Exercise 3.2 - Chapter 3 Integral Calculus II 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $3.2$
Question 1.
The cost of an overhaul of an engine is $₹ 10,000$ The operating cost per hour is at the rate of $2 \mathrm{x}-240$ where the engine has run $x \mathrm{~km}$. Find out the total cost if the engine runs for 300 hours after overhaul.
Solution:
Given that the overhaul cost is Rs. 10,000 .
The marginal cost is $2 \mathrm{x}-240$
$
\begin{aligned}
& M C=2 x-240 \\
& C=\int M C d x+k \\
& C=x^2-240 x+k
\end{aligned}
$
$\mathrm{k}$ is the overhaul cost
$
\Rightarrow \mathrm{k}=10,000
$
So $\mathrm{C}=\mathrm{x}^2-240 \mathrm{x}+10,000$
When $\mathrm{x}=300$ hours, total cost is
$
\begin{aligned}
& \mathrm{C}=(300)^2-240(300)+10,000 \\
& \Rightarrow \mathrm{C}=90,000-72000+10,000 \\
& \Rightarrow \mathrm{C}=28,000
\end{aligned}
$
So the total cost of the engine run for 300 hours after the overhaul is ₹ 28,000 .
Question 2.
Elasticity of a function $\frac{\mathbf{E}_y}{\mathbf{E}_x}$ is given by $\frac{\mathbf{E}_y}{\mathbf{E}_x}=\frac{-7 x}{(1-2 x)(2+3 x)}$ Find the function when $\mathrm{x}=2, \mathrm{y}=$ $\frac{3}{8}$
Solution:

$\text { Given } \begin{aligned}
& \eta=\frac{\mathrm{E}_y}{\mathrm{E}_x}=\frac{-7 x}{(1-2 x)(2+3 x)} \Rightarrow \frac{x}{y} \frac{d y}{d x}=\frac{-7 x}{(1-2 x)(2+3 x)} \\
& \frac{d y}{y}=\frac{-7 x}{(1-2 x)(2+3 x)} \frac{d x}{x} \\
& \int \frac{d y}{y}=7 \int \frac{d x}{(2 x-1)(3 x+2)} \\
& \frac{1}{(2 x-1)(3 x+2)}=\frac{\mathrm{A}}{(2 x-1)}+\frac{\mathrm{B}}{(3 x+2)} \\
& 1=\mathrm{A}(3 x+2)+\mathrm{B}(2 x-1) \\
& \text { Let } x=\frac{1}{2} \Rightarrow 1=\mathrm{A}\left(\frac{3}{2}+2\right) \\
& \text { Let } x=\frac{-2}{3} \Rightarrow 1=\mathrm{A}\left(\frac{7}{2}\right) \Rightarrow \mathrm{A}=\frac{2}{7} \\
& 1=\mathrm{B}\left(\frac{-4}{3}-1\right) \\
& 1=\mathrm{B}\left(\frac{-7}{3}\right) \Rightarrow \mathrm{B}=\frac{-2}{7}
\end{aligned}$

Using these values in (1) we get
$
\begin{aligned}
& \int \frac{d y}{y}=7 \int \frac{\frac{2}{7}}{2 x-1} d x-7 \int \frac{\frac{3}{7}}{3 x+2} d x \\
& \int \frac{d y}{y}=\int \frac{2 d x}{2 x-1}-\int \frac{3 d x}{3 x+2} \\
& \log y=\log (2 x-1)-\log (3 x+2)+\log k
\end{aligned}
$
when $x=2, y=\frac{3}{8} \Rightarrow \frac{3}{8}=\frac{3}{8} k$
$
\Rightarrow k=1
$
Hence the function is $y=\frac{2 x-1}{3 x+2}$

Question 3.
The elasticity of demand with respect to price for a commodity is given by $\frac{(4-x)}{x}$ where $\mathrm{p}$ is the price when demand is $x$. Find the demand function when the price is 4 and the demand is 2. Also, find the revenue function.
Solution:
$
\text { Given } \begin{aligned}
\eta_d & =\frac{4-x}{x} \\
\text { (i.e) } \frac{-\mathrm{p}}{x} \frac{d x}{d p} & =\frac{4-x}{x} \\
\Rightarrow \quad \frac{-d x}{4-x} & =\frac{d p}{p} \\
\int \frac{d x}{x-4} & =\int \frac{d p}{p} \\
\log (x-4) & =\log p+\log k \\
x-4 & =p k
\end{aligned}
$

