Question

2. A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume of the box

Answer 1

Answer:

$2 ft^3$

Step-by-step explanation:

We are given that

Side of square shaped piece= 3ft

Let  square of side x cut from each corner of square piece cardboard.

Length of box=$3-2x$

Breadth of box=$3-2x$

Height of box= x

We have to find the largest volume of box.

Volume of box is given by

$V=x(3-2x)^2$

$V=(4x^3-12x^2+9x)$

Differentiate w.r.t x

$\frac{dV}{dx}=12x^2-24x+9$

Substitute $\frac{dV}{dx}=0$

$12x^2-24x+9=0$

$4x^2-8x+3=0$

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

$2x-1=0\implies x=\frac{1}{2}$

$2x-3=0\implies x=\frac{3}{2}$

$x=\frac{1}{2}, x=\frac{3}{2}$

Again differentiate w.r.t x

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

Substitute $x=\frac{1}{2}$

$\frac{d^2V}{dx^2}=12-24=-12$

Substitute $x=\frac{3}{2}$

$\frac{d^2V}{dx^2}=36-24=12>0$

Therefore, $\frac{d^2V}{dx^2}$ at $x=\frac{1}{2}$

Hence, the volume is maximum at $x=\frac{1}{2}$

Substitute the value $x=\frac{1}{2}$ then we get

$V=\frac{1}{2}(3-1)^2=2 ft^3$