Search any question & find its solution
Question:
Answered & Verified by Expert
A rectangular sheet of tin $45 \mathrm{~cm}$ by $24 \mathrm{~cm}$ 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?
Solution:
1843 Upvotes
Verified Answer
Let each side of the square cut off from each corner be $x \mathrm{~cm}$.
$\therefore$ Sides of the rectangular box are $(45-2 \mathrm{x}),(24-2 \mathrm{x})$ and $x \mathrm{~cm}$.
Then, volume of the box
$V=(45-2 x)(24-2 x)(x)=2\left(2 x^3-69 x^2+540 x\right)$
$\Rightarrow \quad \frac{\mathrm{dV}}{\mathrm{dx}}=12\left(\mathrm{x}^2-23 \mathrm{x}+90\right)$
For maxima and minima $\frac{\mathrm{dV}}{\mathrm{dx}}=0 \Rightarrow \mathrm{x}=5,18$
But $x$ cannot be greater than 12
$\therefore \mathrm{x}=5, \frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{dx}}=12(10-23)=-$ ve $[\because \mathrm{x}=5]$
$\therefore \mathrm{V}$ is maximum at $\mathrm{x}=5$ i.e. square of side $5 \mathrm{~cm}$ is cut off from each corner.
$\therefore$ Sides of the rectangular box are $(45-2 \mathrm{x}),(24-2 \mathrm{x})$ and $x \mathrm{~cm}$.
Then, volume of the box
$V=(45-2 x)(24-2 x)(x)=2\left(2 x^3-69 x^2+540 x\right)$
$\Rightarrow \quad \frac{\mathrm{dV}}{\mathrm{dx}}=12\left(\mathrm{x}^2-23 \mathrm{x}+90\right)$
For maxima and minima $\frac{\mathrm{dV}}{\mathrm{dx}}=0 \Rightarrow \mathrm{x}=5,18$
But $x$ cannot be greater than 12
$\therefore \mathrm{x}=5, \frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{dx}}=12(10-23)=-$ ve $[\because \mathrm{x}=5]$
$\therefore \mathrm{V}$ is maximum at $\mathrm{x}=5$ i.e. square of side $5 \mathrm{~cm}$ is cut off from each corner.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.