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

Given x = 106 - 2p; 2p = 106 - x
P = \(\frac { 106-x }{ 2 } \)
(R) = px = \(\left( \frac { 106-x }{ 2 } \right) = 56x -\frac { { x }{ 2 } }{ 2 } \)
\(A C=\frac{C}{x} =5+\frac{x}{50} \)
\(C =5 x+\frac{x^2}{50} \)
\(p=R-C =53 x-\frac{x^2}{2}-5 x-\frac{x^2}{50} \)
\(=48 x-\frac{26 x^2}{50} \)
\(\frac{d p}{d x} =48-\frac{52 x}{50} \)
\(\text {For maximum profit } \frac{d P}{d x} =0 \)
\(48-\frac{52 x}{50} =0 \)
\(x =\frac{48 \times 50}{52} \approx 46 \)
\(\frac{d^2 p}{d x^2} =-\frac{52}{50}<0\)
\(\therefore\)  At x  = 46, Profit is maximum
Maximum profit  = 48(46) - \(\frac { 26\left( 46 \right) ^{ 2 } }{ 50 } \) [From (1)]
= 2208 - 1100.32
= Rs.1107.68
P is maximum when x ≈ 46, maximum  profit =  Rs.1107.68

We know that profit is maximum when Marginal Revenue
(MR) = Marginal Cost (MC)
x = 6000 - 30p
30p = 6000 -x 
P = \(\frac { 6000-x }{ 30 } \)
\(\therefore\) R = px \(=\frac { 6000x-{ x }^{ 2 } }{ 30 } \)
\(\mathrm{MR} =\frac{d R}{d x}=\frac{6000-2 x}{30} \)
\(C =72000+60 x \)
\(\mathrm{MC} =\frac{d C}{d x}=60 \)
\(\mathrm{MR} =\mathrm{MC} \)
\(\frac{6000-2 x}{30} =60 \)
\(6000-2 x =1800 \)
\(2 x =4200 \)
\(x =2100 \text { units (output) } \)
\(p =\frac{6000-2100}{30} \)
\(= 130 \text { (price) }\)

We know that average cost [AC] is minimum when average cost [AC] = marginal cost [MC].
Cost: \(C={1\over3}x^3-3x^2+9x\)
AC = \({1\over3}-3x^2+9\) and MC = x² - 6x + 9
Now, AC = MC ⇒ \({1\over 3}x^2-3x+9-x^2-6x+9\)
⇒ 2x² - 9x = 0 ⇒ \(x={9\over 2}\) unit (∵ x > 0)

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.

u = log(x²+y²)
\(\frac { \partial u }{ \partial x } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } (2x)=\frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) . 2-2x.2x }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } =\frac { 2\left( { y }^{ 2 }-{ x }^{ 2 } \right) }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } \)
\(\frac { \partial u }{ \partial y } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } (2y)=\frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }u }{ \partial y^{ 2 } } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) . 2-2y.2y }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } =\frac { 2\left( { y }^{ 2 }-{ x }^{ 2 } \right) }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } \)
\(\therefore \frac { { \partial }^{ 2 }u }{ { 2x }^{ 2 } } +\frac { { \partial }^{ 2 }u }{ \partial { y }^{ 2 } } =0\)

Given that f(x) = x² – 4x + 6
Differentiate with respect to x,
f' (x) = 2x = 4
when  f'(x) = 0 \(\Rightarrow\) 2x - 4 = 0 \(\Rightarrow\) x = 2
Then the real line is divided into two intervals namely (-\(\infty \),2) and (2,\(\infty \))

[to choose the sign of f' (x) choose any values for x from the intervals ans substitute in f' (x) and geth the sign]

\(x =\frac{25}{p^4} \)
\(x =25 p^{-4}\)
\(\frac{d x}{d p} =25(-4) p^{-5}=-\frac{100}{p^5} \)
\(\text {Elasticity of demand } =\eta_d=-\frac{p}{x} \frac{d x}{d p} \)
\(=-\frac{p}{\frac{25}{p^4}}\left(-\frac{100}{p^5}\right) \)
\(\eta_d =\frac{100}{25}=4 \)

\(\eta_d=\frac{A R}{A R-M R} \Rightarrow 2 =\frac{50}{50-M R} \)
\(50-M R =25 \)
\(M R =25\)

dy/dx = 4x + 5
At x = 2, dy/dx = 13

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

R = px = 20x- 3x²
M.R = dR/dx = 20 - 6x