Q.Ramesh, the owner of a sweet selling shop, purchased some rectangular cardboard sheets of dimension $25$ cm by $40$ cm to make container packets without top. Let $x$ cm be the length of the side of the square to be cut out from each corner to give that sheet the shape of the container by folding up the flaps. Based on the above information answer the following questions.

Showing the 12 most recent of 13 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.The least value of $f(x) = x^3 - 12x$, $x \in [0, 3]$ is (A) $-16$ (B) $-9$ (C) $0$ (D) $16$

   ›Reveal solutionSolution

   To find the least value of a continuous function on a closed interval, we evaluate the function at its critical points within the interval and at the interval's endpoints. The smallest of these values is the global minimum. For $f(x) = x^3 - 12x$ on $[0, 3]$, the least value is $\boxed{-16}$.

   When we need to find the maximum or minimum value of a function over a specific interval, especially a closed one, we are looking for its global extrema. For a continuous function on a closed interval, these global extrema are guaranteed to exist. The key insight from calculus is that these extreme values can only occur at two types of points:

   1. Critical points: These are points where the derivative of the function is zero or undefined. At such points, the function might have a local maximum or a local minimum.
   2. Endpoints of the interval: Even if the function is increasing or decreasing throughout the interval, its maximum or minimum value could be at one of the boundaries.

   Therefore, our strategy is to find all potential candidates for the minimum (critical points within the interval and the endpoints) and then compare the function's value at each of these candidates. The smallest value will be the least value of the function on the given interval.

   Let's apply this method to $f(x) = x^3 - 12x$ on the interval $x \in [0, 3]$.

   1. Find the derivative of the function.

      To locate critical points, we first need the first derivative of $f(x)$.

      $f(x) = x^3 - 12x$

      $f'(x) = \frac{d}{dx}(x^3 - 12x) = 3x^2 - 12$

   2. Find the critical points.

      Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. Since $f'(x) = 3x^2 - 12$ is a polynomial, it is defined for all real $x$. So, we only need to set $f'(x) = 0$:

      $3x^2 - 12 = 0$

      $3x^2 = 12$

      $x^2 = 4$

      Solving for $x$, we get $x = \pm 2$.

   3. Identify critical points within the given interval.

      The given interval is $[0, 3]$. We must check which of our critical points fall within this range.

      • $x = 2$ is in $[0, 3]$. This is a candidate for the minimum.
      • $x = -2$ is not in $[0, 3]$. We discard this critical point for this problem, as it's outside our domain of interest.

   4. Evaluate the function at the valid critical points and the interval's endpoints.

      We now have a list of candidate $x$-values where the minimum could occur: $x=0$ (left endpoint), $x=2$ (critical point), and $x=3$ (right endpoint). We calculate $f(x)$ for each of these:

      • At $x = 0$ (left endpoint): $f(0) = (0)^3 - 12(0) = 0 - 0 = 0$
      • At $x = 2$ (critical point): $f(2) = (2)^3 - 12(2) = 8 - 24 = -16$
      • At $x = 3$ (right endpoint): $f(3) = (3)^3 - 12(3) = 27 - 36 = -9$

   5. Determine the least value.

      Comparing the values we found: $0, -16, -9$.

      The smallest among these is $-16$.

   Watch out

   A common mistake is to only check critical points and forget to evaluate the function at the endpoints of the interval. For a closed interval, the global maximum or minimum can often occur at an endpoint, not necessarily at a point where the derivative is zero.

   ✓Final answer

   The least value of $f(x) = x^3 - 12x$ on the interval $x \in [0, 3]$ is $\boxed{-16}$.

2. CBSE 2026Set ANNUAL1 mark
   Q.(Continuing the aluminium-box case study of Q.38) What will be the dimensions of the largest box?

   ›Reveal solutionSolution

   Substitute the optimal square side $x=\dfrac{2}{3}$ m (found by maximising the volume function) into the length, breadth and height expressions.

