Exercise 3.3 - Chapter 3 Integral Calculus II 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $3.3$
Question 1.
Calculate consumer's surplus if the demand function $\mathrm{p}=50-2 \mathrm{x}$ and $\mathrm{x}=20$
Solution:
Given demand function $\mathrm{p}=50-2 \mathrm{x}, \mathrm{x}_0=20$
$
\mathrm{CS}=\int_0^{x_0} p(x) d x-x_0 p_0
$
When $x=20, p_0=50-2(20)=10$
$
\begin{aligned}
C S & =\int_0^{20}(50-2 x) d x-(20)(10) \\
& =\left[50 x-x^2\right]_0^{20}-200 \\
& =[1000-400]-200=400
\end{aligned}
$
Hence the consumer's surplus is 400 units.
Question 2.
Calculate consumer's surplus if the demand function $p=122-5 x-2 x^2$, and $x=6$
Solution:
Demand function $\mathrm{p}=122-5 \mathrm{x}-2 \mathrm{x}^2$ and $\mathrm{x}=6$
when $\mathrm{x}=\mathrm{x}_0=6$
$
\begin{aligned}
& \mathrm{p}_0=122-5(6)-2(36) \\
& =122-30-72 \\
& =20
\end{aligned}
$
$
\begin{aligned}
\mathrm{CS} & =\int_0^6\left(122-5 x-2 x^2\right) d x-(20)(6) \\
& =\left[122 x-\frac{5 x^2}{2}-\frac{2 x^3}{3}\right]_0^6-120 \\
& =(122)(6)-\frac{5}{2}(36)-\frac{2}{3}(216)-120 \\
& =732-90-144-120=378
\end{aligned}
$

Hence the consumer's surplus is 378 units
Question 3.
The demand function $p=85-5 x$ and supply function $p=3 x-35$. Calculate the equilibrium price and quantity demanded. Also, calculate consumer's surplus.
Solution:
Given $\mathrm{p}_{\mathrm{d}}=85-5 \mathrm{x}$ and $\mathrm{p}_{\mathrm{s}}=3 \mathrm{x}-35$
At equilibrium prices $\mathrm{p}_{\mathrm{d}}=\mathrm{p}_{\mathrm{s}}$
$
\begin{aligned}
& 85-5 \mathrm{x}=3 \mathrm{x}-35 \\
& \Rightarrow 8 \mathrm{x}=120 \\
& \Rightarrow \mathrm{x}=15 \\
& \mathrm{p}_0=85-5(15)=85-75=10 \\
& p_0=85-5(15)=85-75=10 \\
& \mathrm{CS}=\int_0^{x_0} p d x-x_0 p_0, x_0=15 \\
& \mathrm{CS}=\int_0^{15}(85-5 x) d x-(15)(10) \\
&=\left(85 x-\frac{5 x^2}{2}\right)_0^{15}-150 \\
&=85(15)-\frac{5(225)}{2}-150 \\
&=562.5
\end{aligned}
$
The equilibrium price is ₹ 10 , the quantity demanded is 15 . The consumer surplus is $562.50$ units.

Question 4.
The demand function for a commodity is $\mathrm{p}=\mathrm{e}^{-\mathrm{x}}$. Find the consumer's surplus when $\mathrm{p}=0.5$.
Solution:
Given demand function $p=e^{-x}$
At $\mathrm{p}=0.5$, (i.e) $\mathrm{p}_0=0.5$,
we have $\mathrm{p}_0=e^{-x_0}$
$
\Rightarrow 0.5=e^{-x_0}
$
Taking $\log _{\mathrm{e}}$ on both sides
$
\begin{aligned}
& \log _e(0.5)=-\mathrm{x}_0 \\
& \log _e\left(\frac{1}{2}\right)=-x_0 \\
&-\log _e 2=-x_0 \\
& \Rightarrow x_0=\log _e 2 \\
& \mathrm{CS}=\int_0^{\log _e 2} e^{-x} d x-\left(\log _e 2\right)(0.5) \\
&=\left[-e^{-x}\right]_0^{\log _e 2}-\frac{\log _e 2}{2} \\
&=\frac{-1}{2}+1-\frac{\log _e 2}{2}=\frac{1}{2}-\frac{\log _e 2}{2} \\
& \mathrm{CS}=\frac{1}{2}\left[1-\log _e 2\right] \text { units }
\end{aligned}
$
Question 5.
Calculate the producer's surplus at $x=5$ for the supply function $p=7+x$.
Solution:
Given supply function is $\mathrm{p}=7+\mathrm{x}, \mathrm{x}_0=5$
$\mathrm{p}_0=7+\mathrm{x}_0=7+5=12$
Producer's surplus

$
\begin{aligned}
\mathrm{PS} & =x_0 p_0-\int_0^{x_0} p(x) d x \\
& =5(12)-\int_0^5(7+x) d x \\
& =60-\left(7 x+\frac{x^2}{2}\right)_0^5=60-35-\frac{25}{2}=\frac{25}{2}
\end{aligned}
$
Hence the producer's surplus is $\frac{25}{2}$ units
Question 6.
If the supply function for a product is $p=3 x+5 x^2$. Find the producer's surplus when $x=4$.

