Q.Total Revenue is $TR = 100Q - 2Q^2$ and Total Cost is $TC = Q^2 + 20Q + 50$. Find the output level $Q$ that maximises profit, and confirm it is genuinely a maximum.

$C'(x)=20x-1000=0\Rightarrow x=50$; $C''(x)=20>0$ (minimum); $C(50)=25{,}000$.

Cost function: $C(x)=10x^2-1000x+50000.$

Differentiate and set to zero:

$C'(x)=20x-1000=0\ \Rightarrow\ x=50.$

Second derivative test: $C''(x)=20>0$, so $x=50$ gives a minimum.

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$R = 30x - \tfrac{x^3}{10}$; $R'(x) = 30 - \tfrac{3x^2}{10} = 0 \Rightarrow x = 10$; $P = 30 - \tfrac{100}{10} = 20$. Demand $= 10$, price $= ₹20$.

GSEB Class-12 Statistics, Differentiation (maxima and minima):

Given the demand function $P = 30 - \dfrac{x^2}{10}$.

Revenue function:

$R = P \cdot x = \left(30 - \frac{x^2}{10}\right) x = 30x - \frac{x^3}{10}$

First-order condition (set $\dfrac{dR}{dx} = 0$):

$\frac{dR}{dx} = 30 - \frac{3x^2}{10} = 0$

$\frac{3x^2}{10} = 30 \quad \Rightarrow \quad x^2 = 100 \quad \Rightarrow \quad x = 10$

(We take $x = 10$ as demand cannot be negative.)

Second-order check:

$\frac{d^2R}{dx^2} = -\frac{6x}{10} = -\frac{6(10)}{10} = -6 < 0$ …

$y'=3x^2-4x-4=0\Rightarrow x=2$ (min, $y=-9$) or $x=-\frac23$ (max, $y=\frac{13}{27}$). OR: profit $P=-0.1x^2+1000x-100$, $P'=0\Rightarrow x=5000$.

Main part: $y=x^3-2x^2-4x-1$.

$\frac{dy}{dx}=3x^2-4x-4=0\;\Rightarrow\;x=\frac{4\pm\sqrt{16+48}}{6}=\frac{4\pm8}{6}.$

So $x=2$ or $x=-\dfrac23$. Second derivative $\dfrac{d^2y}{dx^2}=6x-4$.

• At $x=2$: $6(2)-4=8>0\Rightarrow$ minimum. Value: $y=2^3-2(2)^2-4(2)-1=8-8-8-1=-9$.
• At $x=-\tfrac23$: $6(-\tfrac23)-4=-8<0\Rightarrow$ maximum. Value: $y=\Big(-\tfrac23\Big)^3-2\Big(-\tfrac23\Big)^2-4\Big(-\tfrac23\Big)-1=-\tfrac{8}{27}-\tfrac{8}{9}+\tfrac{8}{3}-1=\frac{-8-24+72-27}{27}=\frac{13}{27}.$

OR part: Selling price $=₹10000$ each, so revenue $R=10000x$; cost $C=0.1x^2+9000x+100$. Profit: …

$R=Px=1200x-10x^2$, $C=200x+15x^2$; $\pi=1000x-25x^2$; $\pi'=1000-50x=0\Rightarrow x=\mathbf{20}$; $\pi''=-50<0$ (max); max profit $=\mathbf{₹10000}$.

Given: cost $C=200x+15x^2$; demand (price) function $P=1200-10x$.

Step 1 — revenue.

$R=P\cdot x=(1200-10x)x=1200x-10x^2.$

Step 2 — profit function.

$\pi(x)=R-C=(1200x-10x^2)-(200x+15x^2)=1000x-25x^2.$

Step 3 — maximise: first-order condition.

$\frac{d\pi}{dx}=1000-50x=0\ \Rightarrow\ x=20.$

Step 4 — second-order check.

$\frac{d^2\pi}{dx^2}=-50<0,$ …

$y'=3x^2-4x-4=0 \Rightarrow x=-2/3$ (max, $y=13/27$) and $x=2$ (min, $y=-9$). OR: cost min at $x=50$ tons, min cost ₹25,000.

Main part — maxima/minima of $y = x^3 - 2x^2 - 4x - 1$.

$y' = 3x^2 - 4x - 4$

Set $y' = 0$:

$3x^2 - 4x - 4 = 0 \implies x = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm 8}{6}$

$\therefore\ x = 2 \quad\text{or}\quad x = -\frac{2}{3}$

Second derivative: $y'' = 6x - 4$.

• At $x = -\dfrac23$: $y'' = 6(-\tfrac23) - 4 = -8 < 0 \Rightarrow$ maximum.

  $y = \left(-\tfrac23\right)^3 - 2\left(-\tfrac23\right)^2 - 4\left(-\tfrac23\right) - 1 = -\tfrac{8}{27} - \tfrac{8}{9} + \tfrac{8}{3} - 1 = \frac{-8 - 24 + 72 - 27}{27} = \frac{13}{27} \approx 0.48$

• At $x = 2$: $y'' = 6(2) - 4 = 8 > 0 \Rightarrow$ minimum.

  $y = 8 - 8 - 8 - 1 = -9$

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