Given the marginal cost function y = 23 + 16x - 3x² ; C = 40
Rs.40 is the fixed cost.
We know that
Total Cost function = \(\int { } \) (Marginal cost function) dx + C
TC=\(\int { } \)ydx+C
=\(\int { } \)(23+16x-3x²)dx+c,where C is a constant,
=\(\int { } \)23dx+\(\int { } \)16x dx - \(\int { } \)3x²dx+C
=23x+16\(\left[ \frac { { x }^{ 2 } }{ 2 } \right] \)-3\(\left[ \frac { { x }^{ 3 } }{ 3 } \right] \)+c
TC=23x+8x²-x³+C
c = 40 given
\(\therefore\) TC = 23x + 8x² - x³ + 40
Average cost function=\(\frac { TC }{ x } \)
=23 + 8x - x² + \(\frac { 40 }{ x } \)

\(TC=20+5{ Q }^{ 2 }+{ 3Q }^{ 3 }\)
\(MC=\frac { d(20) }{ dQ } +\frac { d({ 5Q }^{ 2 }) }{ dQ } +\frac { d(3Q^{ 3 }) }{ dQ } \)
\(=0+5\times 2\quad { Q }^{ 2-1 }+3(3)\quad { Q }^{ 3-1 }\)
\(=0+10Q+9{ Q }^{ 2 }\)
\(MC=10Q+{ 9Q }^{ 2 }\)

y = 3x² + 16x³
Differentiating,
\(\frac{dy}{dx}=3(2)x^{2-1}+16(3)x^{3-1}\)
= 6x¹ + 48x²
\(\frac{dy}{dx}=6x+48x^{2}\)

Given
Q_d = 50 - 5P
5P = 50 - Q_d
5P = 50 when Q_d is zero.
\(p=\frac { 50 }{ 5 } \)
P = 10
When P = 10, Demand is 0
Hence P = 10, which is the maximum price beyond which nobody will demand the commodity.

Given TC(Q) = Q³ - 18Q² + 91Q + 10, To find MC differentiate the function with respect to Q.
\(MC(Q)=\frac{d(TC)}{dQ}=3Q^{3-1}-18(2)Q^{2-1}\)+ 91(1)Q^1-1 + 0
= 3Q² - 36Q¹ + 91Q⁰ + 0
MC(Q) = 3Q² - 36Q + 91   (\(\therefore\) Q⁰ = 1)
When Q = 3
MC(Q) = 3(3²) - 36(3) + 91
= 3(9) - 108 + 91
=27 - 108 + 91
= 118 - 108
=10
To find AVC
GivenTC(Q) = Q³ - 18Q² + 91Q + 10
We know TVC (Q) = Q³ - 18Q²
+ 91Q (\(\therefore\) constant value is fixed cost)
AVC(Q) = TVC(Q)/Q
AVC(Q) = Q² - 18Q + 91
When Q = 3
AVC (Q) = 3² - 18(3) + 9
= 9 - 54 + 91
= 100 - 54 = 46