Solution:
Given the supply function $p_s=3 x+5 x^2$
when $\mathrm{x}=4$, (i.e) $\mathrm{x}_0=4$,
$
\begin{aligned}
\mathrm{p}_0 & =3(4)+5(4)^2=12+80=92 \\
\mathrm{PS} & =x_0 p_0-\int_0^{x_0} p_s(x) d x \\
& =4(92)-\int_0^4\left(3 x+5 x^2\right) d x \\
& =368-\left[\frac{3 x^2}{2}+\frac{5 x^3}{3}\right]_0^4 \\
& =368-\left[\frac{48}{2}+\frac{5}{3}(64)\right]=368-24-106.67 \\
& =237.33
\end{aligned}
$
Hence the producer's surplus is $237.3$ units.
Question 7.
The demand function for a commodity is $p=\frac{36}{x+4}$. Find the consumer's surplus when the prevailing market price is $₹ 6$.
Solution:
Given $\mathrm{p}=\frac{36}{x+4}$
The marker price is ₹ 6 (i.e) $p_0=6$

$
\begin{aligned}
p_o & =\frac{36}{x_0+4} \Rightarrow 6=\frac{36}{x_0+4} \Rightarrow x_o=2 \\
\mathrm{CS} & =\int_0^2\left(\frac{36}{x+4}\right) d x-p_0 x_0 \\
& =36 \int_0^2\left(\frac{1}{x+4}\right) d x-(6)(2) \\
& =36[\log (x+4)]_0^2-12 \\
& =36[\log 6-\log 4]-12 \\
& =36 \log \frac{3}{2}-12
\end{aligned}
$
So the consumer's surplus when the prevailing market price is ₹ 6 is $\left(36 \log \frac{3}{2}-12\right)$ units.
Question 8.
The demand and supply functions under perfect competition are $p_d=1600-x^2$ and $p_s=2 x^2$ $+400$ respectively. Find the producer's surplus.
Solution:
Given demand function $\mathrm{p}_{\mathrm{d}}=1600-\mathrm{x}^2$ and
Supply function $p_s=2 x^2+400$
Perfect competition means there is equilibrium between supply and demand
$
\begin{aligned}
& \mathrm{p}_{\mathrm{s}}=\mathrm{p}_{\mathrm{d}} \\
& \Rightarrow 1600-\mathrm{x}^2=2 \mathrm{x}^2+400 \\
& \Rightarrow 3 \mathrm{x}^2=1200 \\
& \Rightarrow \mathrm{x}^2=400 \\
& \Rightarrow \mathrm{x}=\pm 20
\end{aligned}
$
The value of $x$ cannot be negative. So $x=20$ we take $x_0=20$.
$
\mathrm{p}_0=1600-(20)^2=1600-400=1200
$

$
\begin{aligned}
\text { PS } & =x_0 p_0-\int_0^{x_0} p_s d x \\
& =(20)(1200)-\int_0^{20}\left(2 x^2+400\right) d x \\
& =24000-\left[\frac{2 x^3}{3}+400 x\right]_0^{20} \\
& =24000-\left[\frac{16000}{3}+8000\right] \\
& =16000-\frac{16000}{3}=\frac{32000}{3}
\end{aligned}
$
Hence the producer's surplus is $\frac{32000}{3}$ units.
Question 9.
Under perfect competition for a commodity the demand and supply laws are $p_d=\frac{8}{x+1}-2$ and $p_s=\frac{x+3}{2}$ respectively. Find the consumer's and producer's surplus.
Solution:
Given $p_d=\frac{8}{x+1}-2$ and $p_s=\frac{x+3}{2}$
Here, since there is perfect competition, there is equilibrium, that is $p_d=p_s$
$
\begin{aligned}
\frac{8}{x+1}-2 & =\frac{x+3}{2} \\
\frac{8-2 x-2}{x+1} & =\frac{x+3}{2} \\
\frac{6-2 x}{x+1} & =\frac{x+3}{2} \\
(x+1)(x+3) & =12-4 x \\
x^2+4 x+3 & =12-4 x \\
x^2+8 x-9 & =0 \\
(x+9)(x-1) & =0 \\
x & =-9,1
\end{aligned}
$

