The volume of the box is maximized when a 6 x 6 inch or 10 x 10 inch square is cut from the corners of the rectangular sheet.
Volume of rectangle:
The volume of an object is the amount of space occupied by the object or shape, which is in three-dimensional space. It is usually measured in terms of cubic units.
The formula for Volume of the rectangle is
V = l x b x h
where
l represents the length
b represents the breadth
h represents the height.
Given,
A rectangular sheet of tin measures 20 inches by 12 inches. suppose you cut a square out of each corner and fold up the sides to make an open-topped box.
Here we need to find the volume of the box is maximized when a square is cut from the corners of the rectangular sheet.
Here we have the following values:
h = x
l = 20 - 2x
b = 12 - 2x.
Here we have to subtract two times of x value because in the question they said that we have to cut a square from it.
Apply the values on the volume formula then we get,
=> V = (20 - 2x) x (12 - 2x) x (x)
=> V = [240 - 40x - 24x + 4x²] x (x)
=> V = [240x - 64x² + 4x³]
Now differentiate:
=> V' = 240 - 64x + 4x²
Simplify and order the equation,
Then we get,
=> V' = x² - 16x + 60
When we factorize the equation then we get,
=> V' = x² - 6x - 10x + 60
Take the common term out of it,
=> V' = x (x-6) -10(x -6)
Therefore, the value of x is, either 6 or 10.
Now you need to look at your solutions.
At x=6, the volume is,
=> V = (20 - 2(6)) x (12 - 2(6)) x (6)
=> V = (20 - 12) x (12 - 12) x 6
=> V = 8 x 0 x 6
=> V = 0
our corner squares devour the entire piece of cardboard and you are left with zero volume; so that solution is a minimum.
Now at the value of x = 10,
The volume is
=> V = (20 - 2(10) x (12 - 2(10)) x (10)
=> V = (20 - 20) x (12 - 20) x 10
=> V = 0 x (-8) x 10
=> V = 0
So, in both cases we have 0 as minimum volume. So, we can take any one of the value to get the maximum out of it.
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