Question

A piece of cardboard is 1 meter by 1/21/2 meter. A square is to be cut from each corner and the sides folded up to make an open-top box. What are the dimensions of the box with maximum possible volume?

Answer 1

Answer:

The maximum dimensions of the box:

Length of the box  = 0.788676 m

Breadth of the box  = 0.288676 m

Step-by-step explanation:

Original piece of cardboard is a square with sides of length s.

Length of the card board = l = 1 m

Breadth of cardboard = b = 1/2 m = 0.5 m

Squares with sides of length x are cut out of each corner of a rectangular cardboard to form a box.

Now, length of the box = L = 1 - 2x

And breadth of the box = B = 0.5 - 2x

Height of the box ,H = x

Volume of the box ,V= L × B × H

$\frac{dV}{dx}=\frac{(1 -2x)(0.5-2x)x}{dx}$

$\frac{dV}{dx}=\frac{d(0.5x-3x^2+4x^3)}{dx}$

$\frac{dV}{dx}=0.5-6x+12x^2$

$\frac{dV}{dx}=O$

x = 0.394338  , 0.105662

$\frac{d^2V}{(dx)^2}=-6+24x$

When , $x=0.105662$ , $\frac{d^2V}{(dx)^2}$ (maxima)

The maximum dimensions of the box:

Length of the box = L = 1 - 2x = 1 - 2(0.105662) m = 0.788676 m

Breadth of the box = B = 0.5 - 2x = 0.5 - 2(0.105662) m = 0.288676 m