When $p=4, x=2$ gives
$
\begin{aligned}
2-4 & =4 k \\
\Rightarrow \quad k & =-\frac{1}{2}
\end{aligned}
$
Hence
$
p=\frac{x-4}{\left(-\frac{1}{2}\right)}
$
$p=8-2 x$ is the demand function.
The Revenue function $\mathrm{R}=p x$
So
$
\mathrm{R}=8 x-2 x^2
$

Question 4.
A company receives a shipment of 500 scooters every 30 days. From experience, it is known that the inventory on hand is related to the number of days $\mathrm{x}$. Since the shipment, $\mathrm{I}(\mathrm{x})=500$ $-0.03 x^2$, the daily holding cost per scooter is $₹ 0.3$. Determine the total cost for maintaining inventory for 30 days.
Solution:
Here inventory $\mathrm{I}(\mathrm{x})=500-0.03 \mathrm{x}^2$
Unit holding cost $C_1=₹ 0.3$
$\mathrm{T}=30$ days
So total inventory carrying cost
$
\begin{aligned}
= & C_1 \int_0^T \mathrm{I}(x) d x \\
= & 0.3 \int_0^{30}\left(500-0.03 x^2\right) d x \\
= & 0.3\left(500 x-\frac{0.03 x^3}{3}\right)_0^{30} \\
= & 0.3\left[500(30)-\frac{0.03}{3}(30)^3\right] \\
= & 0.3[15000-270] \\
= & 4419
\end{aligned}
$
Hence the total cost for maintaining inventory for 30 days is ₹ 4,419 .

Question 5.
An account fetches interest at the rate of $5 \%$ per annum compounded continuously. An individual deposits ₹ 1,000 each year in his account. How much will be in the account after 5 years. $\left(\mathrm{e}^{0.25}=1.284\right)$
Solution:
$
\begin{aligned}
\mathrm{p}=1000, \mathrm{~N} & =5, \mathrm{r}=5 \%=0.05 \\
\text { Annuity } & =\int_0^5 1000 e^{0.05 t} d t \\
& =\frac{1000}{0.05}\left(e^{0.05 t}\right)_0^5 \\
& =20000\left[e^{0.25}-\mathrm{e}^0\right] \\
& =20000(1.284-1] \\
& =5680
\end{aligned}
$
After 5 years ₹ 5680 will be in the account
Question 6.
The marginal cost function of a product is given by $\frac{d c}{d x}=100-10 \mathrm{x}+0.1 \mathrm{x}^2$ where $\mathrm{x}$ is the output. Obtain the total and the average cost function of the firm under the assumption, that its fixed cost is ₹ 500 .
Solution:
Given $\mathrm{MC}=\frac{d c}{d x}=100-10 \mathrm{x}+0.1 \mathrm{x}^2$
$
\begin{aligned}
& \mathrm{C}=\int \mathrm{MC} \mathrm{dx}+\mathrm{k} \\
& \mathrm{C}=\int\left(100-10 \mathrm{x}+0.1 \mathrm{x}^2\right) \mathrm{dx}+\mathrm{k} \\
& \mathrm{C}=100 \mathrm{x}-5 \mathrm{x}^2+\frac{0.1 x^3}{3}+\mathrm{k}
\end{aligned}
$
The fixed cost is $500 \Rightarrow \mathrm{k}=500$
Hence total cost function $=100 \mathrm{x}-5 \mathrm{x}^2+\frac{0.1 x^3}{3}+500$
Average cost function $\mathrm{AC}=\frac{c}{x}$
$
=100-5 x+\frac{x^2}{30}+\frac{500}{x}
$

Question 7.
The marginal cost function is $\mathrm{MC}=300 x^{\frac{2}{5}}$ and fixed cost is zero. Find out the total cost and average cost functions.
Solution:
Given
$
\begin{aligned}
& \mathrm{MC}=300 x^{\frac{2}{5}} \\
& \mathrm{C}=\int 300 x^{\frac{2}{5}} d x+k=300 \frac{x^{\frac{7}{5}}}{\frac{7}{5}}+k \\
& \mathrm{ost}=0, k=0
\end{aligned}
$
So
$
C=\frac{1500}{7} x^{\frac{7}{5}}
$
Average cost $=\frac{\mathrm{C}}{x}=\frac{1500}{7} x^{\frac{2}{5}}$
Question 8.
If the marginal cost function of $\mathrm{x}$ units of output is $\frac{a}{\sqrt{a x+b}}$ and if the cost of output is zero. Find the total cost as a function of $\mathrm{x}$.
Solution:
Given
$
\begin{aligned}
\mathrm{MC} & =\frac{a}{\sqrt{a x+b}} \\
\text { Total cost } & =\int \frac{a}{\sqrt{a x+b}} d x+k \\
C & =2 \sqrt{a x+b}+k
\end{aligned}
$
The cost of output is zero $\Rightarrow x=0, \mathrm{C}=0$
$
0=2 \sqrt{b}+k \Rightarrow k=-2 \sqrt{b}
$
So total cost function is $2 \sqrt{a x+b}-2 \sqrt{b}$