   From the case study, cutting a square of side $x$ from each corner of the $3\text{ m}\times 8\text{ m}$ sheet and folding up the sides gives a box of:

   • Length $= (8-2x)$ m
   • Breadth $= (3-2x)$ m
   • Height $= x$ m

   Maximising $V(x)=x(3-2x)(8-2x)=4x^3-22x^2+24x$ using $V'(x)=12x^2-44x+24=0$ (i.e. $3x^2-11x+6=0$) gives roots $x=3$ or $x=\dfrac{2}{3}$. Since $0<x<1.5$ is required for the box to be valid, the admissible root is $x=\dfrac{2}{3}$, and $V''\left(\tfrac{2}{3}\right)=-28<0$ confirms this is the maximum.

   Substituting $x=\dfrac{2}{3}$:

   $\text{Length} = 8 - 2\left(\tfrac{2}{3}\right) = 8 - \tfrac{4}{3} = \tfrac{20}{3}\ \text{m}$

   $\text{Breadth} = 3 - 2\left(\tfrac{2}{3}\right) = 3 - \tfrac{4}{3} = \tfrac{5}{3}\ \text{m}$

   $\text{Height} = x = \tfrac{2}{3}\ \text{m}$

   (Check: $\tfrac{20}{3}\times\tfrac{5}{3}\times\tfrac{2}{3}=\tfrac{200}{27}$ m³, matching the maximum volume found earlier.)

   ✓Final answer

   The dimensions of the largest box are $\dfrac{20}{3}$ m (length) $\times\ \dfrac{5}{3}$ m (breadth) $\times\ \dfrac{2}{3}$ m (height).

3. CBSE 2026Set ANNUAL1 mark
   Q.(Continuing the aluminium-box case study of Q.38) What will be the side of the square removed to form the largest box?

   ›Reveal solutionSolution

   The optimal square side is the critical point of the volume function that lies in the valid domain and satisfies the second-derivative maximum test.

   For a square of side $x$ removed from each corner of the $3\text{ m}\times 8\text{ m}$ sheet, the box volume is:

   $V(x) = x(3-2x)(8-2x) = 4x^3 - 22x^2 + 24x, \qquad 0 < x < 1.5$

   Differentiating and setting $V'(x)=0$:

   $V'(x) = 12x^2 - 44x + 24 = 0 \implies 3x^2 - 11x + 6 = 0$

   $x = \frac{11 \pm \sqrt{121-72}}{6} = \frac{11\pm 7}{6} \implies x = 3 \text{ or } x = \frac{2}{3}$

   Since the breadth $(3-2x)$ must stay positive, only $x<1.5$ is valid, so $x=3$ is rejected and $x=\dfrac{2}{3}$ is the only admissible critical point.

   Confirming it is a maximum:

   $V''(x) = 24x - 44 \implies V''\left(\tfrac{2}{3}\right) = 16 - 44 = -28 < 0$

   A negative second derivative confirms $x=\dfrac{2}{3}$ gives the maximum volume (and $V=0$ at the endpoints $x=0,\,1.5$, so this is the global maximum on the interval).

   ✓Final answer

   The side of the square removed from each corner is $x = \dfrac{2}{3}$ m.

4. CBSE 2025Set 65/1/11 markMCQ
   Q.The absolute maximum value of function $f(x) = x^3 - 3x + 2$ in $[0, 2]$ is: (A) $0$ (B) $2$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   To find the absolute maximum of a continuous function on a closed interval, we evaluate the function at its critical points within the interval and at the interval's endpoints, then pick the largest value. For $f(x) = x^3 - 3x + 2$ on $[0, 2]$, the absolute maximum value is $\boxed{4}$.

   When we need to find the absolute maximum (or minimum) value of a continuous function over a closed interval, we rely on a fundamental concept from calculus called the Extreme Value Theorem. This theorem guarantees that such a maximum and minimum must exist.

