(i) p = (a - bx)²
\({dp\over dx}=2(a-bx)(-b)=-2b(a-bx)\)
Elasticity of demand (\(\eta _d\)) \(={-p\over x}.{dx\over dp}\)
\(\Rightarrow -{(a-bx)^2\over x}({-1\over 2b(a-bx)})={a-bx\over x(2b)}\)
When \(\eta_d=1\)
\(\frac{a-b x}{2 b x} =1 \)
\(a-b x =2 b x \)
\(3 b x =a \)
\(x =\frac{a}{3 b} \text { units }\)
(ii) \(p=a-b x^2\)
\(\frac{d p}{d x}=-2 b x\)
\(\eta_d\) Elasticity of demand \(=-\frac{5}{x} \frac{d x}{d p}\)
\(=-\frac{a-b x^2}{x} \cdot \frac{1}{-2 b x}\)
\(=\frac{a-b x^2}{2 b x^2}\)
When \(\eta_d=1\)
\(\frac{a-b x^2}{2 b x^2}=1\) 
\(a-b x^2=2 b x^2\)
\(3 b x^2=a \Rightarrow x^2=\frac{a}{33}\)
\(N=\sqrt{\frac{a}{3 b}}\)units

\(p={{4\over x}}\)
\(⇒\ x={4\over p}\)
∴ \({dx\over dp}=-{4\over p^2}\)
At p = 4, \({dx\over dp}=-{1\over 4}=-0.25\)
∴ Rate of change of demand with respect to the price at p = Rs.4 is -0.25
Interpretation:
When the price increases by 1% from the level of p = Rs. 4, the demand decreases (falls) by 0.25%.

y = x³ + 19
\({dy\over dx}=3x^2 \ \ \ ...(1)\)
\({dy\over dx}=27\ \ \ ...(2)\) [Given]
From (1) and (2), we get
3x² = 27 ⇒ x = ±3

xpⁿ = k ⇒ x = kp^-n
\({dx\over dp}=-nkp^{-n-1}\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)
\(=-{p\over kp^{-n}}(-nkp^{-n-1})\) = n, which is a constant.

x = 2p²-5p+1
\({dx\over dp}=4p-5\)
Elasticity of supply: \(η_s={P\over x}.{dx\over dp}\)
\(={p\over 2p^2-5p+1}.(4p-5)\) \(={4p^2-5p\over 2p^2-5p+1}\)

\(C(x)={1\over3}x^3+4x^2-25x+7\)
(i) Average cost
AC = \({c\over x}\)\(={1\over 3}x^2+4x-25+{7\over x}\)
(ii) Average variable cost
AVC\(={f(x)\over x}\)\(={1\over 3}x^2+4x-25\)
(iii) Average fixed cost
AFC = \({k\over x}\) = \({7\over x}\)
(iv) Marginal cost
MC  \(={dC\over dx}(or){d\over dx}(c(x))\)
\(={d\over dx}\left[{1\over 3}x^2+4x^2-25x+7\right]\)
= x² + 8x-25
(v) Marginal Average cost
MAC \(={d\over dx}[AC]\)\(={d\over dx}\left[{1\over 3}x^2+4x-25+{7\over x}\right]\)
\(={2\over 3}x+4-{7\over x^2}\)

\(p={(a-bx)^{1\over 2}}\)
Differentiating with respect to the price ‘p’ ,
we get \(1={1\over2}(a-bx)^{1\over2}(-b).{dx\over dp}\)
\(∴\ {dx\over dp}={2(a-bx)^{1\over2}\over-b}\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)
\(=-{(a-bx)^{1\over2}\over x}.{2(a-bx)^{1\over2}\over -b}\)\(={2(a-bx)\over bx}\)
When \(η_d=1,\ {2(a-bx)\over bx}=1\)
2(a - bx) = bx ⇒ output \( x={2a\over 3b}\)units

p = 40–x ⇒ x = 40–p
\({dx\over dp}=-1\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)\(={40-x\over x}\)
Given that η_d = 1
\(∴\ {40-x\over x}=1\) ⇒ x = 20
∴ output x = 20 units.

\(y={2x+1\over 3x+2}\)
\({dy\over dx}={(3x+2)(2)-(2x+1)(2)\over (3x+2)^2}\)\(={1\over (3x+2)^2}\)
Elasticity: \(η={x\over y}.{dx\over dx}\)\(={x\over \left(2x+1\over 3x+2\right)}.{1\over (3x+2)}\)
\(={x\over (2x+1)(3x+2)}\)
When x = 1, \(η={1\over15}\)

At equilibrium, demand = supply
\(⇒\ {1\over2}(5-p)=2p-3\)
5 – p = 4p - 6
\(⇒ p={11\over 5}\)
∴ Equilibrium price: \(P_E=Rs : { 11\over 5}\)
Now, put \(p={11\over5}\) in x=2p-3
We get \(x=2\left(11\over5\right)-3={7\over 5}\)
∴ Equilibrium quantity: \(x_E={7\over5}\) units.

C(x) = 50 + 4x + 3√x
Marginal cost (MC) =\({Dc\over dx}-{d\over dx}[C(x)]\)
\(={d\over dx}\left[50+4x+3\sqrt3\right]=4{3\over 2\sqrt3}\)
When x = 9, \({dC\over dx}=4{3\over 2\sqrt9}\)\(=4{1\over 2}\)(or) Rs.4.50
∴ MC is Rs.4.50 , when the level of output is 9 units.

p = 400 - 2x - 3x²
\( R=400x-2x^2-3x^3\)
LHS = Marginal revenue = MR
\(=\frac{d R}{d x}=400-4 x-9 x^2 \) ...(1)
\(\frac{d p}{d x} =-2-6 x \)
\(\eta_d =\text {Elasticity of demand }=-\frac{p}{x} \frac{d x}{d p} \)
\(=-\frac{\left(400-2 x-3 x^2\right)}{x} \cdot \frac{1}{-(2+6 x)} \)
\(=\frac{400-2 x-3 x^2}{2 x+6 x^2} \)
\(\text {RHS } =p\left[1-\frac{1}{\eta_d}\right] \)
\(=\left(400-2 x-3 x^2\right)\left[1-\frac{1}{\left.\frac{400-2 x-3 x^2}{2 x+6 x^2}\right]}\right] \)
\(=\left(400-2 x-3 x^2\right)\left[1-\frac{2 x+6 x^2}{400-2 x-3 x^2}\right] \)
\(=\left(400-2 x-3 x^2\right)\left[\frac{400-2 x-3 x^2-2 x-6 x^2}{400-2 x-3 x^2}\right] \)
\(=400-4 x-9 x^2 \)...( 2)
\(\text {From (1) and (2) } \)
\(\text {M R}=p\left[1-\frac{1}{\eta_d}\right] \)

We know that profit is maximum when marginal revenue [MR] = marginal cost [MC].
Revenue: R = px
= (497–0.2x)x = 497x–0.2x²
\(MR={dR\over dx}\)\(=497-0.4x\)
Cost: C = 25x + 10000
∴ MC = 25
MR = MC ⇒ 497 - 0.4x = 25
⇒ 472–0.4x = 0
⇒ x = 1180 units.
Now, p = 497–0.2x
at x = 1180, p = 497–0.2(1180) = Rs.261.