Given f(x, y) =3x²+4y³+6xy-x²y³+6
Differentiating 'f' partially w.r.t. 'y' we get
f_y(x,y) = 0+ 12y²+6x(1)-x²(3y²)+0
=12y²+ 6x - 3x²y²
Differentiating again partially w.r.t. 'y' we get,
fy(x, y) =24y + 0 - 3x²(2y) = 24y - 6x²y
\(\therefore\)f_yy(1, 1)= 24(1) - 6(1²)(1) = 24 - 6 = 18

Given f(x, y) =3x²+4y³+6xy-x²y³+6
Differentiating partially w.r.t. 'x' we get,
fx(x, y)=6x + 0 + 6y(1) -y³(2x) + 0
=6x + 6y- 2xy³
\(\therefore f_x(1,-1)=\) 6(1) +6(-1)-2(1)(-1)³
= 6-6+2
= 2

Given y=x-1/x
Differentiating w.r.t.'x' we get,
\({dy\over dx}=1+{1\over x^2}>0\) for all real values of x, except x=0
\(\therefore \) y is a strictly increasing function for all real values of x(x\(\neq\)0)

\(p={{4\over x}}\)
\(⇒\ x={4\over p}\)
∴ \({dx\over dp}=-{4\over p^2}\)
At p = 4, \({dx\over dp}=-{1\over 4}=-0.25\)
∴ Rate of change of demand with respect to the price at p = Rs.4 is -0.25
Interpretation:
When the price increases by 1% from the level of p = Rs. 4, the demand decreases (falls) by 0.25%.

Let y=x³-6x²+7
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=3x^2-12x\)
\({dy\over dx}=0\)
\(\Rightarrow3x^2-12x=0\)
\(\Rightarrow3x(x-4)=0\)
\(\Rightarrow x=0 \ or \ x=4\)
\({d^2y\over dx^2}=6x-12\)
when x=0 \({d^2y\over dx^2}=-12<0\)
\(\therefore \) y is maximum at x=0
\(\therefore \) maximum value=0³-6(0)²+7=7
when x=4, \({d^2y\over dx^2}=-6(4)-12=12>0\)
\(\therefore \) y is minimum at x=4
\(\therefore \) Minimum value =4⁴-6(4)²+7=64-96+7=-25
Hence maximum value is 7 and minimum value is -25.

Let y= 75-12x+6x²-x³
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=0-12+12x-3x^2=-3(x^2-4x+4)=-3(x-2)^2\)
\(\Rightarrow{dy\over dx}\le 0\) for all \(x \in (-\infty , \infty)\)
\(\therefore \) y decreases as x increases.

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)

f(x, y) = 3x² + 4y³ + 6xy - x³y³ + 7
\(\frac{\partial f}{\partial y}=f_y=12 y^2+6 x-3 x^2 y^2 \)
\(\frac{\partial^2 f}{\partial y^2}=f_{y y}=24 y-6 x y^2 \)
\(\therefore f_{yy}(1,1)=24-6=18\)
Hence proved.

At equilibrium, demand = Supply
\(\Rightarrow\)  100 - 2p = 3p - 50
\(\Rightarrow\) 150 = 5p
\(\Rightarrow\) p = 30
Equilibrium price is P_E = 30
x = 100 - 2p
= 100 - 60 = 40
x = 40
\(\therefore\) Equilibrium quantity is x_E = 40 units.

x = 10 - p
\(\frac { dx }{ dp } \) = -1
Elasticity of demand n_d = \(\frac { -p }{ x } .\frac { dx }{ dp } \)
\(\Rightarrow\)  n_d = \(\frac { -p }{ 10-p } (-1)=\frac { p }{ 10-p } \)
At p = 6,  n_d  \(=\frac { 6 }{ 4 } =\frac { 3 }{ 2 } > 1\) = |n_d| = 1 = elastic
Hence the demand is elastic.