Rate of Change, Marginal Cost and Marginal Revenue for CUET Applied Maths

AnswerC · $20 \pi$ cm²/s

1. $A = \pi r^{2}$, so $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.

2. $= 2\pi \times 5 \times 2 = 20\pi$ cm²/s.

AnswerB · $12 \pi$ cm²/cm

1. $\frac{dA}{dr} = \frac{d}{dr}(\pi r^{2}) = 2\pi r$.

2. At $r = 6$: $12\pi$ cm² for each cm of radius.

AnswerC · 240 units a month

1. The rate is $S'(t) = 200 + 10t$.

2. At $t = 4$: $200 + 40 = 240$ units a month.

AnswerC · ₹66

1. $R'(x) = 6x + 36$.

2. $R'(5) = 30 + 36 = 66$ rupees per unit.

AnswerD · 12 cm/s

1. $P = 4a$, so $\frac{dP}{dt} = 4\frac{da}{dt}$.

2. $4 \times 3 = 12$ cm/s, whatever the size of the square.

AnswerB · $16 \pi$ cm³ per cm

1. $\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$.

2. At $r = 2$: $16\pi$ cm³ per cm.

AnswerC · 144 cm³/s

1. $V = a^{3}$, so $\frac{dV}{dt} = 3a^{2} \frac{da}{dt}$.

2. $= 3 \times 16 \times 3 = 144$ cm³/s.

AnswerA · 30 people a year

1. The rate of growth is $P'(t) = 20 + 2t$.

2. $P'(5) = 20 + 10 = 30$ people a year.

AnswerA · Only I

1. I is what a derivative means. True.

2. II: the chain rule multiplies, $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$. False.

AnswerB · 1 cm/s

1. $A = \pi r^2$, so $\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$.

2. $10\pi = 2\pi \times 5 \times \frac{dr}{dt}$, so $\frac{dr}{dt} = 1$ cm/s.

AnswerD · $36 \pi$ cm³/s

1. $V = \frac{4}{3}\pi r^3$, so $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$.

2. $4\pi \times 9 \times 1 = 36\pi$ cm³/s.

AnswerD · A is false, but R is true.

1. R is the chain rule applied to $A = \pi r^2$. True.

2. A is false: by R, $\frac{dA}{dt}$ grows with $r$, so the area grows faster and faster.

AnswerA · $\frac{1}{3}$ cm/s

1. $V = a^3$, so $\frac{dV}{dt} = 3a^2\frac{da}{dt}$.

2. $9 = 3 \times 9 \times \frac{da}{dt}$, so $\frac{da}{dt} = \frac{1}{3}$ cm/s.

AnswerA · $\frac{1}{\pi}$ cm/s

1. $\frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}$, so $100 = 4\pi(25)\frac{dr}{dt}$.

2. $\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}$ cm/s.

AnswerB · 40 cm²/s

1. $\frac{dS}{dt} = 8\pi r\frac{dr}{dt}$.

2. At $r = 5$, $\frac{dr}{dt} = \frac{1}{\pi}$: $8\pi(5)\left(\frac{1}{\pi}\right) = 40$ cm²/s.

AnswerA · $\frac{1}{4 \pi}$ cm/s

1. $100 = 4\pi(100)\frac{dr}{dt}$.

2. $\frac{dr}{dt} = \frac{1}{4\pi}$ cm/s.

AnswerD · It falls, in proportion to $\frac{1}{r^{2}}$

1. $\frac{dr}{dt} = \frac{100}{4\pi r^{2}}$.

2. The same 100 cm³ spreads over a larger surface each second, so the radius grows more slowly, as $\frac{1}{r^{2}}$.

AnswerB · ₹500

1. $R(40) = 1{,}800$ and $C(40) = 500 + 800 = 1{,}300$.

2. Profit $= 1{,}800 - 1{,}300 = 500$ rupees.

AnswerD · 200 units

1. Break-even is where revenue equals cost: $40x = 3{,}000 + 25x$.

2. $15x = 3{,}000$, so $x = 200$ units.

