For market equilibrium, P_d = P_s
25 - Q² = 2Q + 1
0 = 25 + Q² + 2Q + 1
0 = -24 + Q² + 2Q
Q² + 2Q - 24 = 0
Q² + 6Q - 4Q - 24 = 0
Q(Q + 6) - 4(Q + 6) = 0
(Q + 6)(Q - 4) = 0
So, Q = 4 or Q =-6. Since Q cannot be equal to -6,
Q = 4
When Q = 4, P_d = 25 - 4² = 9;
P₂ = 2(4)+ 1 = 9
Consumer's surplus =\(\int _{ 0 }^{ 4 }{ (25-Q^{ 2 })dQ)dQ-(9\times 4) } \)
=\({ \left[ 25Q-\frac { Q^{ 3 } }{ 3 } \right] }_{ 0 }^{ 4 }\)-36
=\({ \left[ (25)(4)-\frac { 1 }{ 3 } (4)^{ 3 } \right] }\)-(0)-36
=\(\left[ 100-\frac { 64 }{ 3 } \right] \)-(0)- 36 = 42.67
Producer's surplus Ps
(P_s) = (9 x 4) - \(\int _{ 0 }^{ 4 }{ (2Q+1)dQ } \)
= 36 -\(\left[ (Q^{ 2 }+Q \right] _{ 0 }^{ 4 }\)
= 36 - (16 + 4) = 16

Given that TC(Q) = 128 + 60Q + 8Q²
We know TC = Fixed cost + variable cost
MC (Q) =\(\frac { d(TC) }{ dQ } \)
= 0 + 60(1)Q^1-1 + 8(2)Q^2-1
= 0 + 60Q⁰ + 16Q¹(Since, Q⁰= 1)
MC = 60 + 16Q
Average Cost =\(\frac { TC }{ dQ } \)
=\(\frac { 128+60Q+8Q^{ 2 } }{ Q } \)
AC = \(\frac { 128 }{ Q } \) + 60 + 8Q
Constant value is known as fixed cost
Fixed cost = 128
FC = 128
Average Fixed cost = \(\frac { 128 }{ Q } \)
AFC = \(\frac { 128 }{ Q } \)
Average Variable cost = 60 + 8Q (total variable cost divided by Q)
∴ AVC = 60 + 8Q

The matrix form of the given equation is written as
\(\begin{bmatrix} 7 & -1 & -1 \\ 10 & -2 & +1 \\ 6 & 3 & -2 \end{bmatrix}\left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right]\)\(=\begin{bmatrix} 0\\ 8\\ 7\\ \end{bmatrix}\)
\(\Delta\)=\(\begin{vmatrix} 7& -1& -1\\ 10& -2& +1\\ 6& 3& -2\\ \end{vmatrix}\)
=7(4-3)-(-1)(-20-6)+(-1)(30+12)
=7(1)+1(-26)-1(42)
=7 - 26 - 42 = - 61
\(\triangle \)x_{1 \(=\begin{bmatrix} 0 & -1 & -1 \\ 8 & -2 & +1 \\ 7 & 3 & -2 \end{bmatrix}\)}
=0(4-3)-(-1)(-16-7)+(-1)(24+14)
=0+1(-23)-1(38)
=-23-38 = - 61
\(\Delta x_2=\begin{vmatrix} 7& 0& -1\\ 10& 8& 1\\ 6& 7& -2\\ \end{vmatrix}\)
=7(-16 - 7) - (0)(-20 - 6)+(-1)(70 - 48)
=7(-23)+0 - 1 (22)
=-161 - 22 = -183
\(\triangle \)x_{3 \(=\begin{bmatrix} 7 & -1 & 0 \\ 10 & -2 & 8 \\ 6 &3 & 7 \end{bmatrix}\)}
=7(-14-24)-(-1)(70-48)+0(30+12)
=7(-38)+1(22)+0(42)
=-266+22+0
\(\triangle \)x₁ = - 244
\(x_{1}=\frac{\Delta x_{1}}{\Delta}=\frac{-61}{-61}=1\)
\(x_{2}=\frac{\Delta x_{2}}{\Delta}=\frac{-183}{-61}=3\)
\(x_{3}=\frac{\Delta x_{3}}{\Delta}=\frac{-244}{-61}=4\)

i) zero
\(\mathrm{P} =60-0.2 \mathrm{Q}  \)
\(0 =60-0.2 \mathrm{Q}  \)
\(0.2 \mathrm{Q} =60  \)
\(\mathrm{Q} =\frac{60}{0.2}=\frac{600}{2}=300\)
ii) Rs. 20
\(\mathrm{P} =60-0.2 \mathrm{Q}  \)
\(20 =60-0.2 \mathrm{Q}  \)
\(0.2 \mathrm{Q} =60-20=40 ; \quad \mathrm{Q}=\frac{40}{0.2}=\frac{400}{2}=200\)
iii) Rs. 40
\(\mathrm{P} =60-0.2 \mathrm{Q}  \)
\(40 =60-0.2 \mathrm{Q}  \)
\(0.2 \mathrm{Q} =60-40  \)
\(\mathrm{Q} =\frac{20}{0.2}=\frac{200}{2}=100\)