Question

. Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 2323 ft by 1313 ft. The resulting piece of cardboard is then folded into a box without a lid. Find the volume of the largest box that can be formed in this way. b. Suppose that in part​ (a) the original piece of cardboard is a square with sides of length s. Find the volume of the largest box that can be formed in this way.

Answer 1

Answer:

(a)Therefore the volume of the box is 361.19 cubic ft.

(b)Therefore the volume of the box is 361.19 cubic ft.

Step-by-step explanation:

Given that,

Squares with sides length x are cut out from of each corner of a rectangular piece of cardboard measuring 23 ft and 13 ft.

Now the length of the box is =(23-2x) ft

The width of the box is =(13-2x) ft

The height of the box is= x

The volume of the box is = Length×width×Height

                                          =(23-2x)(13-2x)x cubic ft

                                         = 299x-72x²+4x³ cubic ft

Let,

V=299x-72x²+4x³

Differentiating with respect to x

V'= 299-144x+12x²

Again differentiating with respect to x

V''= -144+24x

To find the maximum volume, we set V'=0

∴299-144x+12x²=0

Applying quadratic formula $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}$, here a=12,b= -144 and c=299

$\therefore x=\frac{-(-144)\pm\sqrt{(-144)^2-4.12.299}}{2.12}$

$\Rightarrow x= 9.33, 2.67$

For x= 9.33 , the width of the box will negative which is impossible.

So, x= 2.67

$V''|_{x=2.67}=-144+24(2.67)$

So, at x=2.67, the volume of the box will be maximum.

Therefore the volume of the box is

=(299x-72x²+4x³) cubic ft

=361.19 cubic ft

(b)

The value of s will same with the value of x

so the volume of the box also remains same.

Therefore the volume of the box is 361.19 cubic ft