Q.The marginal cost of producing $x$ units is $MC=3x^2-4x+10$, and the fixed cost is ₹500. Find the total cost function and the cost of producing 5 units.

Profit is maximum where marginal revenue equals marginal cost: $MR=MC$, i.e. $MC-MR=0$.

Profit is $p(x)=R(x)-C(x)$, where $R$ is total revenue and $C$ is total cost. For a maximum, the first derivative must vanish:

$p'(x) = R'(x) - C'(x) = MR - MC = 0 \;\Rightarrow\; MR = MC.$

This is exactly the condition $MC-MR=0$. (The second-order condition $p''(x)<0$ ensures it is a maximum, i.e. the slope of $MC$ exceeds that of $MR$ at that output.)

A constant marginal revenue integrates to $R=kx$ (no constant, since $R(0)=0$); the demand price $p=R/x=k$ equals $MR$, option (b).

Integrate the constant $MR$. Let $MR=k$. Then

$R=\int MR\,dx=\int k\,dx=kx+C.$

Fix the constant. Revenue is zero when $x=0$, so $C=0$ and $R=kx$.

Demand function. The demand (price) function is

$p=\frac{R}{x}=\frac{kx}{x}=k=MR.$

So the demand function is itself the constant marginal revenue.

Profit is maximised where its derivative is zero: $MR=MC$, i.e. $MC-MR=0$.

Profit is defined as revenue minus cost:

$P(x)=R(x)-C(x).$

For a maximum, set the first derivative to zero:

$P'(x)=R'(x)-C'(x)=0\;\Rightarrow\;MR-MC=0\;\Rightarrow\;MC-MR=0,$

since $R'(x)=MR$ (marginal revenue) and $C'(x)=MC$ (marginal cost). (The second-order condition $P''(x)<0$ confirms it is a maximum.)

Total revenue is the integral of marginal revenue; dividing $R$ by $x$ gives the demand (price) function $p=100-3x^2$.

Given the marginal revenue $MR=100-9x^2$. Total revenue is:

$R=\int MR\,dx=\int (100-9x^2)\,dx=100x-3x^3+c.$

At $x=0$ revenue is $0$, so $c=0$:

$R=100x-3x^3.$

The demand function is the price $p$, where $R=p\cdot x$:

$p=\frac{R}{x}=\frac{100x-3x^3}{x}=100-3x^2.$

Profit peaks when $MC = MR$, i.e. $MC - MR = 0$.

Profit is total revenue minus total cost: $p(x) = R(x) - C(x)$. For a maximum, differentiate and set the derivative to zero:

$\frac{dp}{dx} = \frac{dR}{dx} - \frac{dC}{dx} = MR - MC = 0.$

Hence the necessary condition is $MR = MC$, equivalently $MC - MR = 0$ (the second-order condition confirms it is a maximum when marginal cost is rising relative to marginal revenue).

The constant $k = -\dfrac{1}{\eta_d}$.

The elasticity of demand is defined as

$\eta_d = -\frac{p}{x}\cdot\frac{dx}{dp}.$

Integrate the given relation $\displaystyle\int \frac{dp}{p} = k\int \frac{dx}{x}$:

$\ln p = k\ln x + \text{constant} \;\Rightarrow\; p = C\,x^{k}.$

Differentiate: $\dfrac{dp}{dx} = Ck\,x^{k-1} = k\cdot\dfrac{p}{x}$, so $\dfrac{dx}{dp} = \dfrac{x}{k\,p}$.

Substitute into the elasticity formula:

$\eta_d = -\frac{p}{x}\cdot\frac{x}{k\,p} = -\frac{1}{k} \;\Rightarrow\; k = -\frac{1}{\eta_d}.$

Integrate $MC$ to get $TC$, fix the constant using $TC = 0$ at $x = 0$, then divide by $x$ to obtain the average cost $AC = \frac{200}{3}x^{1/2}$.

Step 1 — Total cost from marginal cost.

$TC = \int MC\,dx = \int 100\sqrt{x}\,dx = 100\int x^{1/2}\,dx = 100\cdot \frac{x^{3/2}}{3/2} = \frac{200}{3}x^{3/2} + c.$

Step 2 — Fix the constant. Given $TC = 0$ when $x = 0$: $\;0 = \frac{200}{3}(0) + c \Rightarrow c = 0$. So $TC = \frac{200}{3}x^{3/2}$.

Step 3 — Average cost.

$AC = \frac{TC}{x} = \frac{\frac{200}{3}x^{3/2}}{x} = \frac{200}{3}x^{3/2 - 1} = \frac{200}{3}x^{1/2}.$