   The intuition behind finding these extreme values is that they can occur in one of two places:

   1. At a "peak" or "valley" within the interval: These are points where the function changes from increasing to decreasing (local maximum) or decreasing to increasing (local minimum). At such points, if the function is differentiable, its derivative will be zero. These are called critical points.
   2. At the boundaries of the interval: Even if the function is steadily increasing or decreasing throughout the interval, its highest or lowest value might simply be at one of the endpoints.

   Therefore, our strategy is to check all these potential locations: the critical points that fall within our interval, and the two endpoints of the interval. We then compare the function values at all these points to find the absolute maximum.

   Here's how we apply this to $f(x) = x^3 - 3x + 2$ on the interval $[0, 2]$:

   1. Find the derivative of the function. The derivative $f'(x)$ tells us about the slope of the tangent line to the function at any point $x$. Critical points occur where the tangent line is horizontal, meaning $f'(x) = 0$.

   $f(x) = x^3 - 3x + 2$

   $f'(x) = \frac{d}{dx}(x^3 - 3x + 2)$

   $f'(x) = 3x^2 - 3$

   2. Find the critical points by setting the derivative to zero. We solve $f'(x) = 0$ to find the $x$-values where the function might have a local maximum or minimum.

   $3x^2 - 3 = 0$

   $3(x^2 - 1) = 0$

   $x^2 - 1 = 0$

   This is a difference of squares, which factors as $(x-1)(x+1)=0$.
   So, the critical points are $x = 1$ and $x = -1$.
   

   3. Identify which critical points lie within the given interval.

   The given interval is $[0, 2]$. We must only consider critical points that are inside or on the boundary of this interval.

   * $x = 1$ is in $[0, 2]$. This is a relevant critical point.

   * $x = -1$ is not in $[0, 2]$. This critical point is outside our region of interest, so we disregard it for finding the absolute maximum on this specific interval.

   4. Evaluate the function at the relevant critical points and at the endpoints of the interval.

      We need to check the function's value at $x=1$ (the critical point within the interval) and at the endpoints $x=0$ and $x=2$.

      • At the left endpoint, $x=0$:

   $f(0) = (0)^3 - 3(0) + 2 = 0 - 0 + 2 = 2$

   *   At the critical point, $x=1$:
   

   $f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$

   *   At the right endpoint, $x=2$:
   

   $f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4$

   5. Compare these values to find the absolute maximum. The values we found are $f(0)=2$, $f(1)=0$, and $f(2)=4$. Comparing these, the largest value is $4$.

   Watch out

   A common mistake is to only check critical points and forget to evaluate the function at the endpoints of the interval. For a closed interval, the absolute maximum or minimum can very well occur at an endpoint, even if there are no critical points there.

   ✓Final answer

   The absolute maximum value of the function $f(x) = x^3 - 3x + 2$ in the interval $[0, 2]$ is $\boxed{4}$.

5. CBSE 2025Set ANNUAL1 mark
   Q.A gardener plans to plant flowers in a rectangular flower bed in such a way that a rectangle is inscribed in the semi-circular field (as shown in the figure). Radius of the semi-circular field is 30 m. Let the length of the rectangle be PQ = $x$ m. Based on above information answer the following: What will be the breadth of the rectangular flower bed in terms of $x$?

   ›Reveal solutionSolution

   The rectangle's top corners lie on the semicircle of radius 30, so the Pythagorean relation between half the length and the breadth gives the breadth as a function of $x$.

   Place the centre $O$ of the semicircle at the origin, with the diameter along the $x$-axis. Since the rectangle $PQRS$ is symmetric about $O$ with top side $PQ=x$, the top corners $P,Q$ are at horizontal distance $x/2$ from $O$. Let the breadth (height of the rectangle) be $b$. Since $P$ (and $Q$) lie on the semicircle of radius 30:

   $\left(\frac x2\right)^2 + b^2 = 30^2 = 900$

   $b = \sqrt{900-\frac{x^2}{4}}$

   ✓Final answer

   Breadth $b = \sqrt{900-\dfrac{x^2}{4}}$ m.