Question 9.
Determine the cost of producing 200 air conditioners if the marginal cost (is per unit) is $\mathrm{C}^{\prime}(\mathrm{x})$ $=\left(\frac{x^2}{200}+4\right)$
Solution:
Given $\mathrm{MC}=\mathrm{C}^{\prime}(x)=\left(\frac{x^2}{200}+4\right)$
Total cost
$
\begin{aligned}
C & =\int\left(\frac{x^2}{200}+4\right) d x+k \\
C & =\frac{x^3}{600}+4 x+k
\end{aligned}
$
When $x=0, c=0 \Rightarrow k=0$
So
$
\begin{aligned}
& C=\frac{x^3}{600}+4 x \\
& \text { When } x=200, \quad C=\frac{(200)^3}{600}+4(200)=\frac{8,000,000}{600}+800 \\
& \mathrm{C}=14133.33 \\
&
\end{aligned}
$
So the cost of producing 200 air conditioners is $₹ 14133.33$
Question 10.
The marginal revenue (in thousands of Rupees) functions for a particular commodity is $5+$ $3 \mathrm{e}^{-0.03 \mathrm{x}}$ where $\mathrm{x}$ denotes the number of units sold. Determine the total revenue from the sale of 100 units. (Given $\mathrm{e}^{-3}=0.05$ approximately)
Solution:
Given, marginal Revenue $R^{\prime}(x)=5+3 e^{-0.03 x}$
Total revenue from the sale of 100 units is

$\begin{aligned}
& \mathrm{R}=\int_0^{100}\left(5+3 e^{-0.03 x}\right) d x \\
& \mathrm{R}=\left[5 x+\frac{3 e^{-0.03 x}}{-0.03}\right]_0^{100} \\
& \mathrm{R}=\left(500+\frac{3 e^{-0.03(100)}}{-0.03}\right)-\left(0-\frac{3}{0.03}\right) \\
& \mathrm{R}=500-100 e^{-3}+100 \\
& \mathrm{R}=600-100(0.05)=595 \\
& \text { Total revenue }=595 \times 1000=₹ 5,95,000
\end{aligned}$

Question 11.
If the marginal revenue function for a commodity is $M R=9-4 x^2$. Find the demand function.
Solution:
Given, marginal Revenue function $M R=9-4 x^2$
Revenue function, $\mathrm{R}=\int(\mathrm{MR}) \mathrm{dx}+\mathrm{k}$
$
\begin{aligned}
& \mathrm{R}=\int\left(9-4 x^2\right) d x+k \\
& \mathrm{R}=9 x-\frac{4}{3} x^3+k \\
& k=0
\end{aligned}
$
Since $\mathrm{R}=0$ when $x=0, k=0$
$
\begin{aligned}
& \mathrm{R}=9 x-\frac{4}{3} x^3 \\
& \mathrm{P}=\frac{R}{x} \\
& \mathrm{P}=9-\frac{4}{3} x^2
\end{aligned}
$

Question 12.
Given the marginal revenue function $\frac{4}{(2 x+3)^2}-1$, show that the average revenue function is $\mathrm{P}=\frac{4}{6 x+9}-1$
Solution:
Given
Since $\mathrm{R}=0$ when $x=0$
$
\begin{aligned}
\mathrm{MR} & =\frac{4}{(2 x+3)^2}-1 \\
\mathrm{R} & =\int \frac{4}{(2 x+3)^2} d x-\int d x \\
\mathrm{R} & =\frac{4}{-(2 x+3)^2}-x+k
\end{aligned}
$
So
$
0=\frac{2}{-3}+k \Rightarrow k=\frac{2}{3}
$
$\mathrm{R}=\frac{-2}{2 x+3}-x+\frac{2}{3}$
Average revenue function
$\mathrm{P}=\frac{R}{x}$
$P=\frac{-2}{x(2 x+3)}-1+\frac{2}{3 x}$
$=\frac{2}{x}\left[\frac{1}{3}-\frac{1}{2 x+3}\right]-1$
$=\frac{2}{x}\left[\frac{2 x+3-3}{3(2 x+3)}\right]-1$
$=\frac{2}{x}\left(\frac{2 x}{3(2 x+3)}\right)-1$
$P=\frac{4}{6 x+9}-1$
which is the required answer.

