Let y = x³ + 3x² + 3x + 7
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=3x^2 + 6x + 3\)
= 3(x² + 2x + 1)
= 3 (x + 1)²
\(\frac { dy }{ dx } >0\) for all real values of x.
\(\therefore\) y is an increasing function for all real values of x.

u = x³ + 3x² y² + y³
\(\frac{\partial u}{\partial x} =3 x^2 + 6 x y^2 \)
\(\frac{\partial u}{\partial y} =6 x^2 y + 3 y^2 \)
\(L H S =\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial}{\partial x}\left(6 y^2 y+3 y^2\right) =12 x \)
\(R H S =\frac{\partial^2 u}{\partial y \partial x}=\frac{\partial}{\partial y}\left(\frac{\partial x}{\partial x}\right) \)
\(=\frac{\partial}{\partial y}\left(3 x^2 + 6 x y^2\right)\)
From (1) and (2),
\(\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}\)

Given that f(x) = 2x³ + 9x² + 12x + 1.
f'(x) = 6x² + 18x + 12
= 6(x² + 3x + 2)
= 6(x + 2)(x + 1)
f'(x) = 0 \(\Rightarrow\) 6 (x + 2)(x + 1) = 0
\(\Rightarrow\) x + 2 = 0 (or) x + 1 = 0.
x = –2 (or) x = –1
f(x) has stationary points at x = – 2 and x = – 1
Stationary values are obtained by putting x = – 2 and x = – 1
When x = – 2, f(–2) = 2(–8)  +9(4) + 12(–2) + 1 = –3
When x = – 1, f(–1) = 2(–1) + 9(1) + 12(–1) + 1 = –4
The stationary points are (–2, –3) and (–1,–4).

x = 10 - p
\(\frac { dx }{ dp } \) = -1
Elasticity of demand n_d = \(\frac { -p }{ x } .\frac { dx }{ dp } \)
\(\Rightarrow\)  n_d = \(\frac { -p }{ 10-p } (-1)=\frac { p }{ 10-p } \)
At p = 6,  n_d  \(=\frac { 6 }{ 4 } =\frac { 3 }{ 2 } > 1\) = |n_d| = 1 = elastic
Hence the demand is elastic.

Given \(\frac{d y}{d x}=2 x\)
\(y=x^3+10 x^2-48 x+8\)
\(\frac{d y}{d x}=3 x^2+20 x-48\)
\(2 x = 3x^2+20 x-48\)
\(3 x^2+18 x-48=0 \)
Dividing by 3 we get,
 x² + 6x - 16 = 0
\(\Rightarrow\)(x + 8)(x - 2) = 0
\(\Rightarrow\) x = -8,2

\(p={(a-bx)^{1\over 2}}\)
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

\(x={20\over p+1}\)
\({dx\over dp}={-20\over (p+1)^2}\)
Elasticity of demand: \(η_d=-{P\over x}.{dx\over dP}\)
\(=-{P\over \left(20\over (p+1)\right)}.{-230\over (p+1)}\)\(={p\over p+1}\)
When p = 3, \(η_d={3\over 4}\) (or) 0.75
Here |η_d |<1
∴ demand is inelastic.

Given that f(x) = x² + 2x – 5 … (1)
f '(x) = 2x + 2
At stationary points, f'(x) = 0
\(\Rightarrow\)  2x + 2 = 0 \(\Rightarrow\)  x = –1
f(x) has stationary value at x = –1
When x = –1, from (1)
f(–1) = (–1)² + 2(–1) – 5 = – 6
Stationary value of f (x) is – 6
Hence stationary point is (–1,–6)

Given p = 10e^{\(\frac { -x }{ 2 } \)}
\({dp\over dx}=10e^{-x/2}(-1/2)=-5e^{-x\over 2}\)
Elasticity of demand \((\eta_d)\)=\({-p\over x}.{dx\over dp}\) 
\(\eta_d=\frac{-10 e^{-\frac{x}{2}}}{x} \cdot \frac{1}{-5 e^{-\frac{x}{2}}}=\frac{2}{x}\)

\(p={{4\over x}}\)
\(⇒\ x={4\over p}\)
∴ \({dx\over dp}=-{4\over p^2}\)
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%.

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

u(x, y) = (x²+y²)^-1/2
u(tx, ty) = (t²x²+t²y²)^-1/2 = t^-1(x²+y²)^-1/2
∴ u is a homogeneous function of degree –1
By Euler’s theorem \(x.\frac { \partial u }{ \partial x } +y.\frac { \partial u }{ \partial y } =\left( -1 \right) u=-u\)
Verification:
\(u =\left(x^2+y^2\right)^{-\frac{1}{2}} \)
\(\frac{\partial u}{\partial x} =-\frac{1}{2}\left(x^2+y^2\right)^{-\frac{3}{2}} \cdot 2 x=\frac{-x}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(x \cdot \frac{\partial u}{\partial x} =\frac{-x^2}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(\frac{\partial u}{\partial y} =-\frac{1}{2}\left(x^2+y^2\right)^{-\frac{3}{2}} \cdot 2 y=\frac{-y}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(y \cdot \frac{\partial u}{\partial y} =\frac{-y^2}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(\therefore x \cdot \frac{\partial u}{\partial x}+y \cdot \frac{\partial u}{\partial y}=\frac{-\left(x^2+y^2\right)}{\left(x^2+y^2\right)^{-\frac{3}{2}}} \)
\(=(-1) \frac{1}{\sqrt{x^2+y^2}}=(-1) u=-u \)
Hence Euler’s theorem verified

