Answer:
A function that represents the volume of the box:
[tex]V(x)=(4x^2-24x+32)[/tex]
The maximum Volume of the box is [tex]32 inch^2[/tex].
Step-by-step explanation:
Width of the card board= b = x
Length of the card board = l = 2x
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 = 2x - 4
Breadth of the box ,B= x - 4
Height of the box ,H= 2 inches
Volume of the box ,V= L × B × H = [tex](2x-4)(x-4)2=(4x^2-24x+32)[/tex]
[tex]V(x)=(4x^2-24x+32)[/tex]
[tex]\frac{dV}{dx}=\frac{(2x-4)(x-4)2}{dx}[/tex]
[tex]\frac{dV}{dx}=\frac{d(4x^2-24x+32)}{dx}[/tex]
[tex]\frac{dV}{dx}=8x-24[/tex]
Putting ,[tex]\frac{dV}{dx}=0[/tex]
[tex]0=8x-24[/tex]
x = 3 inches
The maximum Volume of the box:
[tex]V = (4x^2-24x+32)=(4(3)^2-24(3)+32)=32 inch^2[/tex]
