QJEEM23Maeda7: JEE Main 2023 Mathematics - Class 12 JEEM23Maeda7 Solution: A square piece of tin of side 30 cm is to be made into a box…

Given, length of square sheet of side is $30 \mathrm{~cm}$ .



Let, side of small square be $x \mathrm{~cm}$ .

Since, a square piece of tin is to be made into a box without top by cutting square from each corner and folding up the flaps to from a box. Then, this shape formed a cuboidal shape.

Thus, volume of the box $(V)$

$\begin{aligned}& =(30-2 x)(30-2 x) x \\\\&\Rightarrow V =x(30-2 x)^2\end{aligned}$

On differentiating both side with respect to $x$ , we get

$\begin{aligned}\frac{d V}{d x} & =(30-2 x)^2+x\{2(30-2 x) \cdot(-2)\} \\\\& =(30-2 x)^2-4 x(30-2 x) \\\\& =(30-2 x)\{30-6 x\}\end{aligned}$

Therefore, maximum of $V, \frac{d V}{d x}=0$

$\begin{array}{ll}&\Rightarrow (30-2 x)(30-6 x)=0 \\\\&\Rightarrow x=15 \text { or } x=5 \quad[x=15 \text { (not taken) }]\end{array}$

Thus, $x=5$

So, the surface area

$\begin{array}{ll}=4 \times(30-2 x) \times x+(30-2 x)^2 & \\\\=4 \times(30-10) \times 5+(30-10)^2 & (\text { Put } x=5) \\\\=400+400=800 &\end{array}$