C = 2000 + 1800x -75 x² + x³
\({dC\over dx}=1800-150x+3x^2\)
\({dC\over dx}=0\)
\(\Rightarrow 3(x^2-50x+600)=0\)
\(\Rightarrow x^2-50x+600=0\) (Divided by 3)
\(\Rightarrow x=30,20\)
The intervals are (0, 20) (20, 30) and (30, \(\infty\))

u = x²y³ cos \(\left( \frac { x }{ y } \right) \)
u (x,y) = x²y³ cos \(\left( \frac { x }{ y } \right) \)
u(tx, ty) = (tx)²(tx)³ cos \(\left( \frac { tx }{t y } \right) \)
= t⁵x²y³ cos\(\left( \frac { x }{ y } \right) \)t⁵.u(x, y)
It is a homogenous function od degree 5.
By Euler's theorem,
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =nu\)
\(\Rightarrow x.\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =5u\)
Hence proved.

Here demand rate R = 48000
Inventory cost C₁ = 10.8% of 2.5 = \(\frac { 10.8 }{ 100 } \times 2.5=0.27\)
Ordering cost C₃ = 45
Economic order quantity q₀ = \(\sqrt { \frac { { 2C }_{ 3 }R }{ { C }_{ 1 } } } \)
\(\sqrt { \frac { 2\times 45\times 48000 }{ 0.27 } } \) = 4000 units
Number of orders per year = \(\frac { R }{ { q }_{ 0 } } \)  = \(\frac { 48000 }{ 4000 } =12\)
Time between orders t₀ = \(\frac { { q }_{ 0 } }{ R } \) = \(\frac { 1 }{ 12 } \) = 0.083 year
At EOQ, carrying cost :
=  \(\frac { { q }_{ 0 } }{ 2 } \) x C₁ \(=\frac { 4000 }{ 2 } \times 0.27\) = Rs. 540
Ordering cost = \(\frac { R }{ q_{ 0 } } \times { C }_{ 3 }\) = \(\frac { 48000 }{ 4800 } \times 45\) = Rs. 540
So at EOQ carrying cost is equal to ordering cost.

 As per given information for producing x units of certain item C(x) = labour cost + material cost + overhead cost
= \(\frac { { x }^{ 2 } }{ 100 } \)+30x+1600
AC = \(\frac { c(x) }{ x } \)\(=\frac { \frac { { x }^{ 2 } }{ 100 } +30x+1600 }{ x } \)
\(=\frac { x }{ 100 } +30+\frac { 1600 }{ x } \)
\(\frac { d(AC) }{ dx } =\frac { 1 }{ 100 } -\frac { 1600 }{ { x }^{ 2 } } \) & \(\frac { d^{ 2 }\left( AC \right) }{ { dx }^{ 2 } } =\frac { 3200 }{ { x }^{ 2 } } \)
\(\frac { d\left( AC \right) }{ dx } =0\Rightarrow \frac { 1600 }{ { x }^{ 2 } } +\frac { 1 }{ 100 } =0\)
\(\Rightarrow \frac { 1 }{ 100 } =\frac { 1600 }{ { x }^{ 2 } } \) \(\Rightarrow\) x² = 160000
∴ x = 400 (–400 is not acceptable)
When x = 400, \(\frac { d^{ 2 }\left( AC \right) }{ { dx }^{ 2 } } =\frac { 3200 }{ { 400 }^{ 3 } } >0\)
AC is minimum at x = 400
Hence 400 items should be produced for minimum average cost.

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)
\(\text {From (1) and (2) } \)
\(\text {M R}=p\left[1-\frac{1}{\eta_d}\right] \)

p = 50 – 3x
\({dp\over dx}=-3\) ⇒ \({dx\over dp}=-{1\over 3}\)
Elasticity of demand: \(η_d=-{p\over x}.{dx\over dp}\)
\(=-{50-3x\over x}\left(-{1\over3}\right)\)\(={50-3x\over 3x}\ \ ...(1)\)
Now, Revenue: R = px
= (50 - 3x)x = 50x - 3x²
Average revenue: AR = p = 50 - 3x
Marginal revenue: \(MR={dR\over dx} = 50 - 6x\)
\({AR\over AR-MR}={50-3x\over (50-3x)-(50-6x)}\)
\(={50-3x\over 3x}\ \ ...(2)\)
From (1) and (2), we get
\(η_d={AR\over AR-MR}\), Hence verified.

\(E O Q=\sqrt{\frac{2 C_3 R}{C_1}}=\sqrt{\frac{2 \times 20 \times 5000}{20}}=1000\)

\(\frac{\partial u}{\partial x} =3 x^2+3 y^2 \)
\(\frac{\partial^2 u}{\partial y \partial x} =6 y \)

\(\frac{d C}{d x}=M C=\frac{5 e^{5 x}}{25}=\frac{1}{5} e^{5 x}\)