Calculus

(c) $\dfrac{P}{Q_d}\dfrac{dQ_d}{dP}$

Price elasticity of demand is the proportionate change in quantity divided by the proportionate change in price, which in calculus form is $\eta = \dfrac{P}{Q_d}\cdot\dfrac{dQ_d}{dP}$ (taken in magnitude).

(b) $\dfrac{x^{n+1}}{n+1} + C$

By the power rule of integration, $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \ne -1$. The constant of integration must be included for an indefinite integral.

(b) $-\tfrac{1}{2}$

For $x \ne 0$, $f(x) = \dfrac{x(x-1)}{2x} = \dfrac{x-1}{2}$, so $\displaystyle\lim_{x\to0} f(x) = -\tfrac12$. Continuity at $x=0$ requires $k = \lim_{x\to 0} f(x) = -\tfrac12$.

First limit

Both numerator and denominator vanish at $x = a$, so rationalise the numerator: $\lim_{x\to a}\frac{(\sqrt{3a-x} - \sqrt{x+a})(\sqrt{3a-x} + \sqrt{x+a})}{4(x-a)(\sqrt{3a-x} + \sqrt{x+a})}$ $= \lim_{x\to a}\frac{(3a-x)-(x+a)}{4(x-a)(\sqrt{3a-x} + \sqrt{x+a})} = \lim_{x\to a}\frac{-2(x-a)}{4(x-a)(\sqrt{3a-x} + \sqrt{x+a})}$ $= \lim_{x\to a}\frac{-1}{2(\sqrt{3a-x} + \sqrt{x+a})} = \frac{-1}{2\cdot 2\sqrt{2a}} = -\frac{1}{4\sqrt{2a}}$

Limit of $f(x)$ at $x = 2$

Left-hand limit ($x<2$): $\displaystyle\lim_{x\to2^-}(2x+2) = 6$.

Right-hand limit ($x\ge 2$): $\displaystyle\lim_{x\to2^+}(x+4) = 6$.

Since both are equal, the limit exists and $\displaystyle\lim_{x\to2} f(x) = 6$ (which also equals $f(2) = 6$).

Given $C = 300Q - 10Q^2 + \tfrac13 Q^3$.

Marginal cost $MC = \frac{dC}{dQ} = 300 - 20Q + Q^2$ $\frac{d(MC)}{dQ} = -20 + 2Q = 0 \Rightarrow Q = 10,\qquad \frac{d^2(MC)}{dQ^2} = 2 > 0$ So MC is minimum at $Q = 10$ (minimum value $300 - 200 + 100 = 200$).

Average cost $AC = \frac{C}{Q} = 300 - 10Q + \tfrac13 Q^2$ $\frac{d(AC)}{dQ} = -10 + \tfrac23 Q = 0 \Rightarrow Q = 15,\qquad \frac{d^2(AC)}{dQ^2} = \tfrac23 > 0$ So AC is minimum at $Q = 15$.

MC = AC at minimum AC (at $Q=15$) $AC = 300 - 150 + \tfrac13(225) = 300 - 150 + 75 = 225$ $MC = 300 - 20(15) + 225 = 300 - 300 + 225 = 225$ Thus $MC = AC = 225$ when average cost is minimum.

(a) Let $u = 2x^3 - 3x^2 + 5x - 7$. Then $du = (6x^2 - 6x + 5)\,dx$, which is exactly the numerator. $\int \frac{du}{\sqrt{u}} = \int u^{-1/2}\,du = 2u^{1/2} + C$ $= 2\sqrt{2x^3 - 3x^2 + 5x - 7} + C$

(b) Integrate by parts with $u = \ln x$, $dv = x\,dx$, so $du = \dfrac{dx}{x}$, $v = \dfrac{x^2}{2}$: $\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac1x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$ $= \frac{x^2}{4}(2\ln x - 1) + C$

Definition: The derivative of $y = f(x)$ with respect to $x$ is $\frac{dy}{dx} = f'(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}$ provided the limit exists.

Derivative of $\dfrac{3x+5}{\sqrt x}$

Write $f(x) = \dfrac{3x+5}{\sqrt x} = 3\sqrt{x} + \dfrac{5}{\sqrt{x}}$ and differentiate each term from the definition.

For $\sqrt x$: $\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h} = \lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt x} = \frac{1}{2\sqrt x}$

For $\dfrac{1}{\sqrt x}$: $\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}-\frac1{\sqrt x}}{h} = \lim_{h\to0}\frac{\sqrt x-\sqrt{x+h}}{h\sqrt x\sqrt{x+h}} = \lim_{h\to0}\frac{-1}{\sqrt x\sqrt{x+h}\,(\sqrt x+\sqrt{x+h})} = -\frac{1}{2x\sqrt x}$

Therefore $f'(x) = \frac{3}{2\sqrt x} - \frac{5}{2x\sqrt x} = \frac{3x-5}{2x\sqrt x}$

Using the quotient rule with $u = e^x$ and $v = 1 + \log x$, where $u' = e^x$ and $v' = \dfrac1x$: $\frac{dy}{dx} = \frac{v\,u' - u\,v'}{v^2} = \frac{e^x(1+\log x) - \dfrac{e^x}{x}}{(1+\log x)^2}$ $\frac{dy}{dx} = \frac{e^x\left(1 + \log x - \dfrac1x\right)}{(1+\log x)^2}$

(c) $4$

Factorising, $\dfrac{x^2-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$ for $x\neq2$, so the limit as $x\to2$ is $2+2=4$.

By the first principle, $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

With $f(x)=3x^2-5x+2$, $f(x+h)=3(x+h)^2-5(x+h)+2=3x^2+6xh+3h^2-5x-5h+2$

$f(x+h)-f(x)=6xh+3h^2-5h=h(6x+3h-5)$

$f'(x)=\lim_{h\to0}\frac{h(6x+3h-5)}{h}=\lim_{h\to0}(6x+3h-5)=6x-5$

Now, $f'(1)=6(1)-5=1$

Hence $f'(x)=6x-5$ and $f'(1)=1$.

(i) Marginal cost function

Marginal cost is the derivative of total cost: $MC=\frac{dC}{dx}=6x^2-18x+12$

(ii) Value of $x$ for minimum marginal cost

Let $MC=f(x)=6x^2-18x+12$. For a minimum, $f'(x)=0$: $12x-18=0 \;\Rightarrow\; x=\frac{18}{12}=1.5$

Also $f''(x)=12>0$, so $MC$ is minimum at $x=1.5$ units.

The minimum marginal cost is $MC(1.5)=6(1.5)^2-18(1.5)+12=13.5-27+12=-1.5$

So marginal cost is minimum when $x=1.5$ units.

Given $p=50-2x$ and $C(x)=3x^2+10x+5$.

(i) Total revenue $R(x)=p\cdot x=(50-2x)x=50x-2x^2$

(ii) Profit function $P(x)=R(x)-C(x)=(50x-2x^2)-(3x^2+10x+5)=-5x^2+40x-5$

(iii) Profit-maximising output

$P'(x)=-10x+40=0 \;\Rightarrow\; x=4$

Since $P''(x)=-10<0$, profit is maximum at $x=4$ units.

(iv) Maximum profit $P(4)=-5(16)+160-5=-80+160-5=75$

Maximum profit $=$ Rs 75 (at $x=4$ units).