CBSE Sample Paper 2024-25 · Section E · 4 marks

Q36: Case study: container from a 25 cm by 40 cm sheet

Sample 2024-25Ch 6 · Application of Derivatives

The question

Ramesh, the owner of a sweet selling shop, purchased some rectangular card board 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.

(i) Express the volume (V) of each container as function of $x$ only. (1 mark)

(ii) Find $\dfrac{dV}{dx}$ (1 mark)

(iii) (a) For what value of $x$ ⁠, the volume of each container is maximum? (2 marks)

(iii) (b) Check whether V has a point of inflection at $x = \dfrac{65}{6}$ or not? (2 marks)

Exam answer

(i) After cutting squares of side $x$ cm, the base is $(40 - 2x)$ cm by $(25 - 2x)$ cm and the height is $x$ cm.

Why?

A square of side $x$ is cut from each of the four corners, so each side of the sheet loses $x$ at both ends: $40 - 2x$ and $25 - 2x$ ⁠.

Folding up the flaps, which are $x$ cm wide, makes the height $x$ ⁠. For a real box, $25 - 2x > 0$ ⁠, so $0 < x < 12.5$ ⁠.

$V = (40 - 2x)(25 - 2x)x$ cm $^3$

Why?

Volume of a box $=$ length $\times$ breadth $\times$ height.

(ii)

\begin{aligned} V &= x(1000 - 80x - 50x + 4x^2) \end{aligned}

\begin{aligned} V &= 4x^3 - 130x^2 + 1000x \\ \frac{dV}{dx} &= 12x^2 - 260x + 1000 \end{aligned}

Why?

Multiply the two brackets term by term:

\begin{aligned} 40 \times 25 &= 1000 \\ 40 \times (-2x) &= -80x \\ (-2x) \times 25 &= -50x \\ (-2x)(-2x) &= 4x^2 \end{aligned}

\begin{aligned} &(40 - 2x)(25 - 2x) \\ &= 1000 - 80x - 50x + 4x^2 \\ &= 1000 - 130x + 4x^2 \end{aligned}

Multiply by $x$ ⁠, then differentiate each term with the power rule: $4x^3 \to 12x^2$ ⁠, $-130x^2 \to -260x$ ⁠, $1000x \to 1000$ ⁠.

\begin{aligned} &x(1000 - 130x + 4x^2) \\ &= 1000x - 130x^2 + 4x^3 \\[4pt] &\frac{d}{dx}(4x^3) = 4 \times 3x^2 = 12x^2 \\ &\frac{d}{dx}(-130x^2) = -130 \times 2x \\ &\qquad = -260x \\ &\frac{d}{dx}(1000x) = 1000 \end{aligned}

\begin{aligned} \frac{dV}{dx} &= 4(3x^2 - 65x + 250) \\ &= 4(3x^2 - 15x - 50x + 250) \\ &= 4(3x - 50)(x - 5) \end{aligned}

Why?

Take out the common factor 4. To factorise $3x^2 - 65x + 250$ ⁠, split $-65x$ into two terms whose coefficients multiply to $3 \times 250 = 750$ ⁠:

(-15) \times (-50) = 750

\begin{aligned} 12 \div 4 &= 3,\ 260 \div 4 = 65, \\ 1000 \div 4 &= 250 \\[4pt] -15 + (-50) &= -65 \end{aligned}

Then group in pairs:

\begin{aligned} 3x^2 - 15x &= 3x(x - 5) \\ -50x + 250 &= -50(x - 5) \end{aligned}

\begin{aligned} &3x(x - 5) - 50(x - 5) \\ &= (3x - 50)(x - 5) \end{aligned}

(Take out the common bracket $(x - 5)$ ⁠.)

$\dfrac{dV}{dx} = 4(3x - 50)(x - 5)$

(iii) (a)

For extreme values,

\frac{dV}{dx} = 4(3x - 50)(x - 5) = 0

\Rightarrow x = \frac{50}{3} \text{ or } x = 5

Why?