AnswerB · ₹800

1. The fixed cost is what must be paid even when nothing is made: $C(0)$.

2. $C(0) = 800$ rupees.

AnswerA · ₹750

1. Profit $P(x) = R(x) - C(x) = 35x - 1{,}000$.

2. $P(50) = 1{,}750 - 1{,}000 = 750$ rupees.

AnswerB · 100 units

1. Break-even is where $R(x) = C(x)$: $75x = 4{,}500 + 30x$.

2. $45x = 4{,}500$, so $x = 100$ units.

AnswerD · $- 2 x^{2} + 60 x$

1. Revenue = price × quantity: $R(x) = x(60 - 2x)$.

2. $R(x) = 60x - 2x^{2}$.

AnswerB · ₹60

1. $C(20) = 800 + 200 + 200 = 1{,}200$ rupees.

2. Average cost $= \frac{C(20)}{20} = \frac{1{,}200}{20} = 60$ rupees per unit.

AnswerA · Only I

1. I: $C(0)$ is what must be paid even at zero output, such as rent. True.

2. II: for $C(x) = 2x^{2} + 10x + 200$ at $x = 20$, the average is 60 but the marginal cost is 90. False.

AnswerA · ₹175

1. $R(15) = 15 \times 35 = 525$ and $C(15) = 150 + 200 = 350$.

2. Profit $= 525 - 350 = 175$ rupees.

AnswerD · $60 - 2 x$

1. Average revenue $= \frac{R(x)}{x}$.

2. $\frac{60x - 2x^{2}}{x} = 60 - 2x$, which is the price of a unit.

AnswerD · $- 4 x^{2} + 100 x$

1. Revenue is price times quantity: $R(x) = x \cdot p$.

2. $R(x) = x(100 - 4x) = 100x - 4x^{2}$.

AnswerC · ₹72

1. Average cost $= \frac{C(x)}{x} = 3x + 12 + \frac{300}{x}$.

2. At $x = 10$: $30 + 12 + 30 = 72$ rupees per unit.

AnswerB · I, II and III only

1. I, II and III are the definitions of profit, average cost and break-even.

2. IV is false: the fixed cost, such as rent, stays the same whatever the output; it is the variable cost that changes.

AnswerC · ₹35

1. Marginal cost is the derivative of total cost: $MC = C'(x) = 3x^{2} - 12x + 20$.

2. $C'(5) = 75 - 60 + 20 = 35$ rupees.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. $C'(x) = 8$ for every $x$: each extra unit costs ₹8. A is true.

2. R is the reason the marginal cost does not change, so R explains A.

AnswerA · Only I

1. I: $C'(x) = 7$, a constant. True.

2. II: a constant's derivative is 0, so the fixed cost never enters the marginal cost. False.

AnswerA · ₹60

1. Marginal cost $MC = C'(x) = 3 x^{2} - 6 x + 15$.

2. At $x = 5$: $60$ rupees per unit.

AnswerC · ₹16

1. Marginal cost $MC = C'(x) = \frac{x}{5} + 6$.

2. At $x = 50$: $16$ rupees per unit.

AnswerC · ₹31

1. $MC = 40 - 1.2x + 0.03x^{2}$.

2. At $x = 10$: $40 - 12 + 3 = 31$ rupees.

AnswerD · ₹85

1. $C(10) = 500 + 400 - 60 + 10 = 850$.

2. Average cost $= \frac{850}{10} = 85$ rupees.

AnswerD · One more unit, the $(x + 1)$th

1. $C'(x) \approx \frac{C(x + 1) - C(x)}{1}$, the extra cost when output rises by one unit.

2. So it is roughly the cost of the $(x + 1)$th unit.

AnswerC · ₹32

1. $C'(x) = 12 + 2x$, so $C'(10) = 32$.

2. The 11th unit costs about ₹32. (Exactly, $C(11) - C(10) = 353 - 320 = 33$; the marginal cost is the quick estimate.)

AnswerC · ₹4

1. $C'(x) = 40 \times \frac{1}{2\sqrt{x}} = \frac{20}{\sqrt{x}}$.

2. At $x = 25$: $\frac{20}{5} = 4$ rupees.