Class 12 Applications of Derivatives: Practice and Worked Solutions

1. 1

   Find the dividing point

   The derivative is zero at x = 3.

   $f'(x)=2x-6=0\quad\Rightarrow\quad x=3$
2. 2

   Read the signs

   For x < 3, the derivative is negative. For x > 3, it is positive. The function decreases on (−∞, 3) and increases on (3, ∞).

3. 3

   Find the value

   A negative-to-positive sign change gives a minimum. Completing the square confirms it is the absolute minimum.

   $f(3)=-4,\qquad f(x)=(x-3)^2-4$

Answer: Minimum −4 at x = 3; decreasing before 3 and increasing after 3.

1. 1

   Differentiate and solve

   The stationary candidates are x = 1 and x = 3.

   $g'(x)=3x^2-12x+9=3(x-1)(x-3)$
2. 2

   Classify using signs

   The derivative signs are +, −, + across the intervals separated by 1 and 3. Thus x = 1 is a local maximum and x = 3 is a local minimum.

3. 3

   Give the coordinates

   Substitute into g, not its derivative.

   $g(1)=4,\qquad g(3)=0$

Answer: Local maximum (1, 4); local minimum (3, 0).

1. 1

   Connect the quantities

   Area depends on radius, and radius depends on time.

   $\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}=2\pi r\frac{dr}{dt}$
2. 2

   Substitute at this instant

   Use r = 6 and dr/dt = 0.5.

   $\frac{dA}{dt}=2\pi(6)(0.5)=6\pi$
3. 3

   Interpret the unit

   We calculated square centimetres of area gained per second, not the area of the circle.

Answer: $6\pi\text{ cm}^2/\text{s}$.

1. 1

   Write the constraint

   If the remaining fenced side is y, then 2x + y = 40. A non-degenerate rectangle needs both lengths positive.

   $y=40-2x,\qquad 0<x<20$
2. 2

   Build the objective

   Area is width times length. Differentiate the one-variable model.

   $A=x(40-2x),\qquad A'=40-4x$
3. 3

   Verify the maximum

   The derivative vanishes at x = 10 and changes from positive to negative there. Also A″ = −4. The third fenced side is 20 m.

   $A(10)=200\text{ m}^2$

Answer: Two 10 m sides and one 20 m fenced side; maximum area 200 m².

1. 1

   Check the derivative

   It is zero at zero and positive for every other real x.

   $h'(x)=3x^2$
2. 2

   Compare both sides

   The derivative is positive immediately before and after zero. There is no positive-to-negative or negative-to-positive change.

3. 3

   Correct the claim

   Zero is a stationary point without a local extremum. The second-derivative test is inconclusive here because h″(0) = 0.

Answer: No local maximum or minimum at zero. A zero derivative alone is insufficient.

1. 1

   Find interior candidates

   The derivative has roots −1 and 1. Only 1 lies inside [0, 3].

   $f'(x)=3(x-1)(x+1)$
2. 2

   Compare all included candidates

   Evaluate the function at the endpoints and the interior stationary point.

   $f(0)=0,\quad f(1)=-2,\quad f(3)=18$
3. 3

   Choose largest and smallest

   The smallest value is −2; the largest is 18. The absolute maximum is at an endpoint.

Answer: Absolute minimum −2 at x = 1; absolute maximum 18 at x = 3.

1. 1

   Locate and classify

   The derivative changes from negative to positive at x = 4.

   $f'(x)=2x-8$
2. 2

   Evaluate and verify

   The square is non-negative for all real x.

   $f(x)=(x-4)^2-9$

Answer: Minimum −9 at x = 4.

1. 1

   Apply the chain rule

   Use the current radius and its growth rate.

   $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$
2. 2

   Substitute

   Attach area-per-time units to the result.

   $2\pi(5)(0.2)=2\pi\text{ cm}^2/\text{s}$

Answer: $2\pi\text{ cm}^2/\text{s}$.