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Application of Derivatives — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplication of Derivatives5 Marks
Question
A rectangular sheet of tin $45 \ cm$ by $24 \ cm$ is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?
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Answer

Let the side of the square to be cut off be $x,$ then, the height of the box is $x$ and the length is $45 - 2x$ and the breadth is $24 - 2x$.
Then, the volume $V(x)$ of the box is given by:
$V(x) = x(45 - 2x)(24 - x)$
$= x(1080 - 90x - 48x + 4x^2)$
$= 4x^3 - 138x^2 + 1080x$
$\therefore V^\prime (x) = 12x^2 - 276x + 1080$
$= 12(x^2 - 23x + 90)$
$= 12(x - 18)(x - 5)$
And, $V^\prime{^\prime}(x) = 24x - 276 = 12(2x - 23)$
Now, $V^\prime (x) = 0$
$\Rightarrow x = 18$ or $5$
It is not possible to cut off a square of side $18 \ cm$ from each corner of the rectangular sheet.
So $,x$ cannot be equal to $18$.
Therefore $, x = 5$
$V^\prime{^\prime} (5) = 12(10 - 23) = -156 < 0$
Then, by second derivative test $,x = 5$ is the point of maxima of $V$.
Therefore, the side of the square to be cut off to make the volume of the box maximum is $5 \ cm$.

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