Question

A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-top box. What is the maximum possible volume of the box? Round your answer to the nearest four decimal places.

Answer 1

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square  at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

And height will be x

$V(x)=x*(26-x)*(13-x)$

To get maximum volume differentiate the above equation,

$V(x)=x*(26*13-26*x-13*x+x^2)$

$V(x)=x^3-39x^2+338x$

$V'(x)=3x^2-78x+338$

$V''(x)=6x-78$

Now ,Solve the Quadratic Equation to get x values,

$3x^2-78x+338$=0

x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

x=[78±Sqrt(3028)]/6

x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17  or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0  i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

$V(x)=x^3-39x^2+338x$

$V(x)=3.8288^3-39*(3.82883)^2+338*3.8288$

$V(x)=56.13-571.73+1294.13$

$V(x)=778.53 C.C$