\(\frac { \partial p }{ \partial { P }_{ 1 } } =2-{ p }_{ 1 }{ p }_{ 2 }\)
\(\frac { \partial p }{ \partial { P }_{ 2 } } =1-{ p }_{ 1 }^{ 2 }\)
(i) \(\frac { Eq }{ { EP }_{ 1 } } =\frac { { p }_{ 1 } }{ q } \frac { \partial p }{ \partial { P }_{ 1 } } =\frac { -p }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } (-2-{ p }_{ 1 }{ p }_{ 2 })\)
= \(\frac { { 2p }_{ 1 }+2{ p }_{ 1 }^{ 2 }{ p }_{ 2 } }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
when p₁ = 3 and p₂ = 7
\(\frac { { E }q }{ { Ep_{ 1 } } } =\frac { 2(3)+2(9)(7) }{ 5-6+7-(9)(7) } =\frac { 132 }{ -57 } =\frac { -132 }{ 57 } \)
\(\frac { Eq }{ { EP }_{ 1 } } =\frac { { p }_{ 2 } }{ q } \frac { \partial p }{ \partial { P }_{ 2 } } =\frac { -p\left( 1-{ p }_{ 1 }^{ 2 } \right) }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
\(\frac { { -p }_{ 2 }+{ p }_{ 2 }{ p }_{ 1 }^{ 2 } }{ 5-2{ p }_{ 1 }+{ p }_{ 2 }-{ p }_{ 1 }^{ 2 }{ p }_{ 2 } } \)
when p₁ - 3 and p₂ = 7
\(\frac { { E }q }{ { Ep_{ 2 } } } =\frac { -7+7(9) }{ 5-6+7-(9)(7) } =\frac { 56 }{ -57 } =\frac { -56 }{ 57 } \)

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.

u = log\(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \)
e^u = \(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \) = f(x, y) ... (1)
Consider f(x, y)  = \(\frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \)
f(tx, ty) = \(\frac { { t }^{ 4 }{ x }^{ 4 }+{ t }^{ 4 }{ y }^{ 4 } }{ tx+ty } ={ t }^{ 3 }\left( \frac { { x }^{ 4 }+{ y }^{ 4 } }{ x+y } \right) ={ t }^{ 3 }f(x,y)\)
\(\therefore\)  f is a homogeneous function of degree 3.
Using Euler’s theorem we get
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3f\)
Consider f(x, y) = e^u
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3e\)^u
\(∴ { e }^{ u }x.\frac { \partial u }{ \partial u } +{ e }^{ u }y.\frac { \partial u }{ \partial y } =3{ e }^{ u }\)
\(x.\frac { \partial u }{ \partial u } +y.\frac { \partial u }{ \partial y } =3\)

\(R =400 \text { units } / \text { week }, C_3=Rs. 75 \)
\(C_1 =7.5 \% \text { of } 50 \text { per year. } \)
\(=\frac{7.5}{100 \times 52} \times 50 \text { per week }=0.072 . \)
\(E O Q =\sqrt{\frac{2 R C_3}{C_1}}=\sqrt{\frac{2 \times 400 \times 75}{0.072}} \)
\(=912.87 \approx 913 \text { units per order. } \)
\(\text {Total } \text { Optimum Cost }=\text { Purchasing Cost + Minimum annual Cost. }\)
\(=400 \times 50+\sqrt{2 R C_1 C_3} \)
\(=20,000+\sqrt{2 \times 400 \times 75 \times 0.07211} \)
\(=20,000+65.75 \)
\(=Rs. 20065.75 \approx Rs. 20,066 \text { per week. } \)

Given
Ordering cost per order : C₃ = Rs. 150 per order.
Number of units per order: q = 500 units
Annual demand = 500 × 4 = 2000 units
∴ Demand rate : R = 2000 per year
Carrying cost : C₁ = 20% of unit value
C₁ = \(\frac { 20 }{ 100 } \times 125=Rs25\)
(i) Total annual cost of due existing inventory policy
= \(\frac { R }{ q } \times { C }_{ 3 }+\frac { q }{ 2 } { C }_{ 1 }\) = \(\frac { 2000 }{ 500 } \times 150+\frac { 500 }{ 2 } \times 25\)
= Rs. 6850
(ii) EOQ =  \(\sqrt { \frac { 2Rc_{ 3 } }{ { c }_{ 1 } } } \)
= \(\sqrt { \frac { 2\times 2000\times 150 }{ 25 } } \)
= \(\sqrt { 12\times 2000 } \)
= 155 units (app.)
(iii) Minimum annual cost \(= \sqrt { { 2Rc_{ 3 } }{ { c }_{ 1 } } } \)
= \(\sqrt { 2\times 2000\times 150\times 25 } \)
= Rs. 3873.
By applying the economic order quantity, money saved by a company = 6850 – 3873  = Rs. 2977.