6. CBSE 2025Set ANNUAL1 mark
   Q.A gardener plans to plant flowers in a rectangular flower bed in such a way that a rectangle is inscribed in the semi-circular field (as shown in the figure). Radius of the semi-circular field is 30 m. Let the length of the rectangle be PQ = $x$ m. Based on above information answer the following: What will be the area of rectangular region as a function of $x$?

   ›Reveal solutionSolution

   Area = length × breadth, using the breadth found in terms of $x$.

   The rectangle has length $PQ=x$ and breadth $b=\sqrt{900-x^2/4}$ (from the semicircle constraint). So the area is

   $A(x) = x\cdot b = x\sqrt{900-\frac{x^2}{4}}$

   ✓Final answer

   $A(x) = x\sqrt{900-\dfrac{x^2}{4}}$ sq. m.

7. CBSE 2025Set ANNUAL1 mark
   Q.A gardener plans to plant flowers in a rectangular flower bed in such a way that a rectangle is inscribed in the semi-circular field (as shown in the figure). Radius of the semi-circular field is 30 m. Let the length of the rectangle be PQ = $x$ m. Based on above information answer the following: Gardener wants maximum area for the rectangular flower bed. For this to happen, what will be the value of $x$?

   ›Reveal solutionSolution

   Maximize $A(x)^2$ (equivalent and algebraically simpler) by setting its derivative to zero.

   From $A(x)=x\sqrt{900-x^2/4}$, consider $A^2 = x^2\left(900-\dfrac{x^2}{4}\right) = 900x^2-\dfrac{x^4}{4}$ (maximizing $A^2$ maximizes $A$ since $A\ge0$).

   $\frac{d(A^2)}{dx} = 1800x - x^3 = x(1800-x^2)$

   Setting this to zero: $x=0$ (rejected, gives zero area) or $x^2=1800 \Rightarrow x=\sqrt{1800}=30\sqrt2$.

   Checking it's a maximum: $\dfrac{d^2(A^2)}{dx^2}=1800-3x^2$, which at $x^2=1800$ gives $1800-5400=-3600<0$, confirming a maximum.

   ✓Final answer

   $x = 30\sqrt2$ m gives the maximum area.

8. CBSE 2025Set ANNUAL1 mark
   Q.A gardener plans to plant flowers in a rectangular flower bed in such a way that a rectangle is inscribed in the semi-circular field (as shown in the figure). Radius of the semi-circular field is 30 m. Let the length of the rectangle be PQ = $x$ m. Based on above information answer the following: What will be the area of remaining field (in sq. m) after having the flower bed of maximum area?

   ›Reveal solutionSolution

   Subtract the maximum rectangle area from the total semicircular field area.

   Total area of the semicircular field: $\dfrac12\pi r^2 = \dfrac12\pi(30)^2 = 450\pi$ sq. m.

   At maximum, $x=30\sqrt2$, so breadth $b=\sqrt{900-\dfrac{(30\sqrt2)^2}{4}} = \sqrt{900-\dfrac{1800}{4}} = \sqrt{900-450}=\sqrt{450}=15\sqrt2$.

   Maximum rectangle area $= x\cdot b = 30\sqrt2\cdot15\sqrt2 = 450\cdot2 = 900$ sq. m.

   Remaining area $= 450\pi - 900$ sq. m.

   ✓Final answer

   Remaining area $= (450\pi-900)$ sq. m $\approx 513.7$ sq. m.

9. CBSE 2024Set A1 mark
   Q.A point $C$ in the domain of a function $f$ at which either $f'(C) = 0$ or $f$ is not differentiable is called a ______ point of $f$.

   ›Reveal solutionSolution

   A point where $f'(c)=0$ or $f$ is not differentiable is called a critical point of $f$.

   In the study of maxima/minima (Application of Derivatives), a point $c$ in the domain of $f$ at which either $f'(c)=0$ (a stationary point) or $f$ fails to be differentiable is called a critical point of $f$. Local maxima/minima can only occur at critical points.

