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 = 15 + 9x - 6x² + x³
\({dC\over dx}=9-12x+3x^2\)
\({dC\over dx}=-12x+6x\)
For minimum cost \({dC\over dx}=0\)
\(\Rightarrow 3x^2-12x+9=0\)
\(\Rightarrow x^2-4x+3=0\) (Divided by 3)
\(\Rightarrow (x-1)(x-3)=0\)
\(\Rightarrow x=1,3\)
At x = 1, \({d^2C\over dx^2}=-12+6 = -6 < 0\)
At x = 3, \({d^2C\over dx^2}=-12+6 = -6 > 0\)
At x = 3 cost is minimum.

y = 4x\(\left( \frac { x+2 }{ x+1 } \right) +6\)
y = \({4x^2+8x\over x+1}+6\)
Marginal Cost = \({dy\over dx}={(x+1)(8x+8)-(4x^2+8x)(1)\over (x+1)^2}\)
\(=\frac{8 x^2+8 x+8 x+8-4 x^2-8 x}{(x+1)^2} \)
\(={4x^2+8x+8\over (x+1)^2}={4(x^2+2x+2)\over (x+1)^2}\)
\(={4(x^2+2x+1+1)\over (x+1)^2}=4({(x+1)^2\over (x+1)^2}+{1\over (x+1)^2})\)
As x increases marginal cost decreases.

q = 80 - \({ p }_{ 1 }^{ 2 }\) + 5p₂ - p₁p₂
\({\partial q\over \partial p_1}=-2p_1-p_2\)
\({\partial q\over \partial p_2}=5 - p_1\)
\(\frac{E q}{E p_1}=\frac{-p_1}{q} \frac{\partial q}{\partial p_1}=-\frac{p_1\left(-2 p_1-p_2\right)}{80-p_1^2+5 p_2-p_1 p_2}=\frac{2 p_1^2+p_1 p_2}{80-p_1^2+5 p_2-p_1 p_2}\)
\(\frac{E q}{E p_2}=\frac{-p_2}{q} \frac{\partial q}{\partial p_2}\)
\(=\frac{-p_2\left(5-p_1\right)}{80-p_1^2+5 p_2-p_1 p_2} \)
\(={-5p_2+p_1p_2\over 80-p^2_1+5p_2-p_1p_2}\)
\(\frac { { E }q }{ { E }p_{ 1 } } ={8+2\over 80-4+5-2}={10\over 79}\)
\(\frac { { E }q }{ { E }p_{ 2 } } ={-5+2\over 80-4+5-2}={-3\over 79}\)

P = 3(L)^0.4(K)^0.6
\({\partial P\over \partial L}=1.2(K)^{0.6}(L)^{-0.6}=1.2({K\over L})^{0.6}\)
Differentiating partially w.r.t. 'K' we get,
\({\partial P\over \partial K}=\) 1.8(L)^0.4(K)^-0.4 \(=1.8({L\over K})^{0.4}\)
If L = 10, K = 6
\({\partial P\over \partial L}=\) 1.2(0.6)^0.6 = 1.2(0.736) = 0.8832
\({\partial P\over \partial K} = \) 1.8(1.67)^0.4 = 1.8 x 1.2267 = 2.20806.