Given P= 5(L)^0.7(K)^0.3
Marginal Productivity of Labour (L) is
\({∂P\over ∂L}=5(0.7)(L)^{0.7-1} (K)^{0.3 }= 3.5(L)^{-o·3} (K)^{0.3} = 3.5\left(K\over L\right)^{0.3}\)
When L = 10 and K = 3,
\(\frac { \partial P }{ \partial L } =3.5\left( \frac { 3 }{ 10 } \right) ^{ 0.3 }=3.5(0.3)^{ 0.3 }=3.5\times 0.6968\)
= 2.438 = 2.44
\({∂P\over ∂L}=5(L)^{0.7} (0.3)(K)^{0.3-1 }= 1.5(L)^{0.7}(K)^-{0.7} = 1.5\left(L\over K\right)^{0.7}\)
When L= 10 and K=3,
\({∂P\over ∂L}=1.5\left(10\over3\right)^{0.7}=1.5(3.33)^{0.7} = 1.5(2.2322) = 3.481 = 3.48\)

Let y = x² + 16/x
Differentiating w.r.t. 'x'we get,
\({dy\over dx}=2x-{16\over x^2}\)
Con diittiion for maxi.ma and rmminmimmaa is \({dy\over dx}=0\)
\(2x-{16\over x^2}=0\)
\(⇒2x={16\over x^2}⇒x^2={16\over 2}\)
⇒ x³ = 8 = 2³
⇒ x=2
Also, \({d^2y\over dx^2}=2+{32\over x^3}\)
when x = 2,  \({d^2y\over dx^2}=2+{32\over 8}=2+4=6>0\)
∴ y is minimum when x = 2
∴ Minimum value                 
= 2² + \(16\over 2\)=4+8=12
and there is no maximum.

Let y = x³+8x²+5x-2
Differentiating w.r.t. 'x' we get
\({dy\over dx}=3x^2+16x+5\)
\({dy\over dx}=0⇒3x^2 + 16x + 5=0\)
⇒ (x+5)(3x+1)=0
⇒ x = - 5, -1/3

The possible intervals are (-∞, - 5), (-5, -1/3) and (-1/3, ∞)

Hence the given function is increasing in the intervals (-∞, - 5), (-1/3, ∞) and decreasing in (-5, -1/3).

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.

Given f(x) = x³-3x²-9x+5
Differentiating w.r.t. 'x' we get,
f'(x) = 3x²-6x-9
At stationary points,f'(x) = 0
∴ 3x² - 6x - 9 = 0
⇒ x²-2x-3 = 0 (Divided by 3)
⇒ (x + 1)(x - 3) = 0
The stationary points are obtained when x = -1, x = 3
when x = -1, f(-1) = (-1)³ - 3(-1)² - 9(-1) + 5 = 10
when x = 3, f(3) = (3)³ - 3(3)² - 9(3) + 5 = - 22
∴ The stationary values are 10 and - 22 and the stationary points are (-1, 10) and (3, - 22).

Given C = x (x²-2) = x³-2x
(i) Average Cost  = \(\frac { C }{ x } =\frac { { x }^{ 3 }-2x }{ x } ={ x }^{ 2 }-2\)
(ii) Marginal Cost (MC) = \(\frac { dc }{ dx } =\frac { d }{ dx } \left( { x }^{ 2 }-2x \right) ={ 3x }^{ 2 }-2\)
(iii) R = 14x - x²
Average Revenue (AR) = \(\frac { R }{ x } =14-x\)
(iv) Marginal Revenue (MR) \(=\frac { d R}{ dx } =14-2x\)

\(x={P\over p+5}\)
\({dx\over dp}={(p+5)-p\over (p+5)^2}\)
\(={5\over (p+5)^2}\)
Elasticity of supply: \(η_S={p\over x}.{dx\over dp}\)
\(=(p+5){5\over (p+5)^2} = {5\over (p+5)}\)
When p = 20, \(η_S={5\over 20+5}\) = 0.2
Interpretation:
i) If the price increases by 1% from p = Rs. 20, then the quantity of supply increases by 0.2% approximately.
ii) If the price decreases by 1% from p = Rs. 20, then the quantity of supply decreases by 0.2% approximately.

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

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)

\(p = 4L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }\) is a homogeneous function of degree 1.
Marginal productivity of labour is
\(\frac { \partial P }{ \partial L } \) = \(4 \times { \frac { 3 }{ 4 } }L^{ \frac { -1 }{ 4 } } K^{ \frac { 1 }{ 4 } }\) = 3 \(\left( \frac { K }{ L } \right) ^{ \frac { 1 }{ 4 } }\)
Marginal productivity of capital is
\(\frac { \partial P }{ \partial L } \) = \(4L^{ \frac { 3 }{ 4 } }\times \frac { 1 }{ 4 } K^{ \frac { -3 }{ 4 } }\) = \(\left( \frac { L }{ K } \right) ^{ \frac { 3 }{ 4 } }\)
\(L\frac { \partial P }{ \partial L } +K\frac { \partial P }{ \partial K } =3L\left( \frac { K }{ L } \right) ^{ \frac { 1 }{ 4 } }+k\left( \frac { L }{ K } \right) ^{ \frac { 3 }{ 4 } }\)
\(= 3L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }+L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }\)
\(= 4L^{ \frac { 3 }{ 4 } }K^{ \frac { 1 }{ 4 } }\) = P
Hence Euler’s theorem is verified.