Since the value of $x$ cannot be negative, $x=1$ we take this value as $x_0$
$
\begin{aligned}
p_0=\frac{8}{x_0+1}-2 & =\frac{8}{2}-2=2 \\
\mathrm{CS} & =\int_0^1 p_d d x-x_0 p_0 \\
& =\int_0^1\left(\frac{8}{x+1}-2\right) d x-(1)(2) \\
& =[8 \log (x+1)-2 x]_0^1-2 \\
& =8 \log 2-8 \log 1-2-0-2 \\
& =8 \log 2-4 \\
& =x_0 p_0-\int_0^{x_0} \mathrm{p}_{\mathrm{s}} d x \\
& =2-\int_0^1 \frac{x+3}{2} d x=2-\frac{1}{2}\left(\frac{x^2}{2}+3 x\right)_0^1 \\
=2-\frac{1}{2}\left(\frac{1}{2}+3\right) & =2-\frac{7}{2}=\frac{1}{4}
\end{aligned}
$
Hence under perfect competition,
(i) The consumer's surplus is $(8 \log 2-4)$ units
(ii) The producer's surplus is $\frac{1}{4}$ units.

Question 10.
The demand equation for a products is $\mathrm{x}=\sqrt{100-p}$ and the supply equation is $\mathrm{x}=\frac{p}{2}-10$. Determine the consumer's surplus and producer's surplus, under market equilibrium.

Solution:
Given demand equation is $\mathrm{x}=\sqrt{100-p}$ and supply equation is $\mathrm{x}=\frac{p}{2}-10$
So the demand law is $x^2=100-p$
$
\Rightarrow \mathrm{p}_{\mathrm{d}}=100-\mathrm{x}^2
$
Supply law is given by $\mathrm{x}+10=\frac{p}{2}$
$
\Rightarrow \mathrm{p}_{\mathrm{s}}=2(\mathrm{x}+10)
$
Under equilibrium $\mathrm{p}_{\mathrm{d}}=\mathrm{p}_{\mathrm{s}}$
$
\begin{aligned}
& \Rightarrow 100-x^2=2(x+10) \\
& \Rightarrow 100-x^2=2 x+20
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \mathrm{x}^2+2 \mathrm{x}-80=0 \\
& \Rightarrow(\mathrm{x}+10)(\mathrm{x}-8)=0 \\
& \Rightarrow \mathrm{x}=-10,8
\end{aligned}
$
The value of $x$ cannot be negative, So $x=8$
When $\mathrm{x}_0=8, \mathrm{p}_0=100-8^2=100-64=36$
$
\begin{aligned}
\mathrm{CS} & =\int_0^8\left(100-x^2\right) d x-(8)(36) \\
& =\left(100 x-\frac{x^3}{3}\right)_0^8-288 \\
& =800-\frac{512}{3}-288=\frac{1024}{3}
\end{aligned}
$
So consumer surplus $=\frac{1024}{3}$ units

$
\begin{aligned}
& \text { PS }=8(36)-\int_0^8 2(x+10) d x \\
&=288-2\left[\frac{x^2}{2}+10 x\right]_0^8 \\
&=288-2\left[\frac{64}{2}+80\right] \\
&=288-2(112) \\
&=64
\end{aligned}
$
So the producer's surplus is 64 units.
Question 11.
Find the consumer's surplus and producer's surplus for the demand function $p_d=25-3 x$ and supply function $p_{\mathrm{s}}=5+2 \mathrm{x}$.
Solution:
Given $p_d=25-3 x$ and $p_s=5+2 x$
At market equilibrium, $\mathrm{p}_{\mathrm{d}}=\mathrm{ps}_{\mathrm{s}}$
$
\begin{aligned}
& \Rightarrow 25-3 \mathrm{x}=5+2 \mathrm{x} \\
& \Rightarrow 5 \mathrm{x}=20 \\
& \Rightarrow \mathrm{x}=4
\end{aligned}
$
When $\mathrm{x}_0=4, \mathrm{p}_0=25-12=13$
$
\begin{aligned}
\mathrm{CS} & =\int_0^4(25-3 x) d x-13(4) \\
& =\left(25 x-\frac{3 x^2}{2}\right)_0^4-52
\end{aligned}
$

$
=100-\frac{3}{2}(16)-52=24
$
So the consumer's surplus is 24 units.
$
\begin{aligned}
\text { PS } & =13(4)-\int_0^4(2 x+5) d x \\
& =52-\left(x^2+5 x\right)_0^4=52-16-20=16
\end{aligned}
$
So the producer's surplus is 16 units.