Question 13.
A firm's marginal revenue function is $\mathrm{MR}=20 e^{\frac{-x}{10}}\left(1-\frac{x}{10}\right)$. Find the corresponding demand function.
Solution:
$
\begin{aligned}
\mathrm{MR} & =20 e^{\frac{-x}{10}}\left(1-\frac{x}{10}\right) \\
\mathrm{R} & =\int 20 e^{\frac{-x}{10}}\left(1-\frac{x}{10}\right) d x+k \\
\mathrm{R} & =20 \int\left(e^{\frac{-x}{10}}-\frac{x}{10} e^{\frac{-x}{10}}\right) d x+k \\
\mathrm{R} & =20 \int d\left(x e^{\frac{-x}{10}}\right)+k
\end{aligned}
$
$
\mathrm{R}=20 x e^{\frac{-x}{10}}+k
$
When $x=0, R=0$, so $k=0$
$
\begin{aligned}
& \mathrm{R}=20 x e^{\frac{-x}{10}} \\
& \mathrm{P}=\frac{R}{x}=20 e^{\frac{-x}{10}}
\end{aligned}
$

Question 14.
The marginal cost of production of a firm is given by $\mathrm{C}^{\prime}(\mathrm{x})=5+0.13 \mathrm{x}$, the marginal revenue is given by $\mathrm{R}^{\prime}(\mathrm{x})=18$ and the fixed cost is $₹ 120$. Find the profit function.
Solution:
$
\begin{aligned}
& \mathrm{MC}=\mathrm{C}^{\prime}(\mathrm{x})=5+0.13 \mathrm{x} \\
& \mathrm{C}(\mathrm{x})=\int \mathrm{C}^{\prime}(\mathrm{x}) \mathrm{dx}+\mathrm{k}_1 \\
& =\int(5+0.13 \mathrm{x}) \mathrm{dx}+\mathrm{k}_1 \\
& =5 \mathrm{x}+\frac{0.13}{2} x^2+\mathrm{k}_1
\end{aligned}
$
When quantity produced is zero, fixed cost is 120
(i.e) When $\mathrm{x}=0, \mathrm{C}=120 \Rightarrow \mathrm{k}_1=120$
Cost function is $5 \mathrm{x}+0.065 \mathrm{x}^2+120$
Now given $\mathrm{MR}=\mathrm{R}^{\prime}(\mathrm{x})=18$
$\mathrm{R}(\mathrm{x})=\int 18 \mathrm{dx}+\mathrm{k}_2=18 \mathrm{x}+\mathrm{k}_2$
When $\mathrm{x}=0, \mathrm{R}=0 \Rightarrow \mathrm{k}_2=0$
Revenue $=18 \mathrm{x}$
Profit $\mathrm{P}=$ Total Revenue $-$ Total cost $=18 \mathrm{x}-\left(5 \mathrm{x}+0.065 \mathrm{x}^2+120\right)$
Profit function $=13 \mathrm{x}-0.065 \mathrm{x}^2-120$
Question 15.
If the marginal revenue function is $R^{\prime}(x)=1500-4 x-3 x^2$. Find the revenue function and average revenue function.
Solution:
Given marginal revenue function
$
M R=R^{\prime}(x)=1500-4 x-3 x^2
$
Revenue function $\mathrm{R}(\mathrm{x})=\int \mathrm{R}^{\prime}(\mathrm{x}) \mathrm{dx}+\mathrm{c}$
$
R=\int\left(1500-4 x-3 x^2\right) d x+c
$
$
R=1500 x-2 x^2-x^3+c
$
When $\mathrm{x}=0, \mathrm{R}=0 \Rightarrow \mathrm{c}=0$
So $R=1500 x-2 x^2-x^3$
Average revenue function $\mathrm{P}=\frac{R}{x} \Rightarrow 1500-2 \mathrm{x}-\mathrm{x}^2$

