Given R = 8x, \(C = 150000 + 60\left(x^2\over 900\right)\)
Profit = Revenue - Cost
\(P=8x- 150000- 60\left(x^2\over900\right)\)
Differentiating w.r.t. 'x' we get,
\({dP\over dx}=8-{120x\over 900}=8-{2x\over 15}\)
Condition for maximum is \(dp\over dx\)and \({d^2P\over dx^2}<0\)
\({dP\over dx}=0⇒8={2x\over 15}⇒{8\times15\over2}=x\)
⇒x=60
Also, \({d^2P\over dx^2}={-2\over 15}<0\)
∴ Profit is maximum when x = 60.
So, to get the maximum profit, we should sell 60 units.

Requirement (R) = 24000 units / year
Ordering Cost (C₃) = Rs.22.50
Holding Co t (C₁) = 5.4% of the value of each unit.
\(={5.4\over100}\times\)1.25= Rs.0.0675 unit / year
(i) \(EOQ=\sqrt{2C_3R\over C_1}=\sqrt{2\times2400\times22.5\over0.0675}=4000units\)
(ii) Time between each order \(t_0={{ q }_{ 0 }\over R}={4000\over 24000}={1\over 6}year\)
(iii) Number of orders per year \({R\over q_0}={24000\over4000}=6\)
At EOQ, carrying cost \(={q_0\over 2}\times C_1\)
\(={4000\over2}\times0.0675=Rs.135\)
Ordering Cost \(={R\over q_0}\times C_3\)
\(={24000\over 4000}\times22.50=Rs.135\)
Hence, at EOQ, Carrying Cost = Ordering Cost

Given \(P={4000x\over 500+x}-x\)
Differentiating w.r.t. 'x' we get
\(\frac { dp }{ dx } =\frac { { (500+x)(4000)-(4000x)(1) } }{ (50+x)^{ 2 } } -1\)
\(={4000(500+x-x)\over (500+x)^2}-1={2000000\over(500+x)^2}-1\)
Profit is maximum when \({dP\over dx}=0\) and \({d^2P\over dx}<0\)
\({dP\over dx}=0⇒{2000000\over (500+x)^2}-1=0\)
\(⇒{2000000\over (500+x)^2}=1⇒(500+x)^2=2000000\)
\(⇒500+x=1000\times\sqrt2=1000\times1.414\)
⇒ x = 1414 - 500 = 914
Also \({d^2P\over dx^2} = -{4000000\over (500+x)^3}\)
When x = 914 \({d^2P\over dx^2}<0\)
∴ Profit is maximum, when x = 914

Given q = 16 -3P₁ - 2P₂²
Differentiating partially w.r.t, 'P₁' and 'P₂' we get,
\({\partial q\over \partial p_1}=-3\ and\ {\partial q\over \partial p_2}=-4p_2\)
Partial elasticities
\({Eq\over Ep_1}={-p_1\over q}{\partial q\over \partial p_2}\)
\(={-p_1\over 16-3p_1-2p_2^2}(-3)\)
\(={3p_1\over 16-3p_1-2p_2^2}\)
and \({Eq\over Ep_2}=-{p_2\over q}{\partial q\over \partial p_2}={-p_2\over 16-3p_1-3p_2^2}(-4p_2)\)
\(={4p_2^2\over 16-{ 3p }_{ 1 }-2p_2^2}\)

Given P = 10K - K² + KL
Differentiating partially w.r.t. 'K' we get,
\({\partial P\over \partial K}=10(1)-2K+L(1)\)
=10-2K+L
When K = 2 and L = 6.
Marginal Productivity of Capital
\({\partial P\over \partial K}=10 - 2(2) + 6 = 10 - 4 + 6 = 12\)
Differentiating 'P' partially w.r.t. 'L' we get,
\({\partial P\over \partial L}=0-0+K=K\)
When K = 2 and L = 6,
Marginal Productivity of Labour
\({\partial P\over \partial L}=2\)