dR}{dx} = 30 - \frac{3x^2}{10} = 0 \implies 300 = 3x^2 \implies x^2 = 100 \implies x = 10$.
2. $\frac{d^2R}{dx^2} = -\frac{6x}{10} = -6$ (Negative, so max).
Demand $x = 10$. Price $P = 30 - \frac{100}{10} = 20$.
View full question & answer→Question 394 Marks
Find the maximum and minimum values of : $f(x)=2x^{3}+3x^{2}-12x-4$
Answer$f'(x) = 6x^{2} + 6x - 12$
Set $f'(x)=0 : 6(x^{2}+x-2)=0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=-2, 1$.
$f''(x) = 12x + 6$.
For $x=-2$ : $f''(-2) = -24+6 = -18 < 0$ (Max).
Max value $f(-2) = 2(-8)+3(4)-12(-2)-4 = -16+12+24-4 = 16$.
For $x=1$ : $f''(1) = 12+6 = 18 > 0$ (Min).
Min value $f(1) = 2(1)+3(1)-12(1)-4 = 2+3-12-4 = -11$.
View full question & answer→Question 404 Marks
A toy is sold at ₹ 20. The total cost of producing x such toys is $C=1000+16.5x+0.001x^{2} $ rupees. How many toys should be produced for maximum profit ?
AnswerRevenue $ R = 20x $.
Profit $ P = R - C = 20x - (1000+16.5x+0.001x^{2}) = 3.5x - 0.001x^{2} - 1000 $.
$ P' = 3.5 - 0.002x $. Set $ P' = 0 \Rightarrow x = 3.5/0.002 = 1750 $.
$ P'' = -0.002 < 0 $. So profit is maximum at $ x=1750 $.
View full question & answer→Question 414 Marks
If the production cost function for a producer is $C =100+0.015 x^2$ and Revenue Function is R = 3x then find the profit function. How many units should be produced by the producer for maximum profit ?
Answer$P(x)=R(x)-C(x)$
$P(x)=3 x-\left(100+0.015 x^2\right)$
$=-0.015 x^2+3 x-100$
$P^{\prime}(x)=\frac{d}{d x}\left(-0.015 x^2+3 x-100\right)$
$P^{\prime}(x)=-0.03 x+3$
$0=-0.03 x+3$
$0.03 x=3$
$x=\frac{3}{0.03}$
$x=100$
$P^{\prime \prime}(x)=-0.03$
$P(x)=-0.015 x^2+3 x-100$
100 units
View full question & answer→