Respuesta :
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 [tex]x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}[/tex], here a=12,b= -144 and c=299
[tex]\therefore x=\frac{-(-144)\pm\sqrt{(-144)^2-4.12.299}}{2.12}[/tex]
[tex]\Rightarrow x= 9.33, 2.67[/tex]
For x= 9.33 , the width of the box will negative which is impossible.
So, x= 2.67
[tex]V''|_{x=2.67}=-144+24(2.67)<0[/tex]
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
The volume of a box is the amount of space in it.
- The maximum value of the box is [tex]\mathbf{361.186in^3}[/tex]
- The volume of the largest box is [tex]\mathbf{22.559in^3}[/tex]
The dimension of the cardboard is given as:
[tex]\mathbf{Length = 23}[/tex]
[tex]\mathbf{Width = 13}[/tex]
Assume the cut-out is x.
So, the dimension of the box is:
[tex]\mathbf{Length = 23 -2x}[/tex]
[tex]\mathbf{Width = 13 - 2x}[/tex]
[tex]\mathbf{Height = x}[/tex]
The volume of the box is:
[tex]\mathbf{V = (23 -2x) (13 - 2x)x}[/tex]
Expand
[tex]\mathbf{V = 299x -72x^2 +4x^3}[/tex]
Differentiate
[tex]\mathbf{V' = 299 -144x +12x^2}[/tex]
Set to 0
[tex]\mathbf{299 -144x +12x^2 = 0}[/tex]
Rewrite as:
[tex]\mathbf{12x^2 - 144x + 299= 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{x = (2.67, 9.33)}[/tex]
9.33 is greater than the dimensions of the box.
So, the possible value of x is:
[tex]\mathbf{x = 2.67}[/tex]
Recall that:
[tex]\mathbf{V = 299x -72x^2 +4x^3}[/tex]
So, we have:
[tex]\mathbf{V = 299(2.67) - 72(2.67)^2 + 4(2.67)^3}[/tex]
[tex]\mathbf{V = 361.186}[/tex]
Hence, the maximum value of the box is [tex]\mathbf{361.186in^3}[/tex]
If the dimension of the cardboard is s,
The dimension of the box would be:
[tex]\mathbf{Length = s -2x}[/tex]
[tex]\mathbf{Width = s - 2x}[/tex]
[tex]\mathbf{Height = s}[/tex]
So, the volume is:
[tex]\mathbf{V = (s - 2x)(s - 2x) s}[/tex]
Substitute [tex]\mathbf{x = 2.67}[/tex]
[tex]\mathbf{V = (s - 2 \times 2.67)(s - 2 \times 2.67) s}[/tex]
[tex]\mathbf{V = (s - 5.34)(s - 5.34) s}[/tex]
Expand
[tex]\mathbf{V = (s - 5.34)(s^2 - 5.34s)}[/tex]
[tex]\mathbf{V = s^3 - 5.34s^2 - 5.34s^2 + 28.5156s}[/tex]
[tex]\mathbf{V = s^3 -10.68s^2 + 28.5156s}[/tex]
Integrate
[tex]\mathbf{V' = 3s^2 -21.36s + 28.5156}[/tex]
Set to 0
[tex]\mathbf{3s^2 -21.36s + 28.5156 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{s = (1,78,5.34)}[/tex]
5.34 is greater than the side lengths. So, we have:
[tex]\mathbf{s = 1.78}[/tex]
Recall that:
[tex]\mathbf{V = s^3 -10.68s^2 + 28.5156s}[/tex]
So, we have:
[tex]\mathbf{V = 1.78^3 - 10.68 \times 1.78^2 + 28.5156 \times 1.78}[/tex]
[tex]\mathbf{V = 22.559}[/tex]
Hence, the volume of the largest box is [tex]\mathbf{22.559in^3}[/tex]
Read more about volumes at:
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