Q.(a) The total cost function of a firm is $C(x) = \dfrac{x^3}{3} - 5x^2 + 28x + 10$, where $x$ is the output. A tax at the rate of ₹ 2 per unit of output is imposed and the producer adds it to his cost. If the market demand function is given by $p = 2530 - 5x$, where $p$ is the price per unit of output, find the profit maximizing the output and price.

(a) Add tax to cost, maximise profit $P=-\tfrac{x^3}{3}+2500x-10$: $x=50$, $p=2280$. (b) Log-differentiate $x^m y^n=(x+y)^{m+n}$ to get $\dfrac{dy}{dx}=\dfrac{y}{x}$.

A 5-mark either/or from the Applications of Differentiation / Differential Calculus units of the Tamil Nadu HSC Class-11 Business Mathematics syllabus. Both alternatives are solved.

(a) Profit-maximising output and price.

Step 1 — Cost including tax. A tax of \u20b92 per unit adds $2x$ to cost:

$C(x)=\frac{x^3}{3}-5x^2+28x+10+2x=\frac{x^3}{3}-5x^2+30x+10.$

Step 2 — Revenue. With $p=2530-5x$,

$R=px=(2530-5x)x=2530x-5x^2.$

Step 3 — Profit function.

$P=R-C=2530x-5x^2-\left(\frac{x^3}{3}-5x^2+30x+10\right)=-\frac{x^3}{3}+2500x-10.$

Step 4 — Maximise. $P'(x)=-x^2+2500=0\Rightarrow x^2=2500\Rightarrow x=50$ (output positive). Since $P''(x)=-2x=-100<0$, this is a maximum.

Step 5 — Price. $p=2530-5(50)=2530-250=2280.$

(b) Show $\dfrac{dy}{dx}=\dfrac{y}{x}$ given $x^m y^n=(x+y)^{m+n}$.

Step 1 — Take logarithms.

$m\ln x+n\ln y=(m+n)\ln(x+y).$

Step 2 — Differentiate w.r.t. $x$.

$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right).$

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