At a maximum or minimum the tangent is horizontal, so $\frac{dV}{dx} = 0$ ⁠. A product is 0 when one factor is 0: $3x - 50 = 0$ or $x - 5 = 0$ ⁠.

\begin{aligned} 3x - 50 = 0 &\Rightarrow 3x = 50 \\ &\Rightarrow x = \frac{50}{3} \\[4pt] x - 5 = 0 &\Rightarrow x = 5 \end{aligned}

(The factor 4 is never 0.)

$x = \frac{50}{3}$ is not possible, since $25 - 2x > 0$ needs $x < 12.5$ ⁠.

Why?

$\frac{50}{3} \approx 16.7$ ⁠. Cutting that much from each end of the 25 cm side would leave a negative width, $25 - \frac{100}{3} < 0$ ⁠.

\begin{aligned} 2 \times \frac{50}{3} &= \frac{100}{3} \\ 25 - \frac{100}{3} &= \frac{75}{3} - \frac{100}{3} \\ &= -\frac{25}{3} < 0 \end{aligned}

\frac{d^2V}{dx^2} = 24x - 260

Why?

Differentiate $12x^2 - 260x + 1000$ once more: $12x^2 \to 24x$ ⁠, $-260x \to -260$ ⁠, and the constant gives 0.

\therefore\ \frac{d^2V}{dx^2} \text{ at } x = 5 \text{ is } -140 < 0

Why?

$24(5) - 260 = 120 - 260 = -140$ ⁠.

Second derivative test: if $V'(c) = 0$ and $V''(c) < 0$ ⁠, then $V$ has a local maximum at $c$ ⁠. Here $x = 5$ is the only critical point in $(0, 12.5)$ ⁠.

$\therefore$ V is maximum when $x = 5$ cm.

(iii) (b)

\frac{dV}{dx} = 4(3x^2 - 65x + 250)

\frac{d^2V}{dx^2} = 4(6x - 65)

Why?

Differentiate inside the bracket: $3x^2 \to 6x$ ⁠, $-65x \to -65$ ⁠, $250 \to 0$ ⁠. This is the same as $24x - 260$ ⁠.

\begin{aligned} 4(6x - 65) &= 4(6x) - 4(65) \\ &= 24x - 260 \end{aligned}

\text{At } x = \frac{65}{6},\ \frac{d^2V}{dx^2} = 4(65 - 65) = 0

Why?

$6 \times \frac{65}{6} = 65$ ⁠. Note $\frac{65}{6} \approx 10.8$ lies inside $(0, 12.5)$ ⁠.

\begin{aligned} 4\left(6 \times \frac{65}{6} - 65\right) &= 4(65 - 65) \\ &= 4(0) = 0 \end{aligned}

$\dfrac{dV}{dx}$ at $x = \dfrac{65}{6}$ exists and $\dfrac{d^2V}{dx^2}$ at $x = \dfrac{65}{6}$ is 0.

$\dfrac{d^2V}{dx^2} < 0$ for $x < \dfrac{65}{6}$ and $\dfrac{d^2V}{dx^2} > 0$ for $x > \dfrac{65}{6}$ ⁠, so $\dfrac{d^2V}{dx^2}$ changes sign at $x = \dfrac{65}{6}$ ⁠.

Why?

A point of inflection is where the curve changes from bending one way to bending the other, that is where the second derivative changes sign.

For example, at $x = 10$ ⁠, $4(60 - 65) = -20 < 0$ ⁠, and at $x = 11$ ⁠, $4(66 - 65) = 4 > 0$ ⁠.

Watch out: $\frac{d^2V}{dx^2} = 0$ alone is not enough. The sign must change too.

$\therefore\ x = \dfrac{65}{6}$ is a point of inflection.

Where this comes from

Chapter 6, Application of Derivatives: Open box: maximum volume; point of inflection.

Question type: (iii)(a) Word problems

How to spot it: click to see

(iii)(a) Type 10. Word problems: the biggest box, the least cost

Spot it: a real object and the words "maximum", "minimum", "largest" or "least".

Here: The volume $V = x(40 - 2x)(25 - 2x)$ is in one variable, with $x$ between 0 and 12.5. $\frac{dV}{dx} = 4(3x - 50)(x - 5)$ is 0 at $\frac{50}{3}$ ⁠, outside that range, or at $x = 5$ ⁠, where $V'' = -140$ gives the maximum.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Application of Derivatives in past papers: question 26 of 37 · see them all