   ✓Final answer

   critical point.

10. CBSE 2024Set ANNUAL1 mark
    Q.[Case study] Let a cone be inscribed in a sphere of radius $R$. The height and radius of the cone are $h$ and $r$ respectively; $x$ denotes the distance from the sphere's centre $O$ to the centre of the cone's base. Write the relation between $r$ and $R$ in terms of $x$.

    ›Reveal solutionSolution

    $r^2 = R^2 - x^2$.

    From the figure, $O$ is the sphere's centre, $C$ is the centre of the cone's circular base, $OC=x$, $CA=r$ (radius of the cone's base), and $OA=R$ (a radius of the sphere, since $A$ lies on the sphere).

    Triangle $OCA$ is right-angled at $C$ (the base is horizontal, $OC$ is vertical). By the Pythagorean theorem:

    $OA^2 = OC^2+CA^2 \implies R^2 = x^2+r^2$

    $\therefore r^2 = R^2-x^2$

    ✓Final answer

    $r^2 = R^2 - x^2$.

11. CBSE 2024Set ANNUAL1 mark
    Q.[Case study, same setup as above — cone of height $h$, radius $r$ inscribed in a sphere of radius $R$, with $x$ the distance from the sphere's centre to the cone's base] Write the volume $V$ of the cone in terms of $R$ and $x$.

    ›Reveal solutionSolution

    $V = \dfrac{\pi}{3}(R+x)^2(R-x)$.

    From the figure, the cone's height is $h=R+x$ (from the base at $C$ up to the apex $D$ at the top of the sphere), and from the previous part, $r^2=R^2-x^2$.

    Volume of a cone: $V=\dfrac13\pi r^2h$.

    $V = \frac13\pi(R^2-x^2)(R+x) = \frac13\pi(R-x)(R+x)(R+x) = \frac{\pi}{3}(R+x)^2(R-x)$

    ✓Final answer

    $V = \dfrac{\pi}{3}(R+x)^2(R-x)$.

12. CBSE 2019Set ANNUAL1 mark
    Q.If $\phi(x) = f(x) + f(1-x)$, $f''(x) = 0$ for $0 \le x \le 1$, then is $x = \frac{1}{2}$ a point of maxima or minima of $\phi(x)$?

    ›Reveal solutionSolution

    $f''(x)=0$ forces $f$ to be linear, which makes $\phi(x)=f(x)+f(1-x)$ constant on $[0,1]$ — so $x=\tfrac12$ is neither a genuine maximum nor minimum.

    $\phi(x) = f(x)+f(1-x)$.

    $\phi'(x) = f'(x) - f'(1-x)$

    At $x=\tfrac12$: $\phi'\left(\tfrac12\right) = f'\left(\tfrac12\right) - f'\left(\tfrac12\right) = 0$, so $x=\tfrac12$ is a critical point.

    $\phi''(x) = f''(x) + f''(1-x)$

    We are given $f''(x)=0$ for all $x\in[0,1]$, so $f''(1-x)=0$ too, giving $\phi''(x) = 0+0 = 0$ for every $x\in[0,1]$ — the second-derivative test is inconclusive.

    In fact, since $f''(x)=0$ throughout $[0,1]$, $f$ must be a linear function, say $f(x)=mx+c$. Then

    $\phi(x) = (mx+c) + \big(m(1-x)+c\big) = mx+c+m-mx+c = m+2c$,

    which is a constant, independent of $x$. So $\phi$ does not actually have a strict maximum or minimum anywhere on $[0,1]$ — every point, including $x=\tfrac12$, gives the same value.

    ✓Final answer

    Neither a maximum nor a minimum in the strict sense: because $f''=0$ makes $f$ linear, $\phi(x)=f(x)+f(1-x)$ is constant on $[0,1]$, so $x=\tfrac12$ is just one of infinitely many points where $\phi$ attains its (only) value.