Question 16.
Find the revenue function and the demand function if the marginal revenue for $\mathrm{x}$ units is $M R$ $=10+3 \mathrm{x}-\mathrm{x}^2$
Solution:
Given
$
\begin{aligned}
\mathrm{MR} & =10+3 x-x^2 \\
\mathrm{R}(x) & =\int(\mathrm{MR}) d x+k \\
R & =\int\left(10+3 x-x^2\right) d x+k \\
& =10 x+\frac{3}{2} x^2-\frac{x^3}{3}+k
\end{aligned}
$
Revenue function $\quad \mathrm{R}(x)=\int(\mathrm{MR}) d x+k$
When $x=0, R=0, \Rightarrow k=0$
Demand function
$
\begin{aligned}
& \mathrm{R}=10 x+\frac{3}{2} x^2-\frac{x^3}{3} \\
& \mathrm{P}=\frac{R}{x}=10+\frac{3}{2} x-\frac{x^2}{3}
\end{aligned}
$
Question 17.
The marginal cost function of a commodity is given by $\mathrm{MC}=\frac{14000}{\sqrt{7 x+4}}$ and the fixed cost is $₹$ 18,000 . Find the total cost and average cost.

Solution:
Given
$
\begin{aligned}
\mathrm{MC} & =\frac{14000}{\sqrt{7 x+4}} \quad \text { fixed cost }=₹ 18,000 \\
\text { Total cost } & =\int(\mathrm{MC}) d x+k \\
& =\int \frac{14000}{\sqrt{7 x+4}} d x+k \\
& =14000\left(\frac{2}{7} \sqrt{7 x+4}\right)+k \\
& =4000 \sqrt{7 x+4}+k
\end{aligned}
$
Since the fixed cost is $₹ 18,000$, when $x=0, k=18,000$
$
\begin{aligned}
& \Rightarrow \quad \text { Total cost } \quad C=4000 \sqrt{7 x+4}+18000 \\
& \text { Average cost } A \cdot C=\frac{C}{x} \\
& =\frac{4000}{x} \sqrt{7 x+4}+\frac{18000}{x} \\
&
\end{aligned}
$
Question 18.
If the marginal cost (MC) of production of the company is directly proportional to the number of units (x) produced, then find the total cost function, when the fixed cost is ₹ 5,000 and the cost of producing 50 units is ₹ 5,625 .
Solution:
Given that the marginal cost $\mathrm{MC}$ is directly proportional to the number of units $\mathrm{x}$.
That is, $\mathrm{MC} \propto \mathrm{x}$
$\mathrm{MC}=\mathrm{kx}$, where $\mathrm{k}$ is the constant of proportionality

Total cost $C=\int(\mathrm{MC}) d x+c_1$
$
\begin{aligned}
& =\int(k x) d x+c_1 \\
\mathrm{C} & =\frac{k x^2}{2}+c_1
\end{aligned}
$
The fixed cost is given as 5000 . So $c_1=5000$
When $x=50, \mathrm{C}=5625$
So
$
\mathrm{C}=\frac{k x^2}{2}+5000
$
$
\begin{aligned}
5625 & =\frac{k}{2}(50)^2+5000 \\
625 & =\frac{2500}{2} k \\
k & =\frac{1}{2}
\end{aligned}
$
Thus total cost function
$
\begin{aligned}
& C=\frac{1}{2}\left(\frac{x^2}{2}\right)+5000 \\
& C=\frac{x^2}{4}+5000
\end{aligned}
$
Question 19.
If $\mathrm{MR}=20-5 \mathrm{x}+3 \mathrm{x}^2$
Solution:
$
\begin{aligned}
\mathrm{MR} & =20-5 x+3 x^2 \\
\mathrm{R} & =\int(\mathrm{MR}) d x+k \\
& =\int\left(20-5 x+3 x^2\right) d x+k \\
\mathrm{R} & =20 x-\frac{5 x^2}{2}+x^3+k
\end{aligned}
$
Since $R=0$, when $x=0, k=0$
$
\Rightarrow \mathrm{R}=20 x-\frac{5 x^2}{2}+x^3 \text { is the total revenue function }
$

Question 20.
If $M R=14-6 x+9 x^2$, find the demand function.
Solution:
$
\begin{aligned}
& M R=14-6 x+9 x^2 \\
& R=\int\left(14-6 x+9 x^2\right) d x+k \\
& =14 x-3 x^2+3 x^3+k
\end{aligned}
$
Since $\mathrm{R}=0$, when $\mathrm{x}=0, \mathrm{k}=0$
So revenue function $R=14 x-3 x^2+3 x^3$
Demand function $\mathrm{P}=\frac{R}{x}=14-3 \mathrm{x}+3 \